Due to less information provided, The electric field at 1/5 of the way between two charges can be zero if the charges are of equal magnitude, or if the charges are of unequal magnitude but the distance between the two charges is an exact multiple of 5.
What is Electric Field?
A physical quantity associated with an electric charge or a charged particle is known as an electric field. It is a vector field that describes the force exerted by the first particle on another charged particle. The electric potential of the charge, which is often measured in volts, defines the strength and direction of an electric field. A vector field is used to describe an electric field, with the electric field vector E at each point in space proportional to the force acting on a unit positive test charge at that point.
What does Magnatitude means?
Magnatitude is a term used to describe the strength or intensity of something, such as an emotion, a physical force, or an event. It is often used to describe the intensity of an earthquake, hurricane, or other natural disaster. In some cases, the term can also be used to describe the power of a person or group, such as the magnatitude of a leader's influence or the magnatitude of a corporation's reach.
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If the mass of an object is 10 kg and the velocity is 8 m/s, what is the momentum?
A. 8 kgm/s
B. 120 kgm/s
C. 80 kgm/s
D. 40 kgm/s
Answer:
80 kgm/s
Explanation:
Momentum = Mass x Velocity
It can be expressed as [tex]\displaystyle{p = mv}[/tex] where p is momentum, m is mass and v is velocity.
We know that mass is 10 kg and velocity is 8 m/s - therefore, substitute the given information in formula:
[tex]\displaystyle{p=10 \ \times \ 8}\\\\\displaystyle{p=80 \ \ kgm/s}[/tex]
Hence, the momentum is 80 kgm/s.
Please I need help! This is the last question I need for this assignment!
Part A
Compare the temperature change for cold sand and cold water when the same amount of hot water was added. What do you discover?
Answer:
When the same amount of heat is added to cold sand and cold water, the temperature change of sand will be higher because of its lower specific heat capacity.
What is specific heat capacity?
Specific heat capacity is the quantity
of heat required to raise a unit mass of
a substance by 1 kelvin.
Specific heat capacity of water and sand
{refer to the above attachment}
Δθ = Q/mc
Thus, for an equal mass of water and sand, when the same amount of heat is added to cold sand and cold water, the temperature change of sand will be higher because of its lower specific heat capacity.
A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale.
Evaluate the magnitude of the force on the left hand pole.
Answer: A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale. Then, the magnitude of the force on the left-hand pole will be, 0.167N.
Explanation: To find the correct answer, we have to know more about the Basic forces that acts upon a body.
What is force and which are the basic forces that acts upon a body?A push or a pull which changes or tends to change the state or rest, or motion of a body is called Force.Force is a polar vector as it has a point of application.Positive force represents repulsion and the negative force represented attraction.There are 3 main forces acting on a body, such as, weight mg, normal reaction N, and the Tension or pulling force.How to solve the problem?We have given that, the gold chain hangs between the vertical sticks of 30cm and the horizontal distance between then is 30cm.From the given data, we can find the angle [tex]\alpha[/tex] (in the free body diagram, it is given as θ).[tex]tan\alpha =\frac{30}{30}[/tex]
[tex]\alpha =tan^{-1}(1)=45[/tex] degree.
From the free body diagram given, we can write the balanced equations of total force along y direction as,[tex]y- direction,\\T_2sin\alpha =mg\\T_2=\frac{mg}{sin \alpha } =\frac{17.1*10^{-3}kg*9.8m/s^2}{sin 45}=0.236 N[/tex]
From the free body diagram given, we can write the balanced equations of total force along x direction as,[tex]x- direction\\T_1-T_2cos\alpha =0\\T_1=0.236*cos45=0.167N[/tex]
Thus, we can conclude that, the magnitude of force on the left-hand pole will be 0.167N.
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Answer:
0.1426 N
Explanation:
A chain of uniform mass density is suspended between two poles 30 cm apart. The geometry of the problem is such that the left support only supplies a horizontal force on the chain. The right support must both balance that horizontal force and supply a vertical force that balances the weight of the chain.
