Answer:
27.0 milliliters is the nearest mililiter so 27.0 is the answer
Explanation:
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
Answer:
V = (v1 + v2) / 2 = (8 + 6.5) / 2 = 7.25 m/s average speed
t = 7.2 / 7.25 = .993 sec time to cross patch
a = (v2 - v1) / t = (6.5 - 8) / .993 = -1.51 m/s^2 or 1.5 m/s^2
I have no idea what that is, but all of your answers right
Answer:
Part a)
Part b)
Explanation:
As we know that two carts are moving in opposite directions
so here total momentum of two carts must be given by vector subtraction
It is given as
Now we know that total momentum of the system is conserved here as there is no external force on this system
so the speed of the first cart when second cart was at rest is given as
