<span>So we want to know why is there a difference between the force of gravity on the Moon and the force of gravity of the Earth. So the gravitational force between two objects depends on the masses of both objects. That can be seen from Newtons universal law of gravity. F=G*m1*m2*(1/r^2). So lets say we are holding an object of mass m=1kg on a height r=1m on the Moon and we are holding the same object on the Earth also on the same height of r=1m. The Gravitational force on the Earth will be Fg=G*M*m*(r^2) where M is the mass of the Earth. The force between the moon and that object will be Fg=G*n*m*(r^2), where n is the mass of the moon. Since mass of the Moon is much smaller than mass of the Earth, The gravitational force between the Moon and that body will be almost 6 times smaller than the gravitational force between the Earth and that body. So the correct answer is B. </span>
Answer: the photograph will likely show only one star.
Explanation:
Since their angular separation is smaller than the telescope's angular resolution, the picture will apparently show only one star rather than two.
Answer:
the glass contains air bubbles that expands and contracts as the glass is heated or froze. when they expand they may cause the glass to break or even explode
Answer:
Explanation:
Given
side of square shape 
Electric flux 
Permittivity of free space 
Flux is given by

where E=electric field strength
A=area
=Angle between Electric field and area vector



and Electric field by a uniformly charged sheet is given by

where
=charge density


Answer:
a) f ’’ = f₀
, b) Δf = 2 f₀ 
Explanation:
a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.
Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source
f ’= fo
This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.
f ’’ = f’
where c represents the sound velocity in stationary blood
therefore the received frequency is
f ’’ = f₀
let's simplify the expression
f ’’ = f₀ \frac{c+v}{c-v}
f ’’ = f₀
b) At the low speed limit v <c, we can expand the quantity
(1 -x)ⁿ = 1 - x + n (n-1) x² + ...
f ’’ = fo
f ’’ = fo 
leave the linear term
f ’’ = f₀ + f₀ 2
the sound difference
f ’’ -f₀ = 2f₀ v/c
Δf = 2 f₀ 