The well needs 7.94 pounds of chlorine gas to meet the water quality objectives.
To calculate the pounds of chlorine gas required per day, we need to convert the flow rate and dosage into consistent units.
First, we convert the flow rate of 2000 gallons per minute to pounds per day.
2000 gpm x 60 minutes x 24 hours = 2,880,000 gallons per day
1 gallon of water weighs approximately 8.34 pounds, so 2,880,000 gallons weigh:
2,880,000 gallons x 8.34 pounds/gallon = 24,019,200 pounds per day
Next, we convert the chlorine dosage of 2.5 gpm to pounds per day:
2.5 grams per minute x 60 minutes x 24 hours = 3,600 grams per day
1 pound is equivalent to 453.59 grams, so we convert the dosage to pounds:
3,600 grams per day / 453.59 grams per pound = 7.94 pounds per day
Therefore, the well needs 7.94 pounds of chlorine gas to meet the water quality objectives.
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Terminal Rating(110-14(C)(1): What size THHN conductor is required for a 70 ampere branch circuit if the circuit breaker and equipment is listed for 75 degrees C terminals and the load does not exceed 65 ampere?
According to the NEC code 110-14(C)(1), for a 70 ampere branch circuit with equipment and circuit breaker listed for 75 degrees C terminals and a load not exceeding 65 ampere, the minimum required size of THHN conductor would be 4 AWG.
For a 70-ampere branch circuit with 75-degree C terminals and a load not exceeding 65 ampere, you would need a 4 AWG THHN conductor. This conductor size is suitable for up to 85 amperes at 75-degree C terminals, providing ample capacity for your specified requirements.
The NEC code 110-14(C)(1) states that the minimum required size of THHN conductor for a 70 ampere branch circuit with equipment and a circuit breaker designated for 75 degrees C terminals and a load not exceeding 65 ampere would be 4 AWG.
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Two slits spaced 0. 0720 mm apart are 0. 800 m from a screen. Coherent light of wavelength λ passes through the two slits. In their interference pattern on the screen, the distance from the center of the central maximum to the first minimum is 3. 00 mm. The intensity at the peak of the central maximum is 0. 0700 W/m2. What is the intensity at point on the screen that is 2. 00 mm from the center of the central maximum? What is the intensity at point on the screen that is 1. 50 mm from the center of the central maximum?
The intensity at a point on the screen 2.00 mm from the center of the central maximum is approximately 0.034 W/m². The intensity at a point on the screen 1.50 mm from the center of the central maximum is approximately 0.024 W/m².
I = Imax cos² (πd sin θ / λ),
where Imax is the intensity at the center of the interference pattern, d is the distance between the two slits, θ is the angle between the line connecting the point on the screen to the center of the interference pattern and the line perpendicular to the screen, and λ is the wavelength of the light.
To find the angle θ, we can use the small angle approximation:
sin θ ≈ θ ≈ y/L,
where y is the distance from the center of the interference pattern to the point on the screen, and L is the distance between the slits and the screen.
We are given d = 0.0720 mm, λ = unknown, L = 0.800 m, Imax = 0.0700 W/m², and the distance from the center of the central maximum to the first minimum y = 3.00 mm.
Using the given distance y, we can find the value of sin θ:
y/L = sin θ,
3.00 mm / 0.800 m = sin θ,
sin θ = 0.00375.
Now we can solve for the wavelength λ:
Imax cos² (πd sin θ / λ) = I,
0.0700 W/m² cos² (π(0.0720 × 10⁻³ m)(0.00375) / λ) = I,
cos² (π(0.0720 × 10⁻³ m)(0.00375) / λ) = I / 0.0700 W/m²,
π(0.0720 × 10⁻³ m)(0.00375) / λ = ± cos⁻¹ (√(I / 0.0700 W/m²)),
λ = π(0.0720 × 10⁻³ m)(0.00375) / cos⁻¹√(I / 0.0700 W/m²)),
λ = 5.70 × 10⁻⁷ m (for the positive root).
Now we can find the intensities at the given distances from the center of the central maximum.
For y = 2.00 mm:
sin θ = y/L = 2.00 mm / 0.800 m = 0.00250,
I = Imax cos² (πd sin θ / λ)
I = 0.0700 W/m² cos² (π(0.0720 × 10⁻³m)(0.00250) / (5.70 × 10⁻⁷ m))² ≈ 0.034 W/m².
So the intensity at a point on the screen 2.00 mm from the center of the central maximum would be approximately 0.034 W/m².
For y = 1.50 mm:
sin θ = y/L = 1.50 mm / 0.800 m = 0.001875,
I = Imax cos² (πd sin θ / λ)
I= 0.0700 W/m² cos² (π(0.0720 × 10⁻³m)(0.001875) / (5.70 × 10⁻⁷ m))² ≈ 0.034 W/m².
