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2 years ago

You exert 500 N of force down on the earth as you jump into the air. How much force does the earth exert back on you?

Physics

1 answer:

zalisa [80]2 years ago

6 0

The same 500N, is the Newton’s Third Law.

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A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the ho

Lady_Fox [76]

Answer:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!

The ball rotates 6.78 revolutions.

Explanation:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!

At the bottom the ball has the following angular speed:

$\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s$

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

$sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m$

To find the revolutions we need the time, which can be found using the following equation:

$v_{f} = v_{0} + at$

$t = \frac{v_{f} - v_{0}}{a}$ (1)

So first, we need to find the acceleration:

$v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}$ (2)

By entering equation (2) into (1) we have:

$t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}$

Since it starts from rest (v₀ = 0):

$t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s$

Finally, we can find the revolutions:

$\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev$

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!

3 0

2 years ago

The maximum oxygen uptake is known as the

grandymaker [24]

The maximum oxygen uptake is known as the VO max.

3 0

3 years ago

What type of wave borders the violet end

wariber [46]

Violet light is at the end of the visible light section of the electromagnetic spectrum. Ultraviolet rays are directly next to violet rays on the EM Spectrum.

3 0

3 years ago

What is the escape speed of an electron launched from the surface of a 1.1-cm-diameter glass sphere that has been charged to 8.0

zimovet [89]

At surface,
v = kq/r

And potential energy of an electron is given by,
PE = -ev = -ekq/r

At escape velocity,
PE + KE = 0.
Therefore,
1/2mv^2 - ekq/r =0
1/2mv^2 = ekq/r
v = Sqrt [2ekq/mr], where v = escape velocity, e = 1.6*10^-19 C, k = 8.99*10^9 Nm^2/C^2, m = 9.11*10^-31 kg, r = 1.1*10^-2 m, q = 8*10^-9 C

Substituting;
v = Sqrt [(2*1.6*19^-19*8.99*10^9*8*10^-9)/(9.11*10^-31*1.1*10^-2)] = 47949357.23 m/s ≈ 4.795 *10^7 m/s

5 0

3 years ago

As the distance between two objects decreases, the pull of gravity

aliya0001 [1]

Answer: increases

Explanation:

Because the relationship between hand d is counter

6 0

3 years ago

You exert 500 N of force down on the earth as you jump into the air. How much force does the earth exert back on you?￼

You exert 500 N of force down on the earth as you jump into the air. How much force does the earth exert back on you?