Homework help asap!!

Tidal forces are gravitational forces exerted on different parts of a body by a second body. Their effects are particularly visi

laiz [17]

Answer:

Explanation:

radius of earth = re

mass of the noon = m

mass of the earth = E

distance between earth and moon = r

acceleration of earth ae

force on earth = GMm / r²

acceleration of the earth

ae = force / mass

= GMm / (r² x M )

= Gm / r²

b ) The point on the earth nearest to moon will be at a distance of r - re

a_near = Gm / ( r - re)²

The point farthest on the earth to moon will be at a distance of r + re

a_ far = Gm / ( r + re )²

6 0

4 years ago

You and Mae are in free-float frames. Mae has just passed you, traveling at a speed of 60 km/hr. You throw a ball toward her at

Zina [86]

The ball will be stationary to her

4 0

3 years ago

Groundwater is desirable for which of the following reasons?

Akimi4 [234]

the answer is B. surface water can seep downward to become groundwater

5 0

3 years ago

According to the general theory of relativity, what are consequences of the curvature of space-time? Check all that apply. The d

Svetlanka [38]

Answer:

The dilation of time.

The falling of objects.

The changing of paths of light.

Explanation:

I have explained in the image attached below.

From the explanation, the correct ones are;

The dilation of time.

The falling of objects.

The changing of paths of light.

4 0

3 years ago

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i

Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

$V = 4\frac{kQ}{r}$

here

r = distance of corner of the square from it center

$r = \frac{a}{\sqrt2}$

$r = \frac{10nm}{\sqrt2} = 7.07 nm$

$Q = e = -1.6 * 10^{-19} C$

now the net potential is given as

$V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}$

$V = 0.815 V$

now potential energy of alpha particle at this position

$U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J$

Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

$r_1 = \frac{a}{2} = 5 nm$

$r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}$

$r_2 = 11.2 nm$

now potential is given as

$V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}$

$V = -0.576 - 0.257 = -0.833 V$

now final potential energy is given as

$U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J$

Now work done in this process is given as

$W = U_f - U_i$

$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

$W = -7 * 10^{-22} J$

8 0

3 years ago