Answer: True
hope this helps!
What is the speed of a giraffe that has a
mass of
75 kg and a kinetic energy of 600 J?
O 2 m/s
O
4 m/s
O 8 m/s
O 16 m/s
Name two units for measuring the diameter of nucleus atom.
The two units for measuring the diameter of nucleus atom are femtometre and metre.
How do you measure the size of the nucleus ?
Nucleus size is expressed in fermi, often known as femtometers. between a lighter and a heavier nucleus. Despite its modest size, the nucleus contains the majority of an atom's mass. The weight or mass of the atom's nucleus and neutrons are determined by neutrons.
femtometre (fm), which equals [tex]10^{-15}[/tex] metre.
A nucleus' diameter largely depends as to how many particles it contains, from about 4 fm for a light nucleus like carbon to 15 fm for a heavy nucleus as lead.
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According to Howard Gardner's theories, who among the following would need to have good linguistic intelligence to be successful?
A psychologist will need to have good linguistic intelligence in other to be successful.
Who is a Psychologist?This is referred to as a professional who specializes in the handling of mental health challenges in individuals.
It is best for such professional to have a good linguistic intelligence as the right words being said to the patient will solve the problem thereby bringing in more success.
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A block of mass
m = 2.50 kg
is pushed
d = 2.10 m
along a frictionless horizontal table by a constant applied force of magnitude
F = 14.0 N
directed at an angle
= 25.0°
below is a photo of the horizontal as shown in the figure below.
(a) Determine the work done by the applied force.
_____J
(b) Determine the work done by the normal force exerted by the table.
_____J
(c) Determine the work done by the force of gravity.
_____J
(d) Determine the work done by the net force on the block.
_____J
(a) The work done by the applied force is 26.65 J.
(b) The work done by the normal force exerted by the table is 0.
(c) The work done by the force of gravity is 0.
(d) The work done by the net force on the block is 26.65 J.
Work done by the applied force
W = Fdcosθ
W = 14 x 2.1 x cos25
W = 26.65 J
Work done by the normal forceW = Fₙd
W = mg cosθ x d
W = (2.5 x 9.8) x cos(90) x 2.1
W = 0 J
Work done force of gravityThe work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.
Work done by the net force on the block∑W = 0 + 26.65 J = 26.65 J
Thus, the work done by the applied force is 26.65 J.
The work done by the normal force exerted by the table is 0.
The work done by the force of gravity is 0.
The work done by the net force on the block is 26.65 J.
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What word means the same thing as force of gravity?
A. Density
B. Weight
C. Mass
D. Volume
Answer:
B. Weight
Explanation:
Weight is the force of gravity exerted on a body
OR
It's the force exerted on a body by the influence of the earth's gravitational force
A pendulum in motion can either swing from side to side or turn in a continuous circle. The point at which it goes from one type of motion to the other is called the separatrix, and this can be calculated in most simple situations. When the pendulum is prodded at an almost constant rate though, the mathematics falls apart. Is there an equation that can describe that kind of separatrix?
The kind of equation that can be used to differentiate the kind of separatrix that shows change on motion is
H = 2g/l.
What is simple pendulum?A simple pendulum can be defined as the equipment that displays an oscillatory motion when a mass is tied on a rope and is suspended from it.
The various movements that occur using a simple pendulum is translational ( side to side) or continuous circle (oscillatory motion).
The equation that show that a change from one type of motion to another is H = 2g/l.
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Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b). By what angle has planet Y rotated through during this time?
According to mathematics, the planet's angle is stated as
dY=704 degrees.
What is the current rotational angle of planet Y?We may demonstrate this by using Kepler's third law, which asserts that a planet's orbit squared is a function of cubed radius.
The equation for the period is often expressed numerically as
[tex](periodX / periodY)^2 = (radius X / radius Y)^3[/tex]
Therefore
(pX / pY)^2 = 4^3
(pX / pY)^2 = 64
[tex]\sqrt{(pX / pY )^2}= \sqrt{64}[/tex]
pX / pY=8
In conclusion, planet Y travels 8 times further than planet X does in the same amount of time since one orbit on planet X takes 8 times longer to complete.
planet Y travels ;
dY=8 * 88.0
dY= 704 degrees
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if you dip your hand in cold water after having dipped in warm water, will you feel the water colder than it actually is?
Answer:
Yes.