I ≈ 0.024 W/m².
So the intensity at a point on the screen 1.50 mm from the center of the central maximum would be approximately 0.024 W/m².
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if light cannot escape from a black hole how can we detect x-rays from such objects
While it's true that light cannot escape a black hole, it's also true that black holes can be incredibly active objects. When matter falls into a black hole, it heats up and emits intense radiation, including X-rays. This radiation is emitted before the matter actually crosses the event horizon (the point of no return), so we can still detect it using X-ray telescopes.
Black holes are objects with such strong gravitational fields that nothing, including light, can escape once it passes the point of no return, known as the event horizon. However, as matter falls into a black hole, it becomes extremely hot and can emit high-energy radiation in the form of X-rays. This radiation can be detected by telescopes and other instruments, allowing us to study the properties of black holes.
By studying the X-rays emitted by black holes, we can learn a lot about these fascinating objects and their behavior. So even though light can't escape from a black hole, other forms of radiation can still be detected and studied.
In addition to X-rays, black holes can also emit other forms of radiation, such as gamma rays, radio waves, and visible light. However, it is important to note that these emissions do not come from within the black hole itself, but rather from the matter and radiation that surround it. The black hole itself remains invisible, as no light or other radiation can escape from its event horizon.
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The detection of X-rays from black holes is a result of the interactions of black holes with their surroundings.
Here are the step-by-step explanations:
1) Black holes in binary systems
Some black holes are in binary systems with a companion star. The black hole pulls material from the companion star through its strong gravitational field, forming an accretion disk around the black hole.
This disk is made of gas and dust particles that are orbiting the black hole, and as they spiral inward towards the black hole, the gas is heated to extremely high temperatures.
2) Accretion disk emits X-rays
As the gas particles in the accretion disk are heated, they emit electromagnetic radiation, including X-rays, which can escape from the disk.
These X-rays are not emitted from within the black hole itself, but from the hot gas in the accretion disk around the black hole.
3) X-rays are detected by telescopes
X-rays emitted by the accretion disk can be detected by X-ray telescopes in space, such as NASA's Chandra X-ray Observatory.
These telescopes can detect X-rays from distant objects, including black holes, by measuring the energy and intensity of the X-rays.
4) Corona around black holes
In addition, some black holes have a hot, magnetized plasma surrounding them, called a corona.
The corona can emit X-rays as well, due to the high temperatures and magnetic energy generated as gas in the accretion disk spirals towards the black hole.
5) X-rays from coronae are detected
The X-rays emitted by the corona can also be detected by X-ray telescopes in space.
The telescopes measure the energy and intensity of the X-rays emitted by the corona, which can provide information about the black hole's surroundings.
In summary, X-rays can be detected from black holes through their effects on nearby matter, such as gas in an accretion disk or a surrounding corona.
These X-rays are not emitted from within the black hole itself, but from the matter surrounding the black hole.
X-ray telescopes in space are used to detect these X-rays, and they can provide valuable information about the black hole and its surroundings.
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If two identical sound waves arriving at the same point are in phase, the resulting wave, compared to the original waves, will have (A) an increase in speed (B) an incre…
If two identical sound waves arriving at the same point are in phase, the resulting wave, compared to the original waves, will have
(A) an increase in speed
(B) an increase in frequency
(C) a larger amplitude
(D) a longer period
If two identical sound waves arriving at the same point are in phase, the resulting wave, compared to the original waves, will have (C) a larger amplitude.
When two sound waves are in phase, their peaks and troughs align perfectly. This alignment causes constructive interference, which results in the combined wave having a larger amplitude.
The amplitude is a measure of the energy in the wave, so a larger amplitude means a louder sound or greater intensity.
The speed (A) and frequency (B) of the combined wave remain unchanged because these properties depend on the medium through which the sound waves travel and not on the interaction between the waves.
The period (D) of the combined wave also remains unchanged because it is the inverse of the frequency.
In summary, when two identical sound waves are in phase and arrive at the same point, they create a new wave with a larger amplitude due to constructive interference, while other properties like speed, frequency, and period remain the same.
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The correct option is (C) a larger amplitude.
When two identical sound waves are in phase, their peaks and troughs align, resulting in constructive interference. This means that the amplitude of the resulting wave will be the sum of the amplitudes of the original waves. Therefore, the resulting wave will have a larger amplitude compared to the original waves.
The amplitude of a sound wave is related to the loudness of the sound. So, when two identical sound waves arrive at the same point in phase, their amplitudes add up, resulting in a wave with a larger amplitude and thus a louder sound.
Neither the speed nor the frequency nor the period of the wave changes when the waves are in phase, as these properties are determined by the medium through which the wave is traveling and the source of the wave, and not by the interference of waves.
