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Romashka-Z-Leto [24]

2 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]2 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 sheet of gold leaf has a thickness of 0.125 micrometer. A gold atom has a radius of 174pm. Approximately how many layers of at

Olegator [25]

Answer:

719

Explanation:

Conversion

1 picometer (pm) is equivalent to $1 × 10^{-12}$ meter

1 micrometer is equivalent to $1 × 10^{-6}$ meter

To find the number of layers, we divide the overal leaf thickness by the thickness of one atom hence dividing tex]0.125 × 10^{-6}[/tex] meter by $174 × 10^{-12}$ meter we get that the number of sheets will be as follows

$\frac {0.125× 10^{-6}}{174\times 10^{-12}}=718.3908045\approx 719$

Therefore, they are approximately 719 sheets

7 0

2 years ago

A 5-kg concrete block is lowered with a downward acceleration of 2.8 m/s2 by means of a rope. The force of the block on the Eart

maksim [4K]

When the body touches the ground two types of Forces will be generated. The Force product of the weight and the Normal Force. This is basically explained in Newton's third law in which we have that for every action there must also be a reaction. If the Force of the weight is pointing towards the earth, the reaction Force of the block will be opposite, that is, upwards and will be equivalent to its weight:

F = mg

Where,

m = mass

g = Gravitational acceleration

F = 5*9.8

F = 49N

Therefore the correct answer is E.

5 0

2 years ago

Based on the article “Will the real atomic model please stand up?,” why did J.J. Thomson experiment with cathode ray tubes? to s

PIT_PIT [208]

Answer:

B.) to determine that electric beams in cathode ray tubes were actually made of particles

Explanation:

This is the right answer i just took the quiz on edge.

3 0

2 years ago

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a

Lapatulllka [165]

Answer:

Amplitude, A = 0.049 meters

Explanation:

Given that,

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a function of time according to the equation :

$y = 0.049 \cos(7t)$ .......(1)

The general equation of a wave is given by :

$y=A\cos(\omega t)$ .......(2)

A is amplitude of wave

On comparing equation (1) and (2) we get :

A = 0.049 meters

So, the amplitude of the wave is 0.049 meters.

3 0

2 years ago

In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of you

olchik [2.2K]

PART A)

horizontal distance that will be moved = 14 m

Height of the fence = 5.0 m

height from which it is thrown = 1.60 m

angle of projection = 54 degree

So here we can say that stone will travel vertically up by distance

$\Delta y = 5 - 1.6 = 3.40 m$

now we will have displacement in horizontal direction

$\Delta x = 14 m$

now we know that

$v_x = vcos54$

$v_y = vsin54$

now we will have

$\Delta x = v_x t$

$14 = (vcos54)t$

also for y direction

$\Delta y = v_y t + \frac{1}{2}at^2$

$3.40 = (vsin54)t - \frac{1}{2}(9.8) t^2$

now from the two equations we will have

$3.40 = (vsin54)(\frac{14}{vcos54}) - 4.9 t^2$

$3.40 = 14 tan54 - 4.9 t^2$

$3.40 = 19.3 - 4.9 t^2$

$t = 1.8 s$

now from above equations

$14 = vcos54 (1.8)$

$v = 13.2 m/s$

So the minimum speed will be 13.2 m/s

Part B)

Total time of the motion after which it will land on the ground will be "t"

so its vertical displacement will be

$\Delta y = -1.60 m$

now we will have

$-1.60 = v_y t + \frac{1}{2}at^2$

$-1.60 = (13.2sin54)t - \frac{1}{2}(9.8)t^2$

$4.9 t^2 - 10.7t - 1.60 = 0$

$t = 2.3 s$

Now the time after which it will reach the fence will be t1 = 1.8 s

so total time after which it will fall on other side of fence

$t_2 = t - t_1$

$t_2 = 2.3 - 1.8 = 0.5 s$

now the displacement on the other side is given as

$\Delta x = (vcos54) t_2$

$\Delta x = (13.2 cos54)(0.5)$

$\Delta x = 3.88 m$

4 0

2 years ago