Explanation:
When you move your hand from the cold water to the “warmer” (room temp) water, one hand feels warm.
As you move your hand from the warm water to the “colder” (room temp) water, that hand feels colder.
Although both hands experience the last bowl of water at the same temperature, your brain senses two
separate sensations. So the water feels “warm” or “cold” relative to the water your hand was in previously.
The greater the difference in temperature, the easier it is to sense a difference.
A satellite is in a circular orbit very close to the surface of a spherical planet. The period of the orbit is 2.35 hours. What is density of the planet? Assume that the planet has a uniform density.
The density of the planet is determined as 1,974.26 kg/m³.
Density of the planet
√(⁴/₃πGρ) = 2π/(2.35 x 3600)
where;
ρ is density of the planetG is universal gravitation constant√(⁴/₃πGρ) = 2π/(2.35 x 3600)
√(⁴/₃πGρ) = 2π/8460
(⁴/₃πGρ) = (2π/8460)²
⁴/₃πGρ = 4π²/(8460)²
ρ = 12π/(8460² x 4G)
ρ = (12π) / (8460² x 4 x 6.67 x 10⁻¹¹)
ρ = 1,974.26 kg/m³
Thus, the density of the planet is determined as 1,974.26 kg/m³.
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D. A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weight to be 7.84 N. She then weighs the crown while it is immersed in water, and now the scale reads 6.86 N. Is the crown made of pure gold? Find the density of the crown and compare it to the it to density of the gold.
Answer:
Wc = 7.84 weight of crown
Ww = 7.84 - 6.86 = .98 weight of water displaced
Density = 7.84 / .98 = 8 crown is 8 X that of water
Since gold has a density of 19.3 that of water the crown is certainly not 100 percent (if any) gold
A boy of mass 30.0 kg is sledding down a 70.0-m slope starting from rest. The slope is angled at 15.0° below the horizontal. After going 28.0 m along the slope, he passes his friend, who hops onto the sled. The friend has a mass of 50.0 kg, and the coefficient of kinetic friction between the sled and the snow is 0.120. Ignoring the mass of the sled, find their speed at the bottom.
In m/s
The speed of the boy and his friend at the bottom of the slope is 16.52 m/s.
Their speed at the bottomApply the principle of conservation of energy,
E(up) - E(friction) = E(bottom)
mg sin(15) + ¹/₂(M + m)u² - μ(M + m)cos 15 = ¹/₂(M + m)v²
[tex]v = \sqrt{2[\frac{mgd \ sin15 \ + \frac{1}{2}(M + m)u^2 \ -\mu (M + m)g cos\ 15 }{M + m}] }[/tex]
where;
u is the speed of the after 28 mu = √2gh
u = √(2gL sin15)
u = √(2 x 9.8 x 28 x sin 15)
u = 11.92 m/s
[tex]v = \sqrt{2[\frac{(30)(9.8)(70) \ sin15 \ + \frac{1}{2}(30 + 50)(11.92)^2 \ - 0.12 (30 + 50)9.8 cos\ 15 }{30 + 50}] }\\\\v = 16.52 \ m/s[/tex]
Thus, the speed of the boy and his friend at the bottom of the slope is 16.52 m/s.
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Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radius of Earth = 6.37×103 km, mass of Earth = 5.98×1024 kg, G = 6.67×10-11 Nm2/kg2.)
The speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.
Speed of the satellitev = √(GM/r)
where;
G is universal gravitation constantM is mass of Earthr is radius of the satellitev = √(6.67 x 10⁻¹¹ x 5.98 x 10²⁴/3.57 x 6.37x 10³)
v = 1.32 x 10⁵ m/s
Thus, the speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.
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A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long as 2.20 m have been recorded. If a frog jumps 2.20 m and the launch angle is 36.5°, find the frog's launch speed and the time the frog spends in the air. Ignore air resistance.
(a)the frog's launch speed (in m/s)
(b)the time the frog spends in the air (in s)
The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
What's the expression of range of a projectile motion?Range = U²× sin(2θ)/gU= initial velocity, θ= angle of projectile and g= acceleration due to gravity U=√{Range×g/sin(2θ)}Here, range= 2.20m, = 36.5°U= √{2.20×9.8/sin(73)}U= √{2.20×9.8/sin(73)} = 22.5m/s
What's the expression of time of flight in projectile motion?Time of flight= (2×U×sinθ)/g So, T= (2×22.5×sin36.5°)/9.8= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
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A treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the figure. The mass of the treasure chest is 75.6 kg.