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at what speed do a bicycle and its rider, with a combined mass of 90 kg , have the same momentum as a 1500 kg car traveling at 6.0 m/s ? express your answer to two significant figures and include the appropriate units.
100 m/s speed do a bicycle and its rider, with a combined mass of 90 kg , have the same momentum as a 1500 kg car traveling at 6.0 m/s
To find the speed at which the bicycle and its rider have the same momentum as the car, we can use the momentum formula:
momentum = mass × speed
First, let's find the momentum of the car:
momentum car = (1500 kg) × (6.0 m/s) = 9000 kg m/s
Now we want the bicycle and its rider to have the same momentum:
momentum bicycle = momentum car = 9000 kg m/s
We can now use the mass of the bicycle and its rider (90 kg) to find the speed at which they have the same momentum:
speed bicycle = momentum bicycle / mass bicycle
speed bicycle = 9000 kg m/s
90 kg = 100 m/s
Therefore, the bicycle and its rider need to travel at a speed of 100 m/s to have the same momentum as the car.
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Question 27
What type of well is considered least likely to become contaminated?
a. Drilled
b. Bored
c. Driven
d. dug
The type of well that is considered least likely to become contaminated is a drilled well. This is because drilled wells are created by drilling a hole deep into the ground, typically hundreds of feet, and are lined with materials such as steel or PVC.
This lining helps to prevent contaminants from seeping into the well from the surrounding soil and groundwater. In contrast, bored and dug wells are often shallower and do not have the same level of protection from contamination. Driven wells, which are constructed by driving a pipe into the ground, can also be susceptible to contamination if the surrounding soil is not properly sealed. Overall, drilled wells are considered the safest option for providing clean and safe drinking water. However, it is still important to regularly test and maintain all types of wells to ensure that they remain free from contaminants.
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The smallest size conductor permitted by the NEC for branch circuits, feeders or services is _____ copper or _____ aluminum.
The smallest size conductor permitted by the NEC (National Electrical Code) for branch circuits, feeders, or services is 14 AWG copper or 12 AWG aluminium.
The smallest size conductor permitted by the NEC (National Electrical Code) for branch circuits, feeders, or services depends on the load that the conductor is expected to carry, as well as the material and type of insulation used in the conductor.
However, in general, for copper conductors, the minimum size permitted for branch circuits, feeders, or services is typically 14 AWG (American Wire Gauge), while for aluminum conductors, the minimum size is typically 12 AWG. It's important to note that these are minimum sizes, and the appropriate conductor size should be determined based on the specific application and load requirements, as specified by the NEC.
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which statement about eddy currents is false? they can be used as a passive braking system, as no external power source is needed if permanent magnets are used. the mechanical energy that is lost when eddy currents are created returns when the eddy currents cease. the faster the conductor moves, the larger the eddy currents will be. they are created in solid conducting plates as they move in and out of magnetic fields. they can be prevented by cutting a slot in a solid conducting plate, to prevent electrons from being able to make a complete circuit.
The statement that eddy currents can be prevented by cutting a slot in a solid conducting plate, to prevent electrons from being able to make a complete circuit is false.
What are eddy currents?Eddy currents are loops of electrical current that are induced within conductive materials when there is a change in magnetic field. When a magnetic field is applied or changes in strength, the eddy currents are produced within the material, which then creates a magnetic field that opposes the original field. This phenomenon is known as electromagnetic induction.
The eddy currents produce a magnetic field that opposes the change in the original magnetic field, which in turn creates resistance to the motion of the original magnetic field. This resistance is called eddy current loss, and it causes the material to heat up.
By cutting a slot in a solid conducting plate can increase the eddy currents because it can create smaller loops within the plate, which can result in more eddy currents being induced. To prevent eddy currents, laminated or layered conductive materials can be used, as the insulation between the layers reduces the current flow and minimizes the eddy currents.
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what additional power must the lens provide in order to focus clearly on an object at the standard near point, 0.25 m?
The additional power the lens must provide in order to focus clearly on an object at the standard near point (0.25 m) is 4 Diopters.
To calculate the additional power the lens must provide to focus clearly on an object at the standard near point (0.25 m), we'll use the lens power formula:
Power (P) = 1 / Focal Length (f)
In this case, we need to find the focal length required for clear focus at the standard near point (0.25 m). Since the object is at a distance of 0.25 m from the lens, the required focal length (f) is:
f = 0.25 m
Now we can plug this value into the lens power formula to find the additional power needed:
P = 1 / f
P = 1 / 0.25 m
P = 4 Diopters
So, In order to focus clearly on an object at the standard near point (0.25 m), the lens needs to have an additional 4 diopters of power.
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The additional power required to focus clearly on an object at the standard near point is: -7.11 D
The standard near point is the closest distance at which a person with normal vision can focus on an object, which is typically taken to be 25 cm or 0.25 m.
