Answer:
554.27N
Explanation:
(a) The max frictional force exerted horizontally on the crate and the floor is,
Substitute the values,
μs=0.5
mass=113kg
g=9.81m/s
Ff=μsN
=μsmg
=(0.5 x 113 x 9.81)
Ff=554.27N
Answer:
(a) 108
(b) 110.500 kW
(c) 920.84 A
Solution:
As per the question:
Voltage at primary,
(rms voltage)
Voltage at secondary,
(rms voltage)
Current in the secondary,
Now,
(a) The ratio of secondary to primary turns is given by the relation:

where
= No. of turns in primary
= No. of turns in secondary
≈ 108
(b) The power supplied to the line is given by:
Power, P = 
(c) The current rating that the fuse should have is given by:



Solar system is nested nearly 2/3 of the way from the center of the galaxy to the outskirt of the galactic disc.