Electric
field strength = Electric force/charge
E = F/q
E = (1.8×10^-1) / (6.4×10^-7)
E = 2.81×10^5 NC^-1
Hope it helped!
Answer:
(a) Workdone = -27601.9J
(b) Average required power = 1314.4W
Explanation:
Mass of hoop,m =40kg
Radius of hoop, r=0.810m
Initial angular velocity Winitial=438rev/min
Wfinal=0
t= 21.0s
Rotation inertia of the hoop around its central axis I= mr²
I= 40 ×0.810²
I=26.24kg.m²
The change in kinetic energy =K. E final - K. E initail
Change in K. E =1/2I(Wfinal² -Winitial²)
Change in K. E = 1/2 ×26.24[0-(438×2π/60)²]
Change in K. E= -27601.9J
(a) Change in Kinetic energy = Workdone
W= 27601.9J( since work is done on hook)
(b) average required power = W/t
=27601.9/21 =1314.4W
Medium stars become dwarf stars, large stars become either neutron (pulsar) stars or black holes
Answer:
W = 1.545E6 J total work
P = W / t = 1.545E6 J / 3 sec = 5.15E5 J/sec = 515,000 J/sec (Watts)
Using definition of power