Magnitude of forcesFor some tension T in the chain at the right support, the vertical force will be ...
vertical force = T·sin(α) = W . . . . . matches the weight (W) of the chain
for some angle α between the horizontal and the chain at the right pole.
The corresponding horizontal force is ...
horizontal force = T·cos(α)
This force balances the horizontal force at the left support pole. In terms of W, this force is ...
horizontal force = W/sin(α)·cos(α) = W/tan(α)
AngleThe curve assumed by a chain of uniform mass density can be demonstrated to be a catenary. For supports 30 cm apart, its equation can be described by ...
y = 30·cosh(x/30)
The diagram shows that y=4 for x=0, so we need to subtract 26 cm from this:
y = 30·cosh(x/30) -26
The slope of the curve at any point is the derivative of this function:
y' = 30(1/30)(sinh(x/30)) = sinh(x/30)
At the right support, the slope of the curve is ...
y' = sinh(30/30) = sinh(1) ≈ 1.1752012
This is the tangent of the angle that the curve makes with the horizontal at the right support.
tan(α) = 1.1752012
Note, you can see from the grid squares on the graph that the slope at the right support is slightly more than 1.
WeightThe weight of the chain is the product of its mass and the acceleration due to gravity:
W = ma = (0.0171 kg)(9.8 m/s²) = 0.16758 N
Force on the PoleThen the force on the left-side pole is ...
horizontal force = W/tan(α) = (0.16758 N)/1.1752012
horizontal force ≈ 0.1426 N
__
Additional comment
The attached graph is a plot of the catenary curve we have assumed for the gold chain. We have attempted to match the vertical height on the left side, but we note that there seems to be a small discrepancy at the right side. The graph in the problem statement seems to show the right attach point at about y=21, not 20.3.
A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?
The vertical component of the force being applied to the beam by the high-tension cable is 36.37 N.
Calculation of the cable's tension-
Utilizing the moment principle, determine the cable's tension:
Torque clockwise = TL sin∅
Counterclockwise torque = 1/2WL
TL sin∅ = 1/2WL
⇒T sin∅ = 1/2W
⇒T = W/2sin∅
⇒T = (29* 9.8)/ (2*sin57)
⇒T = 169.43 N
Calculation of the vertical component of the force-
Forces that operate perpendicular to the surface vertical plane are called the vertical force. Gravity always pulls objects straight down to the earth's core.
T+F = W
⇒F = W-T
⇒F = (21*9.8)-169.43
⇒F = 36.37 N
So, the force on the beam has a vertical component of 36.37 N.
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100gm o2 gas is pressurized to 20 degree Celsius. done Also, how much heat energy will be converted into mechanical energy?
The heat energy that will be converted into mechanical energy is 1.83 kJ.
Heat capacity of the O2 gas
The heat energy that will be converted into mechanical energy is calculated as follows;
Q = mcΔθ
where;
m is mass = 100 g = 0.1 kgΔθ is change in temperaturec is specific heat capacityat 20 ⁰C = 293 K, C = 0.915 kJ/kg K
Q = (0.1 kg)(0.915)(20 )
Q = 1.83 kJ
Thus, the heat energy that will be converted into mechanical energy is 1.83 kJ.
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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?
The tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
Tension in the cableApply the principle of moment and calculate the tension in the cable;
Clockwise torque = TL sinθ
Anticlockwise torque = ¹/₂WL
TL sinθ = ¹/₂WL
T sinθ = ¹/₂W
T = (W)/(2 sinθ)
T = (29 x 9.8)/(2 x sin57)
T = 169.43 N
Vertical component of the forceT + F = W
F = W - T
F = (9.8 x 29) - 169.43
F = 114.77 N
Thus, the tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
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When a 2.50 - kg object is hung vertically on a certain light spring described by Hooke’s law, the
spring stretches 2.76 cm. (a) What is the force constant of the spring? (b) If the 2.50 - kg object is
removed, how far will the spring stretch if a 1.25 - kg block is hung on it? (c) How much work must
an external agent do to stretch the same spring 8.00 cm from its unstretched position?
The work done in the spring is calculated to be 2.8 J
What is Hooke's law?Hooke's law states that, the extension of a given material is directly proportional to the applied force as long as the elastic limit is not exceeded . First, we must bear in mind that the material must remain within the elastic limit for us to apply the Hooke's law in solving the problem.
Now;
From Hooke's law;
F = Ke
F = force applied
K = force constant
e = extension
F = W = mg = 2.50 - kg * 9.8 m/s^2 = 24.5 N
K = 24.5 N/ 2.76 * 10^-2
K = 888 N/m
e = F/K
F = W = 1.25 - kg * 9.8 m/s^2 = 12.25 N
e = 12.25 N/ 888 N/m = 0.014 m or 1.4 cm
Work done by an external agent = 1/2 Kx^2
= 0.5 * 888 * (8 * 10^-2)^2
= 2.8 J
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Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N. Calculate the magnitude of the tension in the string marked A.
The magnitude of the tension in the string marked A is 52.5 N.
Given that the magnitude of the tension in the string labeled C is 56.3 N.
The angle at A is
tan θ = [tex]\frac{3}{8}[/tex]
below the negative x
B= tan Φ
tan Φ = [tex]\frac{5}{4}[/tex]
C = tan ρ
tan ρ = [tex]\frac{1}{6}[/tex]
Considering the Horizontal components only
74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)
A = 78.9 - 0.668B
Considering the Vertical components only
74.9*Sin(9.46) + ASin(20.6) = BSin(51.3)
40.07 = 1.015B
B = 39.5 N
By substituting the value of B in the equation of A
Since, A = 78.9 - 0.668B
A = 78.9 - 0.668( 39.5 N)
A = 52.5 N
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What does it mean to say that two systems are in thermal equilibrium
Answer:
In simple words, thermal equilibrium means that the two systems are at the same temperature.
How does an atom of rubidium-85 a rubidium ion with a +1 charge?
A. The atom loses 1 electron to have a total of 47.
B. The atom gains proton to have a total of 38.
C. The atom loses 1 electron to have a total of 36,
D. The atom gains 1 proton to have a total of 86.
Answer:
C. The atom loses 1 electron to have a total of 36
Explanation:
The number of electrons in a Rubidium atom is 37. Since the atom loses 1 electron, it has 36 left.
Determine the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/60 of its value at the Earth's surface.
The distance from the Earth's center to the point outside the Earth is 55800 Km
How to determine the distance from the surface of the EarthAcceleration due to gravity of Earth = 9.8 m/s²Acceleration due to gravity of the poin (g) = 1/60 × 9.8 = 0.163 m/s²Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²Mass of the Earth (M) = 5.97×10²⁴ KgDistance from the surface of the Earth (r) =?g = GM / r²
Cross multiply
GM = gr²
Divide both sides by g
r² = GM / g
Take the square root of both sides
r = √(GM / g)
r = √[(6.67×10¯¹¹ × 5.97×10²⁴) / 0.163)]
r = 4.94×10⁷ m
Divide by 1000 to express in Km
r = 4.94×10⁷ / 1000
r = 4.94×10⁴ Km
How to determine the distance from the center of the EarthDistance from the surface of the Earth (r) = 4.94×10⁴ KmRadius of the Earth (R) = 6400 Km Distance from the centre of the Earth =?Distance from the centre of the Earth = R + r
Distance from the centre of the Earth = 6400 + 4.94×10⁴
Distance from the centre of the Earth = 55800 Km
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An electron gun shoots electrons at a metal plate that is 4.0 mm away in a vacuum. The plate is 5.0 V lower in potential than the gun. How fast must the electrons be moving as they leave the gun if they are to reach the plate?
