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3 years ago

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35 POINTSS!!! PLSSSS HELLPPP!!!

1 answer:

Crazy boy [7]3 years ago

6 0

Answer:

T

beacuse:

Energy can be transferred from one object to another by doing work. ... When work is done, energy is transferred from the agent to the object, which results in a change in the object's motion (more specifically, a change in the object's kinetic energy).

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Cientists have changed the model of the atom as they have gathered new evidence. One of the atomic models is shown below.

scZoUnD [109]

Answer:

A few of the positive particles aimed at a gold foil seemed to bounce back.

Explanation:

3 0

3 years ago

What is the approximate value of the gravitational force between a 73 kg astronaut and a 7.1×104 kg spacecraft when they're 89 m

Luden [163]

Answer:

$F = 4.3671 * 10^{-8}\ Newtons$

Explanation:

The gravitational force between two corpses is given by the following equation:

$F = GMm/d^2$

Where F is the force, G is the gravitational constant

( $G = 6.67408*10^{-11}\ m^3kg^{-1}s^{-2}$ ), M and m are the masses of the corpses and d is the distance between them.

So we have that:

$F = 6.67408*10^{-11} * 7.1*10^4 * 73/89^2$

$F = 4.3671 * 10^{-8}\ Newtons$

5 0

3 years ago

Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.

Fudgin [204]

Answer:

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$

Explanation:

We know that the gravity on the surface of the moon is,

$g'=\frac{g}{6}$

$g'=1.63\ m.s^{-2}$

<u>Gravity at a height h above the surface of the moon will be given as:</u>

$g'_h=\frac{G.m}{(r+h)^2}$ ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

$G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}$

$m=7.35\times 10^{22}\ kg$

$r=1.74\times 10^6\ m$

$h=384.4\times 10^6\ m$ is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

$g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}$

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$ is the gravity on the moon the earth-surface.

4 0

3 years ago

A projectile is fired at v 0 = 381.0 m/s at an angle of θ = 73.5 ∘ , with respect to the horizontal. Assume that air friction wi

Ipatiy [6.2K]

Answer:

Explanation:

velocity of projection, vo = 381 m/s

angle of projection, θ = 73.5°

The formula for the range is

$R=\frac{u^{2}Sin2\theta }{g}$

$R=\frac{381^{2}Sin147 }{9.8}$

R = 8067.4 m

Range in shorten by 34.1 %

So, the new range is

R' = 8067.4 - 34.1 x 8067.4/100

R' = 5316.4 m

5 0

3 years ago

A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/

Elena-2011 [213]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

$KE_i + PE_i = KE_f + PE_f$

$\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0$

$\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2$

So by solving above equation we will have

$v_f = 84.3 m/s$

now in order to find the momentum we can use

$P = mv$

$P = 1.25 \times 84.3$

$P = 105.4 kg m/s$

6 0

3 years ago