If the applied force is in the same direction as the object's displacement, the work done on the object is:
W = Fd
W = work, F = force, d = displacement
Given values:
F = 45N
d = 12m
Plug in and solve for W:
W = 45(12)
W = 540J
I'm pretty sure it is coal
hope this helps!
Yes, H<span>₂0 Is literally the subscript.</span>
Answer:
carrier power is 16.1 kW
Explanation:
given data
power transmit P = 20 kW
modulation index m = 0.7
to find out
carrier power
solution
we will apply here power transmit equation that is express as
...............1
put here all value in equation 1 we get carrier power
![20 = 1 + \frac{0.7^2}{2} * carrier power](https://tex.z-dn.net/?f=20%20%3D%201%20%2B%20%5Cfrac%7B0.7%5E2%7D%7B2%7D%20%2A%20carrier%20power)
solve it we get
carrier power = 16.1
so carrier power is 16.1 kW
In a projectile, the horizontal acceleration is zero. The velocity remains constant at all times. However, the <u>vertical acceleration</u> is -9.81m/s^2.
Hope this helps!