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3 years ago

The impulse experienced by a body is equivalent to the body’s change in

Physics

1 answer:

nika2105 [10]3 years ago

3 0

<h2>Answer: </h2>

Momentum

<h2>Explanation: </h2>

The momentum of a particle is defined as the product of the particle mass and the particle velocity as follows:

$\overrightarrow{p}=m\overrightarrow{v}$

On the other hand, the impulse of a constant force is defined as:

$\overrightarrow{J}=\varSigma\overrightarrow{F}(t_{2}-t_{1})=\varSigma\overrightarrow{F}\Delta t$

We also know that the net force acting on a particle equals the rate of change of the particle’s momentum, so:

$\varSigma\overrightarrow{F}=m\overrightarrow{a}=m\frac{d}{dt}(\overrightarrow{v})=\frac{d}{dt}(m\overrightarrow{v})=\frac{d\overrightarrow{p}}{dt}$

If the force is constant, then $\frac{d\overrightarrow{p}}{dt}$ equals the total change in momentum over a period of time:

$\varSigma\overrightarrow{F}=\frac{\overrightarrow{p_{2}}-\overrightarrow{p_{1}}}{t_{2}-t_{1}} \\ \\ \varSigma\overrightarrow{F}(t_{2}-t_{1})=\overrightarrow{p_{2}}-\overrightarrow{p_{1}} \\ \\ \boxed{\overrightarrow{J}=\Delta \overrightarrow{p}}$

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An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2

Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

$M=-\frac{q}{p}$

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

$M=-\frac{200 cm}{50 cm}=-4.0$

7 0

3 years ago

What happens to rocks as water combines

insens350 [35]

The answers is
D. The acid creates cracks in the rocks, which
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3 years ago

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

Anni [7]

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.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is: