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Romashka-Z-Leto [24]

3 years ago

8

The Earth has mass ME and average radius RE. The Moon has mass MM and the average distance from the center of mass of the moon t

o the center of mass of the Earth is DM. For this problem use the following constants.
Mass of the Earth: 5.97E24 kgMass of the Moon: 7.35E22 kgRadius of the Earth: 6,371 km
Find an expression for the moment of inertia IEof the Earth on its axis in terms of the given parameters.

1 answer:

marusya05 [52]3 years ago

3 0

Answer:

Moment of inertia of Earth about its own axis is given as

$I = 9.7 \times 10^{37} kg m^2$

Explanation:

Since Earth is considered as solid sphere

So we will have

$I = \frac{2}{5}M_eR_e^2$

so we will have

$I = \frac{2}{5}(5.97 \times 10^{24})(6.371 \times 10^6)^2$

so we have

$I = 9.7 \times 10^{37} kg m^2$

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a baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby weighs 12 Kg. how much energy does

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8 0

3 years ago

A proton is projected in the positive x direction into a region of a uniform electric field E S 5 126.00 3 105 2 i^ N/C at t 5 0

Dafna11 [192]

Answer:

(a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). The time is $0.493\times10^{-7}\ sec$ .

Explanation:

Given that,

Electric field $E=-6.00\times10^{5}i\ N/C$

Time = 5.0 sec

Distance 7.00 cm

(a). We need to calculate the acceleration

Using formula of force

$F=F_{e}$

$ma=qE$

$a=\dfrac{Eq}{m}$

Where, E = electric field

m = mass of proton

q = charge of proton

Put the value into the formula

$a=\dfrac{-6.00\times10^{5}\times1.6\times10^{-19}}{1.67\times10^{-27}}$

$a=-5.74\times10^{13}\ m/s62$

The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). We need to calculate the initial peed

Using equation of motion

$v^2-u^2=2as$

Where, s = distance

Put the value into the formula

$0-u^2=2\times(-5.74\times10^{13})\times7.00\times10^{-2}$

$u=\sqrt{2\times5.74\times10^{13}\times7.00\times10^{-2}}$

$u=2834783.94$

$u=2.83\times10^{6}\ m/s$

The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). We need to calculate the time interval over which the proton comes to rest

Using formula

$t=\dfrac{u}{a}$

Where, u = initial velocity

a = acceleration

Put the value into the formula

$t=\dfrac{2.83\times10^{6}}{5.74\times10^{13}}$

$t=0.493\times10^{-7}\ sec$

Hence, (a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). The time is $0.493\times10^{-7}\ sec$ .

5 0

3 years ago