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DanielleElmas [232]

3 years ago

6

If you went to a planet with an acceleration due to gravity of 20 m/s^2 and twirled a bucket of water, what happens to the angul

ar speed you need to twirl the bucket to keep the water in?
a - angular speed increases
b - angular speed decreases
c - angular speed stays the same

1 answer:

wel3 years ago

7 0

Answer:

B

Explanation:

It's Scientifically proven

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Water that flows from behind a large dam can cause machines to produce electricity. What change takes place?

lyudmila [28]

The mechanical energy in the falling water is used to spin the generator, and gets transformed into electrical energy. That's the first choice on the list.

8 0

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

Two bodies of masses 1000kg and 2000kg are separated 1km which is the gravitational force between them

denpristay [2]

Answer:

1.33×10⁻¹⁰ N

Explanation:

F = GMm / r²

where G is the gravitational constant,

M and m are the masses of the objects,

and r is the distance between them.

F = (6.67×10⁻¹¹ N/m²/kg²) (1000 kg) (2000 kg) / (1000 m)²

F = 1.33×10⁻¹⁰ N

3 0

4 years ago

Please help ASAP! I will give the first answer brainliest!

Natalija [7]

Answer:

Sorry for being late. Hope this helps

Explanation:

3 0

3 years ago

2. What are the 3<br> types of matter?

Lapatulllka [165]

Answer:

the three types of matter are solid, liquid, gas

8 0

3 years ago