The energy an object possesses is equal to the work spent on the object. true false

1 answer:

5 0

That's not necessarily true. The object very well could have had
quite a bit of energy ... particularly thermal (heat) energy ... before
anybody came along and gave it MORE energy by doing work on it.

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The moon has a mass of 7.35x1022 kg and is a lot farther away than is shown in textbooks. The mass of the

anyanavicka [17]

Answer:

$F=1.98\times 10^{20}\ N$

Explanation:

Given that,

The mass of a Moon, $M_m=7.35\times 10^{22}\ kg$

The mass of the Earth, $M_e=5.98\times 10^{24}\ kg$

The moon's mean orbit distance around the earth is, $r=3.84\times 10^8\ m$

We need to find the gravitational force exerted on the moon by the Earth.

The formula of gravitational force is given by :

$F=G\dfrac{M_mM_e}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{7.35\times 10^{22}\times 5.98\times 10^{24}}{(3.84\times 10^8)^2}\\\\F=1.98\times 10^{20}\ N$

So, the required force is $1.98\times 10^{20}\ N$ .

3 0

3 years ago

What is the linear speed of a point on the equator, due to the earth's rotation?

kvv77 [185]

The equatorial radius of the earth is
r = 6378 km = 6378 x 10³ m

The earth makes 1 revolution in 24 hours.
The angular velocity is
ω = (2π rad)/(24*3600 s) = 7.2722 x 10⁻⁵ rad/s

The tangential velocity (linear velocity) at a point on the equator is
v = rω
= (6378 x 10³ m)*(7.2722 x 10⁻⁵ rad/s)
= 463.8 m/s

Answer: 463.8 m/s

8 0

4 years ago

Find the net rate of heat transfer by radiation in watts from a skier standing in the shade, given the following. She is complet

fomenos

Answer:

-34.5 W

Explanation:

rate of heat transfer = Stefan-Boltzmann constant x emissivity x surface area x temperature^4

$P = \alpha eA(T_{f} ^{4}- T_{i} ^{4}) = 5.67*10^{-8} *0.205*1.5(256^{4} - 284.4^{4} ) = -34.5 W$