Answer:
5.78amps
Explanation:
Given data
Time t= 57 seconds
Charge Q= 330C
Current I= ??
The expression for the electric current is given as
Q= It
Substituting we have
330= I*57
I= 330/57
I=5.78 amps
Hence the current is 5.78amps
Answer: D
Explanation:
Kinetic energy = 1/2mV^2
From the formula above, we can deduce that kinetic energy is proportional to the square of speed. That is,
K.E = V^2
Graphically, the relationship isn't linear but a positive exponential. Therefore, option D is the correct answer.
Answer:
3.2 m/s²
Explanation:
Acceleration can be calculated as:
v = u + at (where v is final velocity, u is initial velocity, a is acceleration and t is time)
25 m/s = 9 m/s + a(5 s) (a is unknown)
16 m/s = a(5 s)
a = 3.2 m/s²
We assume that this is a uniform acceleration (meaning that the velocity increases at an equal rate for those 5 seconds).
Answer:
The magnitude of the induced electric field at a point 2.5 cm from the axis of the solenoid is 8.8 x 10⁻⁵ V/m
Explanation:
given information:
radius, r = 2.0 cm
N = 700 turns/m
decreasing rate, dI/dt = 9.0 A/s
the magnitude of the induced electric field at a point 2.5 cm (r = 2.5 cm = 0.025 m) from the axis of the solenoid?
the magnetic field at the center of solenoid
B = μ₀nI
where
B = magnetic field (T)
μ₀ = permeability (1.26× 10⁻⁶ T.m/A)
n = the number turn per unit length (turn/m)
I = current (A)
dB/dt = μ₀n dI/dt (1)
now we calculate the induced electric field by using
E = 
= 2E/r (2)
where
E = the induced electric field (V/m)
we substitute the firs and second equation, thus
dB/dt = μ₀n dI/dt
2E/r = μ₀n dI/dt
E = (1/2) r μ₀n dI/dt
= (1/2) (0.025) (1.26× 10⁻⁶) (700) (8)
= 8.8 x 10⁻⁵ V/m
Larry's average speed is
(total distance) / (total time).
Speed = (80 km) / (60 min) = 80 km/hr .
Larry's average velocity is
(distance between start- and end- points),
in the direction from the start-point to the end-point.
Velocity = (40 km east) / (60 min) = 40 km/hr east .
