Answer:
A
Explanation:
Straight line with a negative slope
On a velocity_time graph
Answer:
0.20kg-m^2
Explanation:
Let the linear velocity of the rope(=of pulley) is v m/s
Using kinematic equation
=> v = u + at
=>v = 0 + 4.9a
=>v = 4.9a ------------ eq1
By v^2 = u^2 + 2as
=>v^2 = 0 + 2 x v/4.9 x 1.2
=>4.9v^2 - 2.4v = 0
=>v(4.9v - 2.4) = 0
=>v = 2.4/4.9 = 0.49 m/s
Thus by v = r x omega
=>omega = v/r = 0.49/0.02 = 24.49 rad/sec
BY W = F x s = 50 x 1.2 = 60 J
=>KE(rotational) = W = 1/2 x I x omega^2
=>60 = 1/2 x I x (24.49)^2
=>I = 0.20 kg-m^2
M= ?
g=9.8 m/s (2)
h=20 m
Eg=362,600 J
Eg/mg
362,600 J/9.8 m/s (2) x 20 m
=1,850 m
Answer:
7.8 m/s
Explanation:
Here object is falling with a gravitational acceleration there for we can take acceleration = 10 m/ s² and its constant through out the motion there for we can use motion equation
V = U + at
V - Final velocity
U - Initial velocity
a - acceleration
t - time
V=U+at
107.8=U + 10×10
= 7.8 m/s
Answer:
Explanation:
A proton of charge
q=+1.609×10^-19C
Orbit a radius of 12cm
r=0.12m
Magnetic Field of 0.31T
Angle between velocity and field is 90°
a. Because the magnetic force F supplies the centripetal force Fc.
The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by
F = qvB sin θ
And the centripetal force is given as
Fc=mv²/r
Where m is mass of proton
m=1.673×10^-27kg
Then, F=Fc
qvB sin θ=mv²/r
qBSin90=mv/r
rqB=mv
Then, v=rqB/m
v=0.12×1.609×10^-19×0.31/1.673×10^-23
v=3577692.78m/s
v=3.58×10^6m/s
b. Since,
F=qVBSin90
F=1.609×10^-19×3.58×10^6×0.31
F=1.785×10^-13 N.