The tension in Rope A is 7.42x10^2 N, and the tension Rope B carries is 7.52x10^2 N.
What is the tension in rope C?
Answer: A treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the figure. The mass of the treasure chest is 75.6 kg. The tension in Rope A is 7.42x10^2 N, and the tension Rope B carries is 7.52x10^2 N. Then, the tension in rope C is 376N
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?Given that,
[tex]T_A=7.42*10^2N\\T_B=7.54*10^2N[/tex]
From the free body diagram, we get Tension in the rope C as,[tex]T_C=T_B sin30\\thus,\\T_C=7.52*10^2*0.5=376N[/tex].
Thus, we can conclude that the tension in the rope c will be 376N.
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376N is the tension in rope C.
In order to determine the right response, we must have a better understanding of the fundamental forces that affect a body.
What exactly is force, and what are the fundamental forces that affect a body?Force is a push or a pull that modifies or tends to modify the condition, rest, or motion of a body.Given that it has a point of application, force is a polar vector.Repulsion is represented by positive force, and attraction by negative force.A body is subject to three main forces: weight mg, normal response N, and tension or pulling force.How can the issue be resolved?Tension in the rope C is determined from the free body diagram as,[tex]T_c=T_Bsin30\\T_c=7.52*10*2*0.5=376N[/tex]
Thus, we can infer that the rope's c tension will be 376N.
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A 67-kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 23∘ hill. The skier is pulled a distance x = 300 m along the incline and it takes 2.0 min to reach the top of the hill.
If the coefficient of kinetic friction between the snow and skis is μk = 0.10, what horsepower engine is required if 30 such skiers (max) are on the rope at one time?
Express your answer using two significant figures.
The required horsepower engine is 32 horsepower.
What is the force of the 30 skier?The force of the 30 skiers is calculated as follows:
Force = mass * accelerationMass of the skiers = 30 * 67kg = 2010 kg
Net force acting on the skiers along the x-axis
Fx = mgsinθ + f --- (1)where f is the frictional force
The kinetic frictional force, f = μN
where
μ = The coefficient of the kinetic friction
N = normal reaction
Net force acting on the skiers along y axis, the
Fy = ma
N = mg cos θ
Substituting for N above
f = μk mg cos θ
Substituting for f in (1)
F = mg sin θ + μk mg cos θ
F = mg(sinθ + μk cos θ)
Work done by the engine in pilling up the skiers, W = Fx
W = mg ( sinθ + μk cos θ)x
x = 300 m
W = (2010 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (300 m)
Work done, W = 2.86 * 10⁶ J
Time taken, t = 2.0 * 60sec = 120 s
Power = Work done/time taken1 horsepower = 746 W
Power = 2.86 * 10⁶/ 120 * 1/746
Power = 31.9 horsepower
In conclusion, the power of the engine is the ratio of the work done and time taken.
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A rigid, nonconducting tank with a volume of 4 m3 is divided into two unequal parts by a thin membrane. One side of the membrane, representing 1/3 of the tank, contains nitrogen gas at 6 bar and 100oC, and the other side, representing 2/3 of the tank, is evacuated. The membrane is ruptured and the gas fills the tank. (a) What is the final temperature of the gas? How much work is done? Is the process reversible? (b) How much work is done if the gas is returned to its original state by a reversible process? Assume nitrogen ideal gas for which Cp = (7/2) R and Cv = (5/2)R.
The final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.
To find the answer, we need to know about the thermodynamic processes.
How to find the final temperature of the gas?Any processes which produce change in the thermodynamic coordinates of a system is called thermodynamic processes.In the question, it is given that, the tank is rigid and non-conducting, thus, dQ=0.The membrane is raptured without applying any external force, thus, dW=0.We have the first law of thermodynamic expression as,[tex]dU=dQ-dW[/tex]
Here it is zero.[tex]dU=0[/tex],
As we know that,[tex]dU=C_pdT=0\\\\thus, dT=0\\\\or , T=constant\\\\i.e, T_1=T_2[/tex]
Thus, the final temperature of the system will be equal to the initial temperature,[tex]T_1=T_2=100^0C=373K[/tex]
How much work is done?We found that the process is isothermal,Thus, the work done will be,[tex]W=RT*ln(\frac{V_2}{V_1} )=373R*ln(\frac{4}{\frac{4}{3} })\\ \\W=409.8R J[/tex]
Where, R is the universal gas constant.
