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Olegator [25]
2 years ago
15

On an ice skating rink, a girl of mass 50 kg stands stationary, face to face with a boy of mass 80 kg. The children push off of

one another, and the boy moves away with a velocity of +3 m/s. What is the final velocity of the girl?
–1.9 m/s
+1.9 m/s
–4.8 m/s
+4.8 m/s
Physics
2 answers:
irakobra [83]2 years ago
6 0
Use conservation of momentum.

Initial momentum = 0

Final momentum = momentum of the girl + momentum of the boy = inifial momentum = 0.

Momentum of the girl = - momentum of the boy

mass of the girl * velocity of the girl = - mass of the boy * velocity of the boy

50 kg * velocity of the girl = - 80 kg * 3m/s

=> velocity of the girls = -80 kg * 3m/s / 50 kg = - 4.8 m/s

Answer: - 4.8 m/s
GaryK [48]2 years ago
3 0

Answer:

c. –4.8 m/s

Explanation:

:D :) :(

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lora16 [44]

Since the system itself is giving off heat, this is a reduction in the internal energy.

heat = - 25,000 J

 

Since work is being done on the system, therefore it is an additional energy to the system. Work is given as:

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work = - 1.50 atm (6 L – 12 L)

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energy due to work = 911.7 J

 

Therefore the total change in internal energy is the sum of heat and energy due to work:

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<span>Answer:</span>

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8 0
1 year ago
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A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising th
kobusy [5.1K]

Answer:

24,267.6 watts

Explanation:

from the question we are given the following:

mass (m) = 810 kg

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time (t) = 3.5 s

final height (h₁) = 8.2 m

initial height (h₀) = 0 m

acceleration due to gravity (g) = 9.8 m/s^{2}

find the power

power = \frac{work done}[time}

and

work done = change in kinetic energy (K.E) + change in potential energy (P.E)

work done = (0.5 mv^{2} - 0.5 mu^{2} ) + ( mgh₁ - mgh₀)

since u and h₀ are zero the work done now becomes

work done = (0.5 mv^{2}) + ( mgh₁ )                    

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work done = 84, 936.6 joules

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power = 24,267.6 watts

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1 year ago
Starting from equilibrium at point 0, what point on the pv diagram will describe the ideal gas after the following process? lock
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A measuring microscope is used to examine the interference pattern. It is found that the average distance between the centers of
diamong [38]

Answer:

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Explanation:

The interference by regency in thin films uses two rays mainly the one reflected on the surface and the one reflected on the inside of the film.

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The thickness of a hair is the thickness of the film t

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3 0
2 years ago
The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th
Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em>1.83\times 10^{7}\,kg<em>. The wheels of the locomotive push back on the tracks with a constant net force of </em>7.50\times 10^{5}\,N<em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force (F), measured in newtons, then acceleration (a), measured in meters per square second, must be constant and can be found by the following expression:

a = \frac{F}{m} (1)

Where m is the mass of the freight train, measured in kilograms.

If we know that F = 7.50\times 10^{5}\,N and m = 1.83\times 10^{7}\,kg, then the acceleration experimented by the train is:

a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}

a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}

Now, the time taken to accelerate the freight train from rest (t), measured in seconds, is determined by the following formula:

t = \frac{v-v_{o}}{a} (2)

Where:

v - Final speed of the train, measured in meters per second.

v_{o} - Initial speed of the train, measured in meters per second.

If we know that a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}, v_{o} = 0\,\frac{m}{s} and v = 22.222\,\frac{m}{s}, the time taken by the freight train is:

t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s}  }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }

t = 542.265\,s

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

6 0
1 year ago
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