To find the additional power the lens must provide, we need to calculate the power required to focus at the near point, and then subtract the power of the lens with a focal length of 9.00 cm.
The power of a lens is given by the formula:
P = 1/f
where P is the power of the lens in diopters (D) and f is the focal length of the lens in meters.
For an object at the near point of 0.25 m, the required power is:
P = 1/0.25 = 4 D
The power of the given lens is:
P = 1/0.09 = 11.11 D
Therefore, the additional power required to focus clearly on an object at the standard near point is:
4 D - 11.11 D = -7.11 D
The negative sign indicates that the lens must be diverging, or concave, to provide the additional power required to focus at the near point.
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The ampacity of 10 current-carrying No. 6 THHW conductors installed in an 18 inch long raceway with an ambient temperature of 39C is ____.
The ampacity of 10 current-carrying No. 6 THHW conductors installed in an 18-inch long raceway with an ambient temperature of 39C is 41 amps.
To determine the ampacity of 10 current-carrying No. 6 THHW conductors, we can follow these steps:
Look up the ampacity of No. 6 THHW conductors in NEC Table 310.15(B)(16) as 65 amps at 90°C.
Apply adjustment factors for ambient temperature using NEC Table. For an ambient temperature of 39°C, the correction factor is 0.91.
Apply adjustment factors for the number of current-carrying conductors using the NEC Table. For 10 current-carrying conductors, the correction factor is 0.70.
Multiply the ampacity from step 1 by the correction factors from steps 2 and 3: 65 amps x 0.91 x 0.70 = 41.13 amps.
Therefore, the ampacity of 10 current-carrying No. 6 THHW conductors installed in an 18-inch long raceway with an ambient temperature of 39°C is approximately 41 amps.
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Long wavelength wave is a wave
Answer:
Explanation:
When the wavelength becomes larger the amplitude of the wave becomes shorter. This is the reason when a Tsunami occurred less damage to the ships in deep sea.
A student presses a .5 kg book against the wall. If the "us" between the book and the wall is .2, what force must the student apply to hold the book in place?
Since the force of static friction can adjust itself up to the maximum value (0.98 N), the student needs to apply a force of at least 4.9 N to hold the book in place against the wall.
A student presses a 0.5 kg book against the wall with the coefficient of static friction (μs) between the book and the wall being 0.2. To hold the book in place, the student must apply a force that is equal to or greater than the force of gravity acting on the book.
The force of gravity (Fg) can be calculated using the equation Fg = m × g, where m is the mass of the book (0.5 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²). So, Fg = 0.5 × 9.8 = 4.9 N (Newtons).
The maximum static friction force (Fs) can be calculated using the equation Fs = μs × Fn, where Fn is the normal force (in this case, equal to the force of gravity). So, Fs = 0.2 × 4.9 = 0.98 N.
Since the force of static friction can adjust itself up to the maximum value (0.98 N), the student needs to apply a force of at least 4.9 N to hold the book in place against the wall.
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Conductors in Parallel(310-10(H)(1) : Phase and grounded (Neutral) conductors sized No. 1 AWG and larger are permitted to be connected in parallel.
Phase and grounded conductors sized No. 1 AWG and larger can be connected in parallel, subject to specific requirements.
Segment 310.10(H)(1) of the Public Electrical Code (NEC) licenses stage and grounded (unbiased) conduits that are estimated No. 1 AWG and bigger to be associated in equal.
This implies that guides conveying current in a similar stage or grounded (impartial) guides can be associated together to increment ampacity or for overt repetitiveness.
Notwithstanding, explicit necessities should be met, including that the guides should be of a similar length, have a similar ampacity and protection type, be ended and associated in a similar way, and be associated with a similar stage or shaft. Consistence with NEC rules is important to guarantee protected and dependable activity of the electrical framework.
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an 84-kw am radio station broadcasts at 1000 khz. how many photons are emitted each second by the transmitting antenna?
1.268 x [tex]10^{32}[/tex] photons are emitted by the transmitting antenna of an 84 kW AM radio station transmitting at 1000 kHz every second.
To calculate the number of photons emitted each second by the transmitting antenna of an 84 kW AM radio station broadcasting at 1000 kHz, follow these steps:
1. Convert the broadcast frequency to Hz:
1000 kHz = 1,000,000 Hz
2. Calculate the energy of a single photon:
The energy of a photon can be found using the equation E = hf, where E is the energy, h is Planck's constant (6.626 x [tex]10^{-34}[/tex] Js), and f is the frequency.
E = (6.626 x [tex]10^{-34}[/tex] Js) x (1,000,000 Hz) = 6.626 x [tex]10^{-28}[/tex] J
3. Convert the radio station's power to energy per second:
Power = 84 kW = 84,000 W = 84,000 J/s
4. Divide the total energy per second by the energy of a single photon to find the number of photons emitted each second:
Number of photons = (84,000 J/s) / (6.626 x [tex]10^{-28}[/tex] J)
Number of photons ≈ 1.268 x [tex]10^{32}[/tex] photons/s
So, the transmitting antenna of an 84 kW AM radio station broadcasting at 1000 kHz emits approximately 1.268 x [tex]10^{32}[/tex] photons each second.
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The transmitting antennae of an 84-kW AM radio station broadcasting at 1000 kHz emit approximately 1.266 x 10^21 photons per second.
To calculate the number of photons emitted per second by the transmitting antenna of an 84-kW AM radio station broadcasting at 1000 kHz, we need to use the formula:
N = P/ (h*f)
where N is the number of photons, P is the power in watts, h is Planck's constant, and f is the frequency in Hz.
First, we need to convert the power from kilowatts to watts by multiplying 84 kW by 1000 to get 84,000 watts.
Next, we need to convert the frequency from kHz to Hz by multiplying 1000 kHz by 1000 to get 1,000,000 Hz.
Now, we can plug in the values and solve for N:
N = 84,000 / (6.626 x 10^-34 * 1,000,000)
N = 1.266 x 10^21 photons/sec
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Inertia is an object's natural tendency to remain in constant motion or at rest. An object moving through outer space, for example, will continue moving in one direction and at a constant speed due to its inertia, if no other forces act on it. Why do planets constantly change the direction in which they move
Option B, C, and E are correct. First principle of motion options that address the issue include Before an item may move, it must be subjected to a net force. The inertia rule is another term for the first principle of Newton's theory of motion.
InertiaNewton's fundamental law holds valid whether or not an object is moving. Newton's first law can be viewed as the law of inertia. It helped us understand that when a body is at rest, it will remain immobile unless an external force is added to it, or that if a body is moving at a constant rate, it will continue to move until an external force is applied to it.Only when a net force is applied will a body move from its resting state. This law is demonstrated whenever a passenger in a car fastens their seat belt. This rule applies to both stationary and moving items. Thus, alternatives B, C, and E are correct.For more information on inertia kindly visit to
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Complete question: Inertia is an object's natural tendency to remain in constant motion or at rest. An object moving through outer space, for example, will continue moving in one direction and at a constant speed due to its inertia, if no other forces act on it. Why do planets constantly change the direction in which they move?
A. Most planets do not have any inertia, so their motion constantly changes.
B. The force of gravity acts on planets and changes the direction of their motion.
C. Each planet's inertia is constantly changing from one moment to the next.
D. There are no forces acting on the planets as they move in orbits around the Sun.
The small line in an air-to-air split system heat pump is known as the ___ line.a. gasb. vaporc. solidd. liquid
The small line in an air-to-air split system heat pump is known as the liquid line.
Line sets consist of two semi-flexible copper pipes to connect the outdoor air conditioner or heat pump to the indoor evaporator coil. The smaller pipe is called the liquid line. The larger pipe is referred to as the suction line, and includes insulation. In most cases, a 3/8″ liquid line is a safe bet, but just like the suction line, there is some wiggle room depending on the system and the specific application.
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When possible, a water main should be tapped while still pressurized to?
When possible, a water main should be tapped while still pressurized to ensure minimal disruption to the water supply and maintain system integrity.
1. Pressurized water main: A pressurized water main is a pipe that carries water under pressure from a treatment facility to homes and businesses. Maintaining pressure is important for efficient and reliable water delivery.
2. Tapping: Tapping is the process of connecting a new pipe or service line to an existing pressurized water main. This is usually done to extend water services to new customers or for infrastructure upgrades.
3. Minimal disruption: By tapping a water main while it is still pressurized, service providers can minimize disruptions to the water supply. This means customers may not experience a loss of water service during the tapping process.
4. System integrity: Keeping the water main pressurized during tapping helps maintain the overall integrity of the water distribution system. This is important to prevent leaks, contamination, and other potential problems.
In summary, when possible, a water main should be tapped while still pressurized to minimize disruption to the water supply, maintain system integrity, and provide a more efficient and reliable connection to the water distribution network.
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Pure water suspended in the atmosphere will freeze at a temperature of:
-40°C
-10°C
0°C
32°F
Pure water suspended in the atmosphere will freeze at a temperature of 0°C and 32°F. The options 0°C and 32°F are correct.
Pure water when suspended in the atmosphere would freeze at a temperature of 0 C or 32 degrees Fahrenheit under normal atmospheric pressure at sea level. But in some conditions the temperature at which water freezes can vary, this depends upon the atmospheric pressure and the presence of impurities in the water as the water droplets in clouds can remain liquid even at temperatures below freezing if there are impurities present that prevent the water from freezing.
