Answer:
Woke done, W = 4156.92 Joules
Explanation:
The work done by the force can be calculated as :


is the angle between force and the displacement
It is assumed to find the work done for the given parameters i.e.
Force, F = 30 N
Distance travelled, s = 160 m
Angle between force and displacement, 
Work done is given by :


W = 4156.92 Joules
So, the work done by the object is 4156.92 Joules. Hence, this is the required solution.
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:

Answer:
final displacement lf = 0.39 m
Explanation:
from change in momentum equation:
![\delta p = m \sqrt(2g * y/x)* [\sqrt li + \sqrt lf]](https://tex.z-dn.net/?f=%5Cdelta%20p%20%3D%20m%20%5Csqrt%282g%20%2A%20y%2Fx%29%2A%20%5B%5Csqrt%20li%20%2B%20%5Csqrt%20lf%5D)
given: m = 0.4kg, y/x = 19/85, li = 1.9 m,
\delta p = 1.27 kg*m/s.
putting all value to get the final displacement value
![1.27 = 0.4\sqrt(2*9.81 *(19/85))* [\sqrt 1.9 + \sqrt lf]](https://tex.z-dn.net/?f=1.27%20%3D%200.4%5Csqrt%282%2A9.81%20%2A%2819%2F85%29%29%2A%20%5B%5Csqrt%201.9%20%2B%20%5Csqrt%20lf%5D)
final displacement lf = 0.39 m
Answer:
The new speed is 56.25 miles/hour.
Explanation:
Since speed = distance/time;
time = distance/speed.
While driving at 50 miles/hour, time taken for one to complete 1 mile is (1/50) hour
(1/50) hour = (1/50) × 3600s = 72 seconds.
So, if this time to complete 1 mile (72 seconds) is reduced by 8 seconds,
New time to complete 1 mile will be = 72 - 8 = 64 seconds = (64/3600) hour = 0.0178 hour
New speed would be = (1 mile/64 seconds) = (1 mile/0.0178 hour) = 56.25 miles/hour.
Hope this Helps!!!