Answer:
induced emf = 28.65 mV
Explanation:
given data
diameter = 7.3 cm
magnetic field = 0.61
time period = 0.13 s
to find out
magnitude of the induced emf
solution
we know radius is diameter / 2
radius = 7.3 / 2
radius = 3.65 m
so induced emf is dπ/dt = Adb/dt
induced emf = A × ΔB / Δt
induced emf = πr² × ΔB / Δt
induced emf = π (0..65)² × ( 0.61 - (-0.28)) / 0.13
induced emf = 0.0286538 V
so induced emf = 28.65 mV
Answer:
+7.0 m/s
Explanation:
Let's take rightward as positive direction.
So in this problem we have:
a = -2.5 m/s^2 acceleration due to the wind (negative because it is leftward)
t = 4 s time interval
v = -3.0 m/s is the final velocity (negative because it is leftward)
We can use the following equation:
v = u + at
Where u is the initial velocity
We want to find u, so if we rearrange the equation we find:
and the positive sign means the initial direction was rightward.
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2
Explanation:
(1) Given mass = 0.125 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2
(2) Given mass = 0.250 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2
(3) Given mass = 0.375 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2
(4) Given mass = 0.500 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