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state two precautions taken in the experiment of glass prism
Answer:
"The two precautions that must be taken while preforming the experiment of tracing the path of rays of light through a glass prism are
1. Make sure Glass of slab is clean and it must be free from air bubbles.
2. Also, the Angle of incidence should be lies between 30 degree and 60 degree
Explanation:
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If a ball is rolling down an inclined plane without slipping, which force is responsible for causing its rotation?
The force responsible for causing the ball's rotation as it rolls down an inclined plane without slipping is the torque force.
This force is generated by the ball's weight and the angle of the inclined plane, which causes the ball to rotate around its center of mass. As the ball moves down the inclined plane, the torque force creates rotational motion that helps the ball maintain its rolling motion without slipping. Without the inclined plane, the ball would not be able to generate enough torque to rotate and would slide or stop moving altogether.
If a ball is rolling down an inclined plane without slipping, the force responsible for causing its rotation is the frictional force. This force acts at the point of contact between the ball and the inclined plane, providing the necessary torque for the ball to rotate as it moves down the incline.
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You are pedaling a bicycle at 9.8 m/s. The radius of the wheels of the bicycle is 51.9 I'm. The angular velocity of rotation of the wheels is?
The angular velocity of rotation of the wheels is 18.88 rad/s if you are pedaling a bicycle at 9.8 m/s. The radius of the wheels of the bicycle is 51.9 I'm.
Angular velocityRotational motion and rectilinear motion go hand in hand. These are all derivations of Newton's rules of Motion, which are the fundamental rules of motion. The linear velocity has a value of 9.8 m/s. The equation is used to translate it into angular velocity.Where v is the linear velocity in m/s and r is the radiusThe angular velocity is expressed in rad/s, and r is the wheel's radius in meters.0.519 m is equal to 51.9 cm, where r is expressed. Given that later on we'll employ velocity in m/s, we make sure to maintain consistency in the units.9.8 m/s x 0.519 m is the formula for v=r. =18.88 rad/sFor more information on angular velocity kindly visit to
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Help Please! Will Mark As Brainliest!
A. The percentage of energy of light bulb given out as light is 5%
B. The percentage of energy wasted by the mixer is 60%
C. Part of the mixer becomes hot because some energy is convert to heat energy
A. How do i determine the percentage of energy given out as light?
We can obtain the percentage of energy given out as light as follow:
Percentage of energy wasted = 95%Total energy inputted = 100%Percentage of energy given out as light =?Total energy = Wasted energy + Useful energy
100 = 95 + Percentage of energy given out as light
Collect like terms
Percentage of energy given out as light = 100 - 95
Percentage of energy given out as light = 5%
B. How do i determine the percentage of energy wasted by the mixer?The percentage of energy wasted by the mixer can be obtain as follow:
Percentage of energy used = 40%Total energy inputted = 100%Percentage of energy wasted by mixer = ?Total energy = Wasted energy + Useful energy
100 = Wasted energy + 40
Collect like terms
Wasted energy = 100 - 40
Wasted energy by mixer = 60%
C. Why is part of the mixer hot?A mixer is an equipment which converts electrical energy into mechanical energy.
However, as the mixer is working, certain amount of the energy are converted into heat energy because of the moving parts. This accounts for the hotness of some p[art of the mixer.
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which of the following statements are true about entropy process? multiple select question. processes can occur in any direction without any restriction. entropy is a thermodynamic property. the greater the extent of irreversibilities during a process, the smaller the entropy generation. the performance of engineering systems is degraded by the presence of irreversibilities. entropy generation can be used as a quantitative measure of irreversibilities.
The true statements are: 2. entropy is a thermodynamic property, 4. the performance of engineering systems is degraded by the presence of irreversibilities, and 5. entropy used as measure of irreversibilities.
Based on the given terms, here is an analysis of the statements about the entropy process:
1. Processes can occur in any direction without any restriction - False. Entropy is associated with the second law of thermodynamics, which states that for natural processes, entropy tends to increase, indicating a preferred direction.
2. Entropy is a thermodynamic property - True. Entropy is a state function that measures the energy dispersal in a system, and it is an essential property in thermodynamics.
3. The greater the extent of irreversibilities during a process, the smaller the entropy generation - False. The opposite is true. The greater the extent of irreversibilities, the larger the entropy generation.
4. The performance of engineering systems is degraded by the presence of irreversibilities - True. Irreversibilities, such as friction and heat transfer, reduce the overall efficiency of engineering systems.
5. Entropy generation can be used as a quantitative measure of irreversibilities - True. The amount of entropy generated in a process can serve as an indicator of the irreversibilities associated with that process.
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The average distance from Earth to the sun is 9.3 × 107 miles. How many kilometers isthis?A) 1.5 × 108 km D) 1.7 × 10-8 kmB) 1.5 × 105 km E) 1.5 × 1011 kmC) 5.6 × 107 km
The distance from Earth to the sun is approximately 1.5 x 10^8 kilometers.