What is a reversible process?Any process which can be made to proceed in the reverse direction is called reversible process.During which, the system passes through exactly the same states as in the direct process.Thus, we can conclude that, the final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.
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The system's final temperature will be 373K, which is the same as its starting temperature. The system exerts 409.8R Joules of work.
We need to understand the thermodynamic processes in order to locate the solution.
How can I determine the gas's final temperature?Thermodynamic processes are any actions that result in modifications to a system's thermodynamic coordinates.Given that the tank is stiff and non-conducting, the answer to the question is that dQ=0.Without using any external force, the membrane is torn; hence, dW=0.The first law of thermodynamics is expressed as follows:[tex]dU=dQ-dW[/tex] , It is 0 here.
As we are aware,[tex]dU=C_pdT=0\\dT=0\\T=constant\\T_1=T_2=373K[/tex]
As a result, the system's final temperature will be equal to its starting temperature.
How much work is expended?The process is isothermal, as we discovered.As a result, the work will be,[tex]W=RT ln(\frac{V_2}{V_1} )=373R*ln(3 )\\W=409.8R Joules[/tex]
R is the gaseous universal constant.
A reversible process is what?Reversible processes are any operations that have the ability to be reversed.The system goes through the exact same states as it did during the direct procedure throughout this time.Thus, we can draw the conclusion that the system's end temperature will be 373K, the same as its starting temperature. The system exerts 409.8R Joules of work.
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The pulley is assumed massless and frictionless and rotates freely about its axle. The
blocks have masses m1 = 40 g and m2 = 20 g, and block m1 is pulled to the right by a
horizontal force of magnitude F = 0.03 N. Find the magnitude of the acceleration of block m2 and the tension in the cord if the surface is frictionless
The magnitude of the acceleration of block m2 is 0.5 m/s² and the tension in the cord if the surface is frictionless is 0.02 N.
Acceleration of block m2
The acceleration of bock m2 is calculated from Net force exerted by the pulley.
F - T = m2a
F - m1a = m2a
F = m2a + m1a
F = a(m2 + m1)
a = F/(m1 + m2)
a = (0.03) / (0.04 + 0.02)
a = 0.5 m/s²
Tension in the cordT = m1a
T = 0.04 x 0.5
T = 0.02 N
Thus, the magnitude of the acceleration of block m2 is 0.5 m/s² and the tension in the cord if the surface is frictionless is 0.02 N.
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A 1.0 mal sample of an ideal pas is kept at 0.0°c during an expansion from Bradba xa 3.0 litre to 10.0 litre a) How much work is done on the pas during the expansion? b) How much everly transfer by heat occurs with the surroundings in this Process?
The work done during expansion is 2.73×10³J and the heat transferred to the surrounding is -2.73×10³ J
What is isothermal process ?The thermodynamic process which takes place in constant temperature of the system is called isothermal process.
What is work done in isothermal process?Due to change in volume of the system at constant temperature, some work is done .As, the temperature of gas is fixed. This process is isothermal.Now, work done (W) = nRT ln ( V2/V1)where W = work done by the systemn= number of moles
R= universal gas constant
V2= volume of the system after expansion
V1= volume of the system before expansion
W= (1)(8.31)(273)ln ( 10/3)= 2.73×10³ J.
Also in isothermal process the amount of work done by the system = -( the amount of heat transferred to the surrounding)So the amount of heat transferred is -2.73×10³ JThus, we can conclude that a) the amount of work done is 2.73×10³J and b) the amount of heat transferred to the surrounding is -(2.73×10³J).
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A tennis player tosses a tennis ball straight up and then catches it after 1.64 s at the same height as the point of release.
(a) What is the acceleration of the ball while it is in flight?
magnitude ______ m/s2
Which direction?
1. Upward
2. Downward
3. The magnitude is zero
(b) What is the velocity of the ball when it reaches its maximum height?
magnitude _______________ m/s
Which direction?
1. Upward
2. Downward
3. The magnitude is zero
(c) Find the initial velocity of the ball.
______m/s upward
(d) Find the maximum height it reaches.
