On a different day the same car enters a 420-m radius horizontal curve on a rainy day when the coefficient of static friction be
1 answer:
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Answer:0.302
Explanation:
Given
mass of crate m=27 kg
Force required to set crate in motion is 80 N
Once the crate is set in motion 56 N is require to move it with constant velocity
i.e. 80 N is the amount of force needed to just overcome static friction and 56 is the kinetic friction force
thus
where is the coefficient of static friction and N is Normal reaction
Answer:
(a) 4.27 x 10^-4 Telsa
(b) 3.28 x 10^-4 Telsa
Explanation:
side of square, a = 5.49 cm
inner radius, r = 18.1 cm = 0.181 m
number of turns,N = 450
current, i = 0.859 A
(a)
The magnetic field due to a solenoid due to inner radius is
B = 4.27 x 10^-4 Telsa
(b)
The outer radius is R = 18.1 + 5.49 = 23.59 cm = 0.236 m
The magnetic field due to the outer radius is
B = 3.28 x 10^-4 Tesla