To convert miles to kilometers, we can use the conversion factor 1 mile = 1.609344 kilometers.
So, to find the distance from Earth to the sun in kilometers, we can multiply the given distance in miles by the conversion factor:
d (km) = 9.3 x 10^7 miles x 1.609344 km/mile
d (km) = 1.496 x 10^8 km
Therefore, the distance from Earth to the sun is approximately 1.5 x 10^8 kilometers.
The closest answer choice is A) 1.5 x 10^8 km, which is the correct answer.
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the block, starting from rest, slides down the ramp a distance 34 cm before hitting the spring. how far, in centimeters, is the spring compressed as the block comes to momentary rest?
The block, starting from rest, slides down the ramp at a distance of 34 cm before hitting the spring. The distance the spring is compressed is approximately [tex]\sqrt{(0.136 sin(theta))}[/tex] cm
To solve this problem, we can use the principle of conservation of energy. The block starts with gravitational potential energy and converts it into kinetic energy as it slides down the ramp. When it hits the spring, the kinetic energy is converted into potential energy stored in the compressed spring.
First, we need to find the speed of the block when it hits the spring. We can use the equation:
mgh = 1/2 [tex]mv^2[/tex]
Where m is the mass of the block, g is the acceleration due to gravity, h is the height of the ramp, and v is the speed of the block.
We know that the block starts from rest, so its initial speed is 0. The height of the ramp is not given, but we can use the distance it travels (34 cm) to find it. If we assume the ramp is at an angle θ to the horizontal, then the height h can be found using trigonometry:
h = 34 sin(θ)
Substituting this into the equation above and solving for v, we get:
v = [tex]\sqrt{(2gh)}[/tex] = [tex]\sqrt{(2g(34 sin(theta)))}[/tex] = [tex]\sqrt{(68g sin(theta))}[/tex]
Next, we need to find how much the spring compresses when the block comes to momentary rest. We can use the equation:
1/2 [tex]kx^2[/tex] = 1/2 [tex]mv^2[/tex]
Where k is the spring constant and x is the distance the spring compresses.
We know that the mass of the block is given, and the spring constant is not given, but we can assume a value for it (let's say k = 100 N/m). Substituting in the values we have and solving for x, we get:
x = [tex]\sqrt{(2mv^2/k)}[/tex] = [tex]\sqrt{(2(0.1 kg)(68g sin(theta))/100)}[/tex] = [tex]\sqrt{(0.136 sin(theta))}[/tex] cm
Therefore, the distance the spring is compressed is approximately [tex]\sqrt{(0.136 sin(theta))}[/tex] cm. Note that the angle θ is not given, so we cannot find an exact value for x. We would need more information about the ramp and the spring to do so.
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Setting mgh = (1/2)kx^2 and solving for x, we get x = sqrt(2mgh/k). Plugging in the values given, we get x = 4.7 cm. The spring is compressed by 4.7 cm as the block comes to momentary rest. To find the distance the spring is compressed, we can use the conservation of energy principle.
The initial potential energy of the block at the top of the ramp is converted to kinetic energy as it slides down the ramp. When the block hits the spring, the kinetic energy is converted to elastic potential energy stored in the spring. Therefore, we can equate the initial potential energy to the elastic potential energy of the compressed spring.
The initial potential energy is given by mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the ramp. The elastic potential energy stored in the compressed spring is given by (1/2)kx^2, where k is the spring constant and x is the compression distance.
Assuming the ramp is frictionless, we can use the distance the block slides down the ramp, 34 cm, as the height of the ramp. We can also assume that all the kinetic energy is converted to elastic potential energy when the block hits the spring.
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17. A spinning disc rotating at 130 rev/min slows and stops 31 s later. How many revolutions did the disc make during this time?
A) 34
B) 67
C) 8.4
D) 17
E) 4.2
The number of revolutions the spinning disc makes during the given time is 34 revolutions .
To find the number of revolutions the spinning disc makes during the given time, we need to first find the average angular velocity and then multiply it by the time.
Step 1: Calculate the initial angular velocity (ω₁).
Given that the spinning disc rotates at 130 revolutions per minute (rev/min), we first convert it to revolutions per second (rev/s) by dividing by 60:
ω₁ = 130 rev/min / 60 = 2.167 rev/s
Step 2: Calculate the final angular velocity (ω₂).
Since the disc stops, its final angular velocity is 0 rev/s.
Step 3: Calculate the average angular velocity (ω_avg).
The average angular velocity is the mean of the initial and final angular velocities:
ω_avg = (ω₁ + ω₂) / 2 = (2.167 + 0) / 2 = 1.0835 rev/s
Step 4: Multiply the average angular velocity by time to find the number of revolutions.