The speed of the electron is 1.3 * 10^6 m/s
What is the velocity?We know that when the electron gun is shot, the potential energy of the electron is converted into kinetic energy. The mass of the electron is given as 9.11 * 10^-31 Kg.
The energy of the electron is;
eV = 1e * 5V = ev or 8 * 10^-19 J
Given that E = 1/2mv^2
8 * 10^-19 = 0.5 * 9.11 * 10^-31 * v^2
v = √ 8 * 10^-19/0.5 * 9.11 * 10^-31
v = √1.75 * 10^12
v = 1.3 * 10^6 m/s
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john has 4 apples , is train is 7 minutes early calculate te mass of the sun
Answer:
The mass of Sun doesn't change with respective to the conditions.
Michael has 4 Apples, which may increase his own mass or weight but not the Sun's .
His train is 7 minutes, but this doesn't mean the Sun has been made to change. The train coming late affects the time management and delays work.
So, As the per the question, It is evident that Sun's Mass is still the same irrespective of conditions .
Hence, The required answer Sun's Mass is 2*10^30 kg
Explanation:
What is the maximum speed with which a 1200- kg car can round a turn of radius 90.0 m on a flat road if the coefficient of static friction between tires and road is 0.55?
The maximum speed with which a 1200- kg car can make a turn is 22.02 m/s.
Maximum speed of the carv(max) = √μrg
where;
μ is coefficient of frictionr is radius of the curved roadg is acceleration due to gravityv(max) = √(0.55 x 90 x 9.8)
v(max) = 22.02 m/s
Thus, the maximum speed with which a 1200- kg car can make a turn is 22.02 m/s.
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Question 8: Cosmology (8 points)
a. Write 3 - 4 sentences to describe the beginning of the universe according to the big bang theory, and to describe the future of the universe according to the flat model. (4 points)
b. What is cosmic background radiation? How do observations of the cosmic background radiation provide evidence to support the big bang theory? Write 2 - 3 sentences to present your response. (4 points)
Answer in complete sentences. Will mark brainiest
Big bang happened about 13.7 billion years ago in our universe.
Describe the beginning of the universe according to the big bang theory?According to the big bang theory, about 13.7 billion years ago, an explosive expansion began, expanding our universe outwards faster than the speed of light.
Describe the future of the universe according to the flat model?According to the flat model, the universe is infinite and will continue to expand forever because the universe is expanding.
What is cosmic background radiation?Cosmic background radiation is a weak radio-frequency radiation that is traveling through outer space in every direction. It is the residual radiation of the big bang, when the universe was very hot.
How do observations of the cosmic background radiation provide evidence to support the big bang theory?The Big Bang theory predicts that the early universe was a very hot place and that as it expands, the gas within it cools. Thus the universe has all over radiation which is called the “cosmic microwave background".
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In an experiment replicating Millikan’s oil drop experiment, a pair of parallel plates are placed 0.0200 m apart and the top plate is positive. When the potential difference across the plates is 240.0 V, an oil drop of mass 2.0 × 10-11 kg gets suspended between the plates. (e = 1.6 × 10-19 C)
a) Draw a free-body diagram for the charge.
b) What is the charge on the oil drop?
c) Is there an excess or deficit of electrons on the oil drop? How many electrons are in excess or deficit?
Answer: See below
Explanation:
Given:
The potential between plates, V = 240 V
Distance between plates, d = 0.02 m
The mass of drop, m = 2x10^-11
Charge on electron, e = 1.6x10^-19
Part (a)
The free-body diagram is attached below
Part (b)
The electric field is given by,
[tex]E=\frac{V}{d}[/tex]
On applying force balance, the force on oil drop is equal to the weight of the oil,
[tex]$$\begin{aligned}F_{E} &=m g \\q E &=m g \\q \frac{V}{d} &=m g \\q &=\frac{m g d}{V}\end{aligned}$$[/tex]
Substituting the given values in the above equation,
[tex]\begin{aligned}&q=\frac{2 \times 10^{-11} \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2} \times \frac{1 \mathrm{~N}}{1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}} \times 0.02 \mathrm{~m}}{240 \mathrm{~V} \times \frac{1 \mathrm{~N} \cdot \mathrm{m} / \mathrm{C}}{1 \mathrm{~V}}} \\&q=1.63 \times 10^{-14} \mathrm{C}\end{aligned}[/tex]
Therefore, the charge on the oil drop is 1.63x10^-14 C
Part (c)
There will be an excess of electrons on the oil drop.
