(1) You must find the point of equilibrium between the two forces,
<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2
So,
x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.
</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and
R - x = 3.46*10^8 m
from the center of the Earth.
(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
I'd go for D here. It also fits in with the idea of thermal expansion - as something is heated up, molecules vibrate and maybe collide. they vibrate with bigger amplitudes, so taking up more space, so expanding. maybe
Answer:
The tension in the cord is 
Explanation:
Given:
M = mass
b = radius
R = spool of radius
The equation is:
(eq. 1)
The sum of forces in y:
∑Fy = Mg - T = Ma

Replacing in eq. 1

Given :
Initial speed , u = 0 m/s .
Final speed , v = 91 km/h = 25.28 m/s .
To Find :
a) Average acceleration .
b ) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.3 s .
Solution :
a )
We know ,by equation of motion :

b)
Also , by equation of motion :

Hence , this is the required solution .
To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.
In other words the acceleration can be described as

Where
G = Gravitational Universal Constant
M = Mass of Earth
r = Radius of Earth
This equation can be differentiated with respect to the radius of change, that is


At the same time since Newton's second law we know that:

Where,
m = mass
a =Acceleration
From the previous value given for acceleration we have to

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:




But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:


Therefore there is a weight loss of 0.3N every kilometer.