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Naya [18.7K]
3 years ago
6

If an object with a mass of 12 kilograms is moving at 10 meters per second, what is its momentum?

Physics
1 answer:
ValentinkaMS [17]3 years ago
4 0

Answer:

120 kg.m/s

Explanation:

Momentum is defined as the product of mass and velocity of an object. Velocity in this case considers the direction of movement ie whether East to East or West etc. The formula for momentum is p=mv where p is momentum, m is the mass in kgs, v is the velocity in m/s. Substituting 12 kg for m and 10 m/s for v then momentum will be 12*10=120 kg.m/s

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Determine the largest intensity w of the uniform loading that can be applied to the frame without causing either the average nor
jeka94

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3 years ago
Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca
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Answer: a) 4.7 mi/hr.  b) 86,500 lbs. mi²/Hr²

Explanation:

As in any collision, under the assumption that no external forces exist during the very small collision time, momentum must be conserved.

If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

vf = (200,000 lbs. 4 mi/hr + 100,000 lbs. 6 mi/hr) / 300,000 lbs = 4.7 mi/hr

If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

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This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

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3 years ago
Which of the following hypotheses is both falsifiable and testable?
rosijanka [135]
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5 0
3 years ago
Read 2 more answers
Oscilloscope amplitude and frequency problem. Study the above graph. The volts/div dial is set to 2 volts/div and the time/div d
denis-greek [22]

Answer:

Amplitude = 8 Volts

Frequency = 0.067 kHz

Explanation:

Note: The missing picture in question is attached for your review.

Given:

Volts/Div = 2 V/div

Time/Div = 5 msec/div

Finding Amplitude:

Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,

Amplitude = 4 div/volts * 2 volts/ div )\\Amplitude = 8 Volts

(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)

Finding Frequency:

As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,

Time Period = 3 div * 5msec /div\\Time Perod = 15 msec

Since,

Frequency = \frac{1}{Time Period}\\Frequency = \frac{1}{15m}\\Frequency = 0.067 kHz

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3 years ago
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