The given time is 31 seconds:
Number of revolutions = ω_avg × time = 1.0835 rev/s × 31 s ≈ 33.6 revolutions
The closest answer among the given choices is A) 34.
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If all objects have gravity, why do you think we don’t get pulled into the other objects around us all the time?
All objects do have gravity, but the gravitational force between two objects depends on their masses and the distance between them. The force of gravity between two objects decreases rapidly as the distance between them increases.
Gravity and ObjectsIt's also worth noting that objects need to be very massive and very close together for the gravitational force to become noticeable. For example, two people standing next to each other have a very small gravitational force between them, while two planets orbiting each other have a much stronger gravitational force.
In summary, while all objects have gravity, the gravitational force between objects depends on their masses and the distance between them, and the force of gravity between us and nearby objects is usually too small to have a noticeable effect. The force of gravity between us and the Earth is what keeps us in place and gives us weight.
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Question 12
The greatest exposure to human made ionizing radiation for most individuals is through:
a. Eating plants that contain radioactive elements
b. Medical x-rays
c. Nuclear power plant emissions
d. Building made of stone
The greatest exposure to human-made ionizing radiation for most individuals is through medical X-rays.
Therefore the answer is b. Medical x-rays.
Medical X-rays are a common source of ionizing radiation exposure for many people. X-rays use electromagnetic radiation to produce images of the inside of the body, and the radiation can potentially damage living tissue at high doses. While the dose from a single X-ray is usually small, frequent or unnecessary medical imaging can increase an individual's cumulative exposure. Eating plants that contain radioactive elements, living in a building made of stone, and exposure to nuclear power plant emissions can also result in ionizing radiation exposure, but these sources are generally less significant than medical X-rays for most individuals.
Therefore, the greatest exposure to human made ionizing radiation for most individuals is through medical X-rays. Hence the correct answer is option b.
The level of radiation exposure also depends on a variety of factors, such as the type of radiation, the duration of exposure, and the distance from the source. To minimize the risk of radiation exposure, it is important to use medical imaging only when necessary, follow appropriate safety procedures in radiation-related occupations, and limit exposure to other sources of ionizing radiation when possible.
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a microscope with an overall magnification of 750 has an objective that magnifies by 150. (a) what is the magnification of the eyepiece in multiples?
Hello! I'd be happy to help with your microscope question. To find the magnification of the eyepiece, you'll need to use the following formula:
Overall Magnification = Objective Magnification × Eyepiece Magnification
Given:
Overall Magnification = 750
Objective Magnification = 150
Now, we need to find the Eyepiece Magnification:
750 = 150 × Eyepiece Magnification
To find the Eyepiece Magnification, divide the Overall Magnification by the Objective Magnification:
Eyepiece Magnification = 750 / 150
Eyepiece Magnification = 5
So, the magnification of the eyepiece is 5 times.
The magnification of the eyepiece in the given microscope is 5x .
To calculate the magnification of the eyepiece, we need to use the formula:
Total Magnification = Objective Magnification x Eyepiece Magnification
Given that the overall magnification of the microscope is 750 and the objective magnifies by 150, we can plug those values into the formula and solve for the eyepiece magnification:
750 = 150 x Eyepiece Magnification
Eyepiece Magnification = 750 / 150
Eyepiece Magnification = 5
Therefore, the magnification of the eyepiece in multiples is 5x.
The eyepiece, also known as the ocular lens, is located at the top of the microscope and is responsible for further magnifying the image produced by the objective lens.
The eyepiece magnification, when combined with the objective magnification, determines the total magnification of the microscope.
It's important to note that the total magnification of a microscope is not an indicator of the quality or clarity of the image produced.
Other factors such as resolution, field of view, and depth of field also play a crucial role in determining the overall performance of a microscope.
In conclusion, the magnification of the eyepiece in the given microscope is 5x, and understanding how the different components of a microscope work together is important in achieving optimal results.
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5. When the LRC circuit in this experiment is driven at its resonance frequency the voltage across the resistor will be:
When the LRC circuit in this experiment is driven at its resonance frequency, the voltage across the resistor will be maximum. This is because at resonance frequency, the reactance of the inductor and capacitor cancels out, resulting in a minimum impedance in the circuit.
Therefore, the current in the circuit will be maximum, leading to a maximum voltage across the resistor according to Ohm's law. A resistor (R), an inductor (L), and a capacitor (C) are the three parts of an LRC circuit, a sort of electrical circuit. From the initials of these three parts, the word LRC is derived. The interaction of the resistor, inductor, and capacitor controls how an LRC circuit behaves. The capacitor stores energy in an electric field, the inductor stores energy in a magnetic field, and the resistor dissipates energy. The circuit oscillates at the resonant frequency as a result of the interaction between these parts.
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