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arsen [322]
3 years ago
8

A woman of mass 50 kg is swimming with a velocity of 1.6 m/s. If she stops stroking and glides to a stop in the water, what is t

he impulse of the force that stops her?
Physics
1 answer:
SashulF [63]3 years ago
4 0

Answer:

– 80 Ns.

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 50 kg

Initial velocity (u) = 1.6 m/s

Final velocity (v) = 0 m/s

Impulse (I) =?

Impulse is simply defined by the following equation:

I = Ft = m(v – u)

I = 50 (0 – 1.6)

I = 50 (–1.6)

I = – 80 Ns

Thus, the impulse of the force that stopped her is – 80 Ns.

NOTE: The negative sign indicates that the net force is in opposite direction to her movement.

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2. A carpenter tosses a shingle off a 9.4 m high roof, giving it an initial horizontal
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Sounds like the shingle/ball is thrown from the roof horizontally, so that the distance it travels <em>x</em> after time <em>t</em> horizontally is

<em>x</em> = (7.2 m/s) <em>t</em>

The object's height <em>y</em> at time <em>t</em> is

<em>y</em> = 9.4 m - 1/2 <em>gt</em>²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, and its vertical velocity is

<em>v</em> = -<em>gt</em>

(a) The object hits the ground when <em>y</em> = 0:

0 = 9.4 m - 1/2 <em>gt</em>²

<em>t</em>² = 2 * (9.4 m) / (9.80 m/s²)

<em>t</em> ≈ 1.92 s

at which time the object's vertical velocity is

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(b) See part (a); it takes the object about 1.9 s to reach the ground.

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Work out the kinetic energy of a 2.5 kg remote-controlled car that is moving at 2 m/s.
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Answer: 5 joules

Explanation:

mass=m=2.5kg

Velocity=v=2m/s

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Ke=10/2

Ke=5

Kinetic energy=5 joules

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A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha
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Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss} (1)

From the bottom to the top of the second hill

\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss} (2)

Where:

m - Mass of the roller coaster car, in kilograms.

v_{1} - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

g - Gravitational acceleration, in meters per square second.

h_{1} - Height of the first top of the hill with respect to the bottom, in meters.

W_{1, loss} - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

v_{2} - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

h_{2} - Height of the second top of the hill with respect to the bottom, in meters.

W_{2, loss} - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss} (3)

Where W_{loss} is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

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W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]

W_{loss} = 6574.75\,J

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Explanation:

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