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3 years ago

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What are the 3 formulas you can use for vertical motion for a projectile?

1 answer:

klio [65]3 years ago

4 0

I hope this can help you ask me if you need help again

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( Can someone help? )

Murrr4er [49]

Answer:

Answer would be 0.33

Explanation:

Calculations

8 0

3 years ago

Read 2 more answers

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0

3 years ago

Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a

malfutka [58]

Answer:

a) $p_i=1.568\hat{i}+0.752 \hat{j}$

b) $v_{fx}=1.668\ m.s^{-1}$

c) $v_{fy}=0.7999\ m.s^{-1}$

Explanation:

Given masses:

$m_1=0.49\ kg$

$m_2=0.47\ kg$

Velocity of mass 1, $v_1=3.2 \hat{i}\ m.s^{-1}$

Velocity of mass 2, $v_2=1.6 \hat{j}\ m.s^{-1}$

a)

Initial momentum:

$p_i=m_1.v_1+m_2.v_2$

$p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}$

$p_i=1.568\hat{i}+0.752 \hat{j}$

b)

magnitude of initial momentum:

$p_i=\sqrt{1.568^2+0.752 ^2}$

$p_i=1.739\ kg.m.s^{-1}$

From the conservation of momentum:

$p_f=p_i$

$m_f.v_f=1.739$

$v_f=\frac{1.739}{0.49+0.47}$

$v_f=1.85\ m.s^{-1}$ is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

$tan\theta=\frac{0.752 }{1.568}$

$\theta=25.62^{\circ}$

$\therefore v_{fx}=1.85\ cos25.62^{\circ}$

$v_{fx}=1.668\ m.s^{-1}$

c)

Vertical component of final velocity:

$v_{fy}=1.85\ sin 25.62^{\circ}$

$v_{fy}=0.7999\ m.s^{-1}$

6 0

3 years ago

Give two conditions in which an object becomes weightless

Evgesh-ka [11]

Answer:

A body becomes weightless in a zero-gravity scenario and when a force is applied to a body that is equal and opposite to the force of gravity.

4 0

3 years ago

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50 J of work was done in order to move a chair 5 meters. what was the force applied to the chair

kolezko [41]

Using the Definition of Work, we have:

$T=F\DeltaS \\ 50=5F \\ \boxed {F=10N}$

If you notice any mistake in my english, please let me know, because i am not native.

3 0

3 years ago