We can first calculate the net force using the given information.
By Newton's second law, F(net) = ma:
F(net) = 25 * 4.3 = 107.5
We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):
f = F(net) - F(app) = 150 - 107.5 = 42.5 N
Now we can calculate the coefficient of friction, u, using the normal force, F(N):
f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
The frequency of the radio wave is:

The wavelength of an electromagnetic wave is related to its frequency by the relationship

where c is the speed of light and f the frequency. Plugging numbers into the equation, we find

and this is the wavelength of the radio waves in the problem.
Momentum question. This is an inelastic collision, so
m1v1+m2v2=Vf(m1+m2)
Vf=(m1v1+m2v2)/(m1+m2)=[(120kg)(0m/s)+(60kg)(2m/s)] / (120kg+60kg)
Vf=120kg m/s / 180kg
Vf=0.67m/s
0.67m/s
Answer:
t=0.704s
Explanation:
A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How long will it take the car to reach the bottom of the ramp if there is no friction?
from newton equation of motion , we look for the y component of the speed and look for the x component of the speed. we can then find the resultant of the speed

Vy^2=0+2*9.8*1.73sin40.5
Vy^2=22.021
Vy=4.69m/s
Vx^2=u^2+2*9.81*cos40.5
Vy^2=25.81
Vy=5.08m/s
V=(Vy^2+Vx^2)^0.5
V=47.71^0.5
V=6.9m/s
from newtons equation of motion we know that force applied is directly proportional to the rate of change in momentum on a body.
f=force applied
v=velocity final
u=initial velocity
m=mass of the toy, 0.046
f=ma
f=m(v-u)/t
v=u+at
6.9=0+9.8t
t=6.9/9.81
t=0.704s
Total time in between the dropping of the stone and hearing of the echo = 8.9 s
Time taken by the sound to reach the person = 0.9 s
Time taken by the stone to reach the bottom of the well = 8.9 - 0.9 = 8 seconds
Initial speed (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s^2
Time taken (t) = 8 seconds
Let the depth of the well be h.
Using the second equation of motion:

h = 313.6 m
Hence, the depth of the well is 313.6 m