1. 12.75 J
Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is

where
F = 15 N is the magnitude of the force
d = 85 cm = 0.85 m is the displacement of the cart
Substituting in the formula, we get

2. 10.6 N
In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).
Therefore, the work done is

However, in this case the displacement is
d = 120 cm = 1.20 m
Therefore, the magnitude of the force in this case is

Answer:
=0.855V
Explanation:
The induced voltage can be calculated using below expression
E =B x dA/dt
Where dA/dt = area
B= magnetic field = 6.90×10-5 T.
We were given speed of 885 km/h but we will need to convert to m/s for consistency of unit
speed = 885 km/h
speed = 885 x 10^3 m/hr
speed = 885 x 10^3/60 x60 m/s
speed = 245.8 m/s
If The aircraft wing sweep out" an area
at t= 50.4seconds then we have;
dA/dt = 50.4 x 245.8
= 123388.32m^2/s
Then from the expression above
E =B x dA/dt substitute the values of each parameters, we have
E = 6.90 x 10^-5 x 12388.32 V
E =0.855V
Hence, the average induced voltage between the tips of the wings is =0.855V
Answer:
I'm sorry I don't know what is the answer
Explanation:
Given , F = 30N and mass m = 90kg
°•° F = ma
=> a = F/m
=> a = 30/90
=> a = 1/3m/s^2