Answer:
Explanation:
First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:
Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.
Since 
and 
, we can rewrite the first equation as:
Now, we solve for 
and calculate it:
This means that the crate's coefficient of kinetic friction on the floor is 0.18.
Answer:
Explanation:
fundamental frequency, f = 250 Hz
Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.
the formula for the frequency is given by
.... (1)
Now the length is doubled ans the tension is four times but the mass remains same.
let the frequency is f'
.... (2)
Divide equation (2) by equation (1)
f' = √2 x f
f' = 1.414 x 250
f' = 353.5 Hz
Answer:
Take-off velocity = v = 81.39[m/s]
Explanation:
We can calculate the takeoff speed easily, using the following kinematic equation.
where:
a = acceleration = 4[m/s^2]
x = distance = 750[m]
vi = initial velocity = 25 [m/s]
vf = final velocity
![v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D%5Csqrt%7B%2825%29%5E%7B2%7D%2B%282%2A4%2A750%29%20%7D%20%5C%5Cv_%7Bf%7D%3D81.39%5Bm%2Fs%5D)
The cat has a speed. The truck and the bicycle both have velocity. We can't be sure of what the plane has.
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