The coil of a generator has a radius of 0.13 m. when this coil is unwound, the wire from which it is made has a length of 9.50 m. the magnetic field of the generator is 0.40 t, and the coil rotates at an angular speed of 29 rad/s. what is the peak emf of this generator?

7.14 V

Explanation:

radius, r = 0.13 m

length, L = 9.5 m

magnetic field, B = 0.40 t

angular speed, ω = 29 rad/s

Let there are N number of turns in the coil.

Length of each turn  2 π r

So, N = total length / length of one turn

N = 9.5 / ( 2 x 3.14 x 0.13) = 11.6

the peak value of emf is given by

e = N B A ω

e = N B πr^2 ω

e = 11.6 x 0.4 x 3.14 x 0.13 x 0.13 x 29

e = 7.14 Volt

Thus, the peak value of induced emf is 7.14 volt.

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//

answer///isobars;