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professor190 [17]
3 years ago
8

According to Newton’s Law of gravity, which of the following is inversely related to the square of the distance from earth?

Physics
2 answers:
Dima020 [189]3 years ago
3 0
I think it’s D

I hope this helps ❤️
Travka [436]3 years ago
3 0

Answer:

B. Weight

Explanation:

If you look at my crude drawing of two bodies attracted by each other (hey that's gravity, not love!) and to Newton's law of gravity:

F_{1} =F_{2} =G\frac{m_{1} m_{2} }{r^{2} }

For this to work we have to assume that the Forces are equal and the whole system is in equilibrium.

Now if we take a look at the picture, r is the distance between the two bodies, G is the gravitational constant, and m is the mass either of body 1 or 2, so, if we increase the distance, the force of gravity will decrease by the square root of that unit, because it is the denominator of the fraction.

Now emphasis on "Force of Gravity" Force is the only factor inversely proportionate to the square root of the distance, and the only answer that is actually a Force from the options is weight (m*g).

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liubo4ka [24]

Answer:

 ω₂=1.20

Explanation:

Given that

mass of the turn table ,M= 15 kg

mass of the ice ,m= 9 kg

radius ,r= 25 cm

Initial angular speed ,ω₁ = 0.75 rad/s

Initial mass moment of inertia

I_1=\dfrac{M+m}{2}r^2

I_1=\dfrac{15+9}{2}\times 0.25^2\ kg.m^2

I_1=0.75\ kg.m^2

Final mass moment of inertia

I_2=\dfrac{M}{2}r^2

I_2=\dfrac{15}{2}\times 0.25^2\ kg.m^2

I_2=0.468\ kg.m^2

Lets take final speed of the turn table after ice evaporated =ω₂ rad/s

Now by conservation angular momentum

I₁ ω₁ =ω₂ I₂

\omega_2=\dfrac{0.75\times 0.75}{0.468}\ rad/s

ω₂=1.20

7 0
3 years ago
for an ideal monoatomic gas, the internal energy U os due to the kinetic energy and U=3/2RT per mole.show that cv=3/2R per mole
sladkih [1.3K]

Answer:

i. Cv =3R/2

ii. Cp = 5R/2

Explanation:

i. Cv = Molar heat capacity at constant volume

Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT

Differentiating U with respect to T, we have

= d(3/2RT)/dT

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substituting Cv into the equation, we have

Cp = 3R/2 + R

taking L.C.M

Cp = (3R + 2R)/2

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