___________m
A. The acceleration of the ball while it is in flight?
magnitude is 0 m/s² (magnitude is zero)
B. The velocity of the ball when it reaches its maximum height is 0 m/s (magnitude is zero)
C. The initial velocity of the ball 8.036 m/s upward
D. The maximum height reached by the ball is 3.29 m
A. How to determine the acceleration in the flightConsidering that the ball came to rest after 1.64s, it means the entire acceleration of the flight is zero as the ball was not moving in any form again.
B. How to determine the velocity at maximum heightAt maximum height, the velocity of the ball is zero as it no longer has magnitude to keep going upwards. Hence the ball begins to ball down.
C. How to determine the initial velocityAcceleration due to gravity (g) = 9.8 m/s²Final velocity (v) = 0 m/sTime of flight (T) = 1.64 sTime to reach maximum height (t) = T / 2 = 1.64 / 2 = 0.82 sInitial velocity (u) =?v = u - gt (since the ball is going against gravity)
0 = u - (9.8 × 0.82)
0 = u - 8.036
Collect like terms
u = 0 + 8.036
u = 8.036 m/s upward
D. How to determine the maximum height reached by the ballTime to reach maximum height (t) = T / 2 = 1.64 / 2 = 0.82 sAcceleration due to gravity (g) = 9.8 m/s²Maximum height (h)h = ½gt²
h = ½ × 9.8 × 0.82²
h = 3.29 m
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I AM OFFERING 100 POINTS!!! I REALLY NEED HELP!!!!!!
Part C
Now prepare the cold sand and cold water samples from part A:
Fill a 100-milliliter container with 50 grams of sand. Fill a 100-milliliter container with 50 grams of cold tap water. Fill the last 100-milliliter container with 100 grams of cold tap water. Use the scale to measure the masses.
*Image should be there*
2. Pour all the ice cubes into a tub, and fill it with cool tap water to a depth of 2 inches. Place the sand and water samples in the ice water. Cover the entire tub.
*Image should be there*
3. Every 15 minutes, remove the cover and check the temperatures of the samples using the three thermometers. Wait 30 seconds before recording the thermometer reading. Once the temperatures of the three samples are no more than a degree apart, record the temperatures.
*Image should be there
Answer:
Every 15 minutes, remove the cover and check the temperatures of the samples using the three thermometers. Wait 30 seconds before recording the thermometer reading. Once the temperatures of the three samples are no more than a degree apart, record the temperatures.
(three 100-milliliter containers (each with a thermometer) in an ice bath inside an uncovered basin, with one container holding 50 grams of sand, one holding 50 grams of cold water, and one holding 100 grams of cold water)
Next, prepare the hot water.
-Fill each of the three 200-milliliter containers with 100 grams of hot tap water. Measure the mass using the mass scale.
(100 grams of hot water in a 200-milliliter container atop a mass scale)
-Prepare a hot-water bath by boiling water in a pot. Use the heat mitts to pour the hot water from the pot into the second tub. Fill the tub to a depth of about 2 inches. Carefully place the three containers of hot water into the bath without submerging them. Cover and wait for five minutes until the temperatures stabilize.
(three 200 mL containers (each holding 100 grams hot water) in a covered tub containing 2 inches of boiling water)
Prepare to mix the cold samples with hot water:
-Have three empty 300-milliliter mixing containers and three thermometers ready. Timing is important. Uncover the cold-water bath, and pour each cold sample into a different mixing container.
(three 300-milliliter containers (each with a thermometer), one holding 50 grams of sand, one holding 50 grams of cold water, and one holding 100 grams of cold water)
-Uncover the hot-water bath, and use a heat mitt to remove the three containers of hot water from the hot-water bath. Pour 100 grams of hot water into each mixing container.
(three 300 mL containers (each with a thermometer), one holding a 150-gram mixture of sand and hot water, one holding a 150-gram mixture of cold water and hot water, and one holding a 200-gram mixture of cold water and hot water)
Explanation:
A train traveling initially at 16 m/s is under constant acceleration of 2 m/2. How far will it travel in 20 s? What will its final velocity be
Answer: A train traveling initially at 16 m/s is under constant acceleration of 2 m/2. At a distance of 720m it will travel in 20 s, and the final velocity will be 56m/s.
Explanation: To find the answer, we need to know about uniformly accelerated motion.
