Acceleration = (change in speed) / (time for the change)
Change in speed = (end speed) - (start speed) = (15 m/s - 7 m/s) = 8 m/s
time for the change = 2 minutes = 120 seconds
Acceleration = (8 m/s) / (120 seconds)
Acceleration = 0.067 m/s²
Hooke's Law says that F=-kx where k is the spring constant measured in N/m (newtons per meter)
Answer:
B = 0.8 T
Explanation:
It is given that,
Radius of circular loop, r = 0.75 m
Current in the loop, I = 3 A
The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.
When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.
We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :
B is magnetic field
So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.
If the distance around the equator is reduced by half, then the radius is also reduced by half.
Since the acceleration due to gravity is proportional to 1/(radius²),
the acceleration changes by a factor of 1/(1/2)² = 1/(1/4) = <em>4 </em>.
The acceleration due to gravity ... and also the weight of everything on Earth ...
becomes <em>4 times what it is now</em>.