The number of electrons on oil drop can be calculated as,
[tex]\begin{aligned}q &=n e \\1.63 \times 10^{-14} \mathrm{C} &=n \times 1.6 \times 10^{-19} \mathrm{C} \\n &=1.01 \times 10^{5}\end{aligned}[/tex]
Therefore, the number of excess electrons is 1.01x10^5
Find the orbital speed of an ice cube in the rings of Saturn. The mass of Saturn is 5.68 x 1026 kg, and use an orbital radius of 3.00 x 105 km. (G = 6.67 × 10-11 N ∙ m2/kg2)
The orbital speed of an ice cube in the rings of Saturn is determined as 11,237.7 m/s.
What is orbital speed?The orbital speed of an astronomical body or object is the speed at which it orbits around the center of mass of the most massive body.
Orbital speed of ice cube in the rings of SaturnThe orbital speed of ice cube in the rings of Saturn is calculated as follows;
v = √GM/r
where;
G is universal gravitation constantM is mass of Saturnr is the distance of the ice cube = 3 x 10⁸ mv = √(6.67 x 10⁻¹¹ x 5.68 x 10²⁶)/(3 x 10⁸)
v = 11,237.7 m/s
Thus, the orbital speed of an ice cube in the rings of Saturn is determined as 11,237.7 m/s.
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Two builders carry a sheet of drywall up a ramp. Assume that W = 1.80 m, L = 3.30 m, θ = 24.0°, and that the lead builder carries a (vertical) weight of 147.0 N (33.0 lb).
1. What is the (vertical) weight carried by the builder at the rear?
2. The builder at the rear gets tired and suggest that the drywall should be held by its narrow side. What is the weight (in N) he must now carry?
The vertical weight carried by the builder at the rear is 240.89N. The weight he must now carry is 352.26N
1. How to solve for the vertical weightWe have w = 1.8
Then we have L as 3.30
θ = 24.0
FC = 147
We have to find FB
147 (3.3 + 1.8 tan24)/(3.3 - 1.8 tan24)
= 240.896
The vertical weight carried by the builder is 240.896
2. 240.896 + 147
= 387.896
387.896/[1 + (1.8 + 3.3 tan24) /(1.8 - 3.3 tan24)]
= 387.896/10.885
= 35.64
387.896 - 35.64
= 352.26N
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which object has a weight of about 22.5 n the book the rock the box the fish
Answer: The rock
Explanation:
A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .
a) What is the magnitude of the magnetic field at the center of the coil?
b) What is the magnitude of the magnetic field at a point on the axis of the coil a distance of 9.50 cm from its center?
Hi there!
a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}[/tex]
dB = Differential Magnetic field element
μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
R = radius of loop (2.15 cm = 0.0215 m)
i = Current in loop (0.460 A)
For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}[/tex]
Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
[tex]B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}[/tex]
Taking out constants from the integral:
[tex]B =\frac{\mu_0 i}{4\pi R^2} \int ds[/tex]
Since we are integrating around an entire circle, we are integrating from 0 to 2π.
[tex]B =\frac{\mu_0 i}{4\pi R^2} \int\limits^{2\pi R}_0 \, ds[/tex]
Evaluate:
[tex]B =\frac{\mu_0 i}{4\pi R^2} (2\pi R- 0) = \frac{\mu_0 i}{2R}[/tex]
Plugging in our givens to solve for the magnetic field strength of one loop:
[tex]B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T[/tex]
Multiply by the number of loops to find the total magnetic field:
[tex]B_T = N B = 0.00631 = \boxed{6.318 mT}[/tex]
b)
Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}[/tex]
In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}[/tex]
Using the diagram, if 'z' is the point's height from the center:
[tex]r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}[/tex]
Substituting this into our expression:
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }[/tex]
Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
[tex]B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds[/tex]
Evaluate:
[tex]B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}[/tex]
Multiplying by the number of loops:
[tex]B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}[/tex]
Plug in the given values:
[tex]B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ = 0.00006795 = \boxed{67.952 \mu T}[/tex]
If Earth were a perfect sphere, would you weigh more or less at the equator than at the poles? Explain
Answer:
You would weigh the same.
Explanation:
At the moment, since Earth is not a perfect sphere, the Earth "bulges out" at the equator, so you're further from the centre of the Earth. Since gravity acts through a body's center of mass, the further you are from the centre the weaker the gravitational acceleration you will feel, because gravity weakens over distance.
So, you're actually lighter at the equator than you'd be at the poles.
However, if the Earth was a perfect sphere, this "bulge" at the equator would not happen, and so you would weigh the same at the poles and at the equator.
Hope this makes sense.
P = Patm + pgh is which law
The law that relates the absolute pressure to atmospheric pressure and gauge pressure is Pascal law.
What is Pascal law?
Pascal's law states that when an object is immersed in a fluid, it experiences equal pressure on all surfaces.
P = Patm + pgh
where;
P is absolute pressurepgh is gauge pressurePatm is atmospheric pressureThus, the law that relates the absolute pressure to atmospheric pressure and gauge pressure is Pascal law.
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A baseball (m=145g) traveling 39 m/s moves a fielder's glove backward 23 cm when the ball is caught.
What was the average force exerted by the ball on the glove?
Express your answer to two significant figures and include the appropriate units.
The average force exerted by the ball on the glove is 480 N.
What is the force exerted by the ball on the glove?
The average force exerted on the glove by the ball is equal in magnitude to the force on the ball.
Force = mass * accelerationmass = 145 g = 0.145 kg
Acceleration of the baseball, a = (v² - u²)/2s
where:
v is final velocity = 0
u is initial velocity = 39 m/s
s is distance = -23 cm = 0.23 m
a = (0 - 39²)/2(-0.23)
a = 3306.52 m/s²
Force = 0.145 * 3306.52
Force = 479.4 N
Average force = 480 N
In conclusion, force is derived from the product of mass and acceleration.
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Two stretched copper wires both experience the same stress. The first wire has a radius of 3.9×10-3 m and is subject to a stretching force of 450 N. The radius of the second wire is 5.1×10-3 m. Determine the stretching force acting on the second wire.
The stretching force acting on the second wire, given the data is 588 N
Data obtained from the questionRadius of fist wire (r₁) = 3.9×10⁻³ mForce of first wire (F₁) = 450 NRadius of second wire (r₂) = 5.1×10⁻³ mForce of second wire (F₂) =?How to determine the force of the second wireF₁ / r₁ = F₂ / r₂
450 / 3.9×10⁻³ = F₂ / 5.1×10⁻³
cross multiply
3.9×10⁻³ × F₂ = 450 × 5.1×10⁻³
Divide both side by 3.9×10⁻³
F₂ = (450 × 5.1×10⁻³) / 3.9×10⁻³
F₂ = 588 N
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) Define magnetic flux density
how we will measure centimeters
Answer:
.Explanation:
——»To measure centimeters, we can use ruler.
Use a ruler with the side marked either cm or mm. Align the edge of the object with the first centimeter line on the ruler, then find the length in whole centimeters, or the larger numbers on the ruler.Which statement is a postulate of general relativity?
The speed of light is constant for all observers.
Observers will see the same laws of physics whether at rest or in uniform motion.
A gravitational field is the same as an object moving at the speed of light.