How to solve the problem?Given that,[tex]u=16m/s\\a=2 m/s^2\\t=20s[/tex]
We have to find the distance travelled by the train.As we have,[tex]S=ut+\frac{1}{2}at^2[/tex]
Substituting values, we get,[tex]S=(16*20)+\frac{2*20^2}{2} =720 m.[/tex]
We have the equation for final velocity as,[tex]v^2=u^2+2aS\\thus,\\v=\sqrt{u^2+2aS} =\sqrt{16^2+(2*2*720)} =56 m/s.[/tex]
Thus, we can conclude that, a train traveling initially at 16 m/s is under constant acceleration of 2 m/2. At a distance of 720m it will travel in 20 s, and the final velocity will be 56m/s.
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A train experiencing constant acceleration of 2 m/s^2 is moving at an initial speed of 16 m/s. It will go 720 meters in 20 seconds, with a final velocity of 56 m/s.
Understanding uniformly accelerated motion is necessary in order to determine the solution.
How can the issue be resolved?We need to determine how far the train has traveled. We have,[tex]S=ut+\frac{1}{2}at^2 \\S=720m\\where,\\u=16m/s, a=2m/s^2,t=20s[/tex]
The formula for final velocity is as follows:[tex]v^2-u^2=2aS\\v=\sqrt{u^2+2aS} \\v=56m/s[/tex]
Thus, we may say that a train moving at 16 m/s initially experiences constant acceleration of 2 m/2. It will go 720 meters in 20 seconds, with a final velocity of 56 meters per second.
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The paths of two small satellites, M1 = 4.00 kg and M2 = 1.00 kg, are shown below, drawn to scale, with M1 corresponding to the circular orbit. They orbit around a massive star, also shown below. The orbits are in the plane of the paper.
The period of M1 is T1 = 34.0 years. Calculate the period of M2, in years.
The period of M2, in years. is mathematically given as
T2= 134.3968years
What is the period of M2, in years.?
M1 = 4.00 kg
M2 = 1.00 kg
T1 = 34.0 years.
Generally, the equation for is mathematically given as
T2 = T1 (a2/a1)3/2
T2= 34.0 * (5/2)^{3/2}
T2= 134.3968years
In conclusion, the period of M2, in years
T2= 134.3968years
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A blue train of mass 50 kg moves at 4 m/s toward a green train of 30 kg initially at rest. What is the initial momentum of the blue and green train combined?
A. 20 kgm/s
B. 50 kgm/s
C. 0 kgm/s
D. 200 kgm/s
The correct option is D. The initial momentum of the blue and green train combined during the collision is 200 kgm/s.
Initial momentum of the blue and green trainApply the principle of conservation of linear momentum as follows;
Pi = m1v1 + m2v2
where;
m1 is mass of blue trainm2 is mass of green trainv1 is velocity of blue trainv2 is velocity green trainPi is the initial momentum of the two trainsPi = (50 x 4) + 30(0)
Pi = 200 kgm/s
Thus, the initial momentum of the blue and green train combined is 200 kgm/s.
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"You can't see the forest for the trees" might seem an appropriate analogy for astronomers attempting to determine the shape of the Milky Way galaxy when we are in fact located inside the galaxy. Discuss techniques used by astronomers to determine the type of galaxy in which we live, why it is so difficult to determine the shape of our galaxy, and where our Sun is located in our galaxy?
Some of the techniques used by astronomers to determine the type of galaxy in which we live are:
radio, optical, infraredx-ray astronomyWhat is Astronomy?This refers to the study of heavenly bodies and space and other things that space is made up of.
Hence, we can see that the reason why it is so difficult to determine the shape of our galaxy is that astronomers can only infer its presence from the motions of stars in the galaxy, and a precise shape is difficult to be determined.
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Find the minimum diameter of an l = 19.3 m long steel wire that will stretch no more than 7.32 mm when a mass of 350 kg is hung on the lower end. (Hint: The Young's modulus of steel is 200.0 GPa.)
The minimum diameter of the steel wire is 7.58 mm.
