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3 years ago

This whole worksheet. I am not good at this and I am completely lost

Physics

1 answer:

Ket [755]3 years ago

3 0

1.) potential energy
2.)potential and kinetic
3.)The roller coaster car has the most kinetic energy at point X i know this because the car is moving and kinetic energy has the power to move or change things therefore point X is when the roller coaster car has the most energy.
4.)potential energy
5.)kinetic energy
6.) potential and kinetic energy

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What describes a sound wave as it travels through a medium

Softa [21]

Sound waves in air (and any fluid medium) are longitudinal waves because particles of the medium through which the sound is transported vibrate parallel to the direction that the sound wave moves.

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3 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,

$t_{tot}=2\dfrac{v_0}{g}+t_0$

where $t_0$ is the time it takes the rock to go from the roof of the building to the ground, and it is given by

$H=v_0t_0+\dfrac{1}{2}gt_0^2$

we solve for $t_0$ using the quadratic formula and take the positive value to get:

$t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

Therefore the total time is

$t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

$\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH} }{g}}$

Now the average velocity is

$v_{avg}=\dfrac{y_{tot}}{t}$

$v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }$

$\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }$

5 0

3 years ago

The most precise measurement of the speed of light in vacuum is 299,792,458 m/s. Which of the following values would you use to

Luda [366]

Answer:

v = 3.00 x 10⁸ m/s

Explanation:

given,

speed of light in vacuum = 299,792,458 m/s

speed of light in scientific notation to three significant figures

v = 2.99792458 x 10⁸ m/s

by rounding off the speed to three significant figure.

v = 3.00 x 10⁸ m/s

On the fourth place the value is greater than 5 so, on the third place 1 will be added.

now, the speed with three significant figure comes out to be

v = 3.00 x 10⁸ m/s

3 0

3 years ago

In a swimming meet, the swimmers swim a total of 8 laps of a 50-meter-long swimming pool. What is the distance traveled by a swi

N76 [4]

Answer:

The swimmer has a distance traveled of 800 meters.

The final displacement of the swimmer is 0 meters.

Explanation:

A lap is a round trip made by a swimmer in the pool, so that the distance traveled by swimmer is sixteen times the length of the swimming pool. That is:

$s = \left(2\,\frac{travels}{lap} \right)\cdot \left(8\,laps \right)\cdot \left(50\,\frac{m}{travel}\right)$

$s = 800\,m$

A swimmer has a distance traveled of 800 meters.

The displacement is the distance between swimmer and a reference point, let suppose that reference point is located at the beginning of the first lap. Hence, the final displacement of the swimmer is 0 meters.

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3 years ago

A force of 8480 N is applied to a cart or decelerate is at a rate of 32.0 m/s2. What is the mass of the cart?

Otrada [13]

I believe it to be 265kg.

4 0

3 years ago