Observers will see the same laws of physics in any frame of motion whether accelerated or not
Answer:
The speed of light is constant for all observers
Explanation:
As per general postulate of relativity
Lorentz covariance of special relativity becomes a local Lorentz covariance in the presence of matter.So
light speed c is independent of States of matter
The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×10^3 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10^-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×10^5 m around the planet Glob. Calculate the speed of the satellite.
Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
Explanation: To find the answer, we need to know about the different equations of planetary motion.
How to find the initial speed of the rock as it left the astronaut's hand?We have the expression for the initial velocity as,[tex]v=\sqrt{2gh}[/tex]
Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,[tex]g_g=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2}= 0.132[/tex]
Now, the velocity will become,[tex]v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s[/tex]
How to find the speed of the satellite?As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,[tex]v=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10^{-11}*7.88*10^{18}}{1.45*10^5} } =3.624km/s[/tex]
Thus, we can conclude that,
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
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The tiny planet known as "Glob" has a radius of 6.32× 10^4 meters and a mass of 7.88× 10^18 kg. On Glob's surface, an astronaut launches a rock straight upward. Before falling back down, the rock rises to a maximum height of 1.44×10^3 m above the planet's surface.
1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.
2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.
Understanding the planetary motion equations is necessary in order to determine the solution.
How to determine the rock's original speed when it left the astronaut's hand?The starting velocity's expression is as follows:[tex]V=\sqrt{2gh}[/tex]
So, in order to determine v, we must determine the acceleration of glob caused by gravity. We already have,[tex]a=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2} \\a=0.132m/s^2[/tex]
The velocity will now change to,[tex]V=\sqrt{2*0.132*1.44*10^3} =19.46m/s[/tex]
How can I determine the satellite's speed?As we are aware, the centripetal force and gravitational force are equivalent, and thus leads to the following satellite speed equation:[tex]v=\sqrt{\frac{GM}{r} } =3,624km/s\\where,\\M=7.88*10^{18}kg[/tex]
Consequently, we can say that
1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.
2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.
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Which of the following accurately describes the behavior of water when subjected to temperature change? Question 9 options: A) The volume of water will decrease if heated from 6°C to 7°C. B) The volume of water will increase if cooled from 3°C to 2°C. C) A mass of water will contract if cooled from 1°C to 0°C. D) A mass of water will expand if heated from 0°C to 2°C.
The behavior of water when subjected to temperature change is that the volume of water will increase if cooled from 3° to 2°C.
The chemical compound water, which can exist in the gaseous, liquid, and solid phases, is made up of the elements hydrogen and oxygen in the ratio 2: 1 i.e. 2 atoms of hydrogen and 1 atom of oxygen.
In general
Volume of water depends on the temperature and is directly proportional to it.
Thus, as the temperature rises, the molecules of water gain energy and move more quickly, which causes the molecules to spread apart and increase the volume of the liquid.
When water cools, it initially contracts (decreases in volume) until a temperature of about four degrees Celsius (4°C).
But at temperatures below 4.0° C, water undergoes an abnormal expansion that causes its volume to start to rise.
This ability is related to the formation of hexagonal structures, which take up a lot of room and increase the volume of the water, as a result of strong hydrogen bonding between water molecules at a lower temperature.
Hence, the volume of water will increase if cooled from 3° to 2°C
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A car is traveling 30 m/s around a curve of radius 100 m. What is the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from skidding?
The minimum value of the coefficient of static friction is equal to 0.92.
How to determine the minimum value of the coefficient of static friction?First of all, we would derive an expression for the horizontal and vertical component of forces acting acting on the car.
For the vertical component, we have:
∑Fy = 0
Fn + Fg = 0
Fn - mg = 0
Fn = mg .....equation 1.
For the horizontal component, we have:
∑Fx = mAc
uFn = m(V²/r) .....equation 2.
Substituting eqn. 1 into eqn. 2, we have:
umg = m(V²/r)
u = 1/g(V²/r)
u = (V²/gr)
u = (30²/9.8 × 100)
u = 900/980
u = 0.92.
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