We are given that,

We need to find
when 
The equation that relates x and
can be written as,


Differentiating each side with respect to t, we get,



Replacing the value of the velocity


The value of
could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to



Search light is rotating at a rate of 0.96rad/s
im not gonna write a research paper but this is the really easy way write global warming talk about animals the polar ice caps and water levels then for what causes it burning fossil fuels and energy plants. then finish off with its awful and we should use solar or geothermic or wind or when the time comes fusion not fission fusion makes helium from hydrogen then burylliam from helium then oxygen and silicon so on so forth instead of fissions uranium,plutonium and thorium and with radioactive waste
Answer:
T = 692.42 N
Explanation:
Given that,
Mass of hammer, m = 8.71 kg
Length of the chain to which an athlete whirls the hammer, r = 1.5 m
The angular sped of the hammer, 
We need to find the tension in the chain. The tension acting in the chain is balanced by the required centripetal force. It is given by the formula as follows :

So, the tension in the chain is 692.42 N.
Answer:
300 cos 30 = 40 a + 40 * .2 * 10
Total force = mass * acceleration + frictional force
260 = 40 a + 80
a = 180 / 40 = 4.5 m/s^2
Check:
15 a + 15 * 10 * .2 = T acceleration of 15 kg block (assuming a = 4.5)
T = 15 (4.5) + 30 = 97.5 force required to accelerate 15 kg block
260 - 97.5 = 162.5 net force on 25 kg block
162.5 = 4.5 (25) + 25 * 10 * .2
162.5 = 112.5 + 50 = 162.5
4.5 m/s^2 checks out as correct
I think it’s D................