Answer:
4 A
Explanation:
V = IR, where V=voltage, I=current, R=resistance. This is Ohm's Law. (remember that for units V = volts, Ω = ohms, A = amperes.)
V = IR
12 V = I * 3 Ω
12/3 = I
<u>I = 4 A</u>
The speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.
The angular momentum(L) of an electron moving in a circular path is given by the formula,
L = mvr ........(i)
We know that the radius of the path of an electron in a magnetic field is
r = mv/qB
Putting this value in equation (i),
L = mv x mv/qB
or L = (mv)^2/qB
Putting the given values in the above equation,
4 x 10^-25 = (9.1x10^-31)^2 x v^2/ 1.6 x 10^-19 x 1 x 10^-3
v comes out to be 8.88 x 10^7 m/s.
Hence, the speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.
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(a)
The formula is:
∑ F = Weight + T = mass * acceleration
as the elevator and lamp are moving downward, I choose downward forces to be
positive.
Weight is pulling down = +(9.8 * mass)
Tension is pulling up, so T = -63
Acceleration is upward = -1.7 m/s^2
(9.8 * mass) + -63 = mass * -1.7
Add +63 to both sides
Add (mass * 1.7) to both sides
(9.8 * mass) + (mass * 1.7) = 63
11.5 * mass = 63
mass = 63 / 11.5
Mass = 5.48 kg
(b)
Since the elevator and lamp are going upward, I choose upward forces to be
positive.
Weight is pulling down = -(9.8 * 5.48) = -53.70
Acceleration is upward, so acceleration = +1.7
-53.70 + T = 5.48 * 1.7
T = 53.70 + 9.316 = approx 63 N
The Tension is still the same - 63 N since the same mass, 5.48 kg, is being accelerated
upward at the same rate of 1.7 m/s^2
Answer:
Coin is dropped from a height of 19.62 m
Explanation:
We have given time t = 2 sec
As coin is drop means its initial velocity u = 0 m/sec
We have to find the height from which coin is dropped
From second equation of motion we know that

So height 
So coin is dropped from a height of 19.62 m