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.
Perimeter=26in
Area would be to multiply the units together
Answer:
1.when it is closest to the sun
2.when it is midway between its farthest
Explanation:
According to the law of Kepler's
T ² ∝ r³
T=Time period
r=semi major axis
We also know that time period T given as
v=Speed
So we can say that ,when r is more then the speed will be minimum and when r is low then speed will be maximum.
Answer:
This is because, the flywheel has a very large moment of Inertia and hence sudden piston torques have negligible effect on the flywheel, but every piston combined has a significant torque. This smoothens out the vibrations.
Explanation:
