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.
60 N because 98N=mg (here g= 9.8 on earth) thus mass can be calculated which is 98/9.8 = 10kg
Now,new weight with g = 6m/s^2
=m×g' (here g' is new acceleration of the new planet)
= 10×6=60N
Answer:
unbalanced force
Explanation:
this is a guess so just look it up
From Newton's second law, we know F = ma, where a is the acceleration and m is the mass in kg.
F = 1000kg * 9.8m/s = 9800N
F = 9800 N
Hope this helps!