Stress applied to the steel wireYoung's modulus = stress/strain
strain = e/l = (7.32 x 10⁻³ m) / (19.3 m) = 3.79 x 10⁻⁴
stress = Young's modulus x strain
stress = 200 x 10⁹ N/m² x 3.79 x 10⁻⁴ = 7.59 x 10⁷ N/m²
Area of the wirestress = Force/Area
Area = Force/stress
Area = mg/stress
Area = (350 x 9.8) / (7.59 x 10⁷)
Area = 4.519 x 10⁻⁵ m²
Minimum diameter of the wireArea = πd²/₄
πd²/₄ = 4.519 x 10⁻⁵ m²
πd² = 4(4.519 x 10⁻⁵)
d² = (4 x 4.519 x 10⁻⁵)/π
d² = 5.75 x 10⁻⁵
d = √(5.75 x 10⁻⁵)
d = 7.58 x 10⁻³ m
d = 7.58 mm
Thus, the minimum diameter of the steel wire is 7.58 mm.
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Solve the following numerical problems
Answer:
See below
Explanation:
To change C to F°
F = 9/5 C + 32
= 9/5 ( 45) + 32 = 113° F
To change F to C° <= you could use the same equation...or re-earrange to:
C = 5/9 (F-32)
= 5/9 ( 98.6 - 32) = 37° C
Similarly
100 C = 212 ° F
to change C ° to K add 273.15 = 373.15 K
Why is air resistance friction not useful for an airplane?
A. Causes turbulence
B. Speeds it up
C. Slows it down
D. Creates heat
The correct option is C. Air resistance friction is not useful for an airplane because it slows it down.
What is air resistance?Air resistance is the opposition to motion of an object caused by air flow.
High air resistance may cause turbulence during the motion of an air plane. This can lead to engine failure or some mishap during flight.
Air resistance is also a type of friction between air and another material such as airplane.
Effect of air resistance on airplaneFriction between the air and the plane slows the airplane down. This is known as air resistance, or drag.
The faster an object travels through the air, the more it has to fight against drag.
Thus, air resistance friction is not useful for an airplane because it slows it down.
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You apply a force on an object that is 100 times its mass. The acceleration of the object will be 100m/s^2. After a while, it perpendicularly hits another object with a mass 10 times smaller than the first object. When they collide, in theory, the momentum of the first object will all get transferred into the small object. And that means, the heavy object will completely be at rest, while the smaller object will have a force applied to it by the big object equal to ma. In real life, it seems that when two objects collide, the bigger object doesn’t completely stop and just pushes the smaller or the same massed object. In real life, it seems that when the collision happens, the first object continues moving towards the other object, and doesn’t completely get at rest right after the collision. Why is that?
The bigger object will move some distance because the initial and final momentum of the colliding particles are not zero.
What is momentum?The term momentum has to do with the product of mass and velocity. We know that during a collision, momentum is conserved. This implies that the momentum before collision is equal to the momentum after collision. Thus, the total momentum of the system is constant.
Given the fact that the initial and final momentum of the colliding particles are not zero, the the big object is not going to stop immediately but must move some distance towards the smaller object.
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A tennis player tosses a tennis ball straight up and then catches it after 1.77 s at the same height as the point of release.
(a) What is the acceleration of the ball while it is in flight?
magnitude ______ m/s2
Which direction?
1. Upward
2. Downward
3. The magnitude is zero
(b) What is the velocity of the ball when it reaches its maximum height?
magnitude _______________ m/s
Which direction?
1. Upward
2. Downward
3. The magnitude is zero
(c) Find the initial velocity of the ball.
______m/s upward
(d) Find the maximum height it reaches.
___________m
(a) The acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.
(b) The velocity of the ball when it reaches its maximum height is zero.
(c) The initial velocity of the ball is 17.36 m/s.
(d) The maximum height it reaches is 15.36 m.
Acceleration of the ball
The acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.
Velocity of the ball at maximum heightThe velocity of the ball decreases as the ball moves upwards and eventually becomes zero at maximum height.
Initial velocity of the ballv = u - gt
at maximum height, final velocity, v = 0
0 = u - gt
u = gt
u = 9.81 x 1.77
u = 17.36 m/s
Maximum height reached by the projectileh = ut - ¹/₂gt
h = 17.36(1.77) - ¹/₂(9.81)(1.77²)
h = 15.36 m
Thus, the acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.
The velocity of the ball when it reaches its maximum height is zero.
The initial velocity of the ball is 17.36 m/s.
The maximum height it reaches is 15.36 m.
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What is the equation used to find the angle of refraction? Identify each variable. (1 point)
Answer:
pictures please
Explanation:
I need a picture so I can tell you