Answer:
123.30 m
Explanation:
Given
Speed, u = 22 m/s
acceleration, a = 1.40 m/s²
time, t = 7.30 s
From equation of motion,
v = u + at
where,
v is the final velocity
u is the initial velocity
a is the acceleration
t is time
V = at + U
using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane
where m is the slope
Comparing equation (1) and (2)
a = m
Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.
a = - 1.40 m/s²
Sunstituting a = - 1.40 m/s² and u = 22 m/s
The speed of the train at 7.30 s is 11.78 m/s.
The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.
Area of triangle + Area of rectangle
![[\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%2822%20-%2011.78%29%20%2A%20%287.30%29%5D%20%20%2B%20%5B%2811.78%20-%200%29%20%2A%20%287.30%29%5D)
= 37.303 + 85.994
= 123. 297 m
≈ 123. 30 m
Answer:
Rotating the loop until it is perpendicular to the field
Explanation:
Current is induced in a conductor when there is a change in magnetic flux.
The strength of the induced current in a wire loop moving through a magnetic field can be increased or decreased by the following methods:
By increasing the strength of the magnetic field there will be increased in the induced current. If the strength of the magnetic field is decreased then there is a decrease in induced current.
By increasing the speed of the wire there will be increased in the induced current. When the speed of the wire is decreased then there is a decrease in induced current.
By increasing the number of turns of the coil the strength of the induced current can be increased. when there is less number of turns in coils then there is a decrease in induced current.
Rotating the loop until it is perpendicular to the field will not increase the current induced in a wire loop moving through a magnetic field.
Therefore, the option is (c) is correct.
Answer:
Explanation:
Mass of ball Is m=96.1g=0.0961kg
Height above spring is 59.1cm
L=0.591m
Extension of the spring is 4.75403cm
e=0.0475403m
Then the distance the ball traveled is H=L+e
H=0.591+0.0475403
H=0.6385403m
Then, the potential energy of the ball is given as
P.E=mgh
P.E=0.0961×9.81×0.6385403
P.E=0.602J
From conservation of energy, energy cannot be created nor destroy but can be transferred from one form to another
Then, the P.E is transferred to the work done by the spring
Then, Work done by spring is given as
W=½ke²
W=P.E=½×k×0.0475403²
0.602=½×k×0.0475403²
k=0.602×2/0.0475403²
k=532.72N/m
The spring constant is 532.72 N/m
Answer:
speed of the charge electric is v = - (Eo q/m) cos t
Explanation:
The electric charge has a very small mass so it follows the oscillations of the electric field. We force ourselves on the load,
F = q Eo sint
a) To find the velocity of the particle, let's use Newton's second law to find the acceleration and of this by integration the velocity
F = ma
q Eo sint = ma
a = Eo q / m sint
a = dv / dt
dv = adt
∫ dv = ∫ a dt
v-vo = I (Eoq / m) sin t dt
v- vo = Eo q / m (-cos t)
We evaluate the integral from the initial point, as the particle starts from rest Vo = 0, for t = 0
v = - (Eo q / m) cos t
b) Kinetic energy
K = ½ m v2
K = ½ m (Eoq / m)²2 (sint)²
K = ¹/₂ Eo² q² / m sin² t
c) The average kinetic energy over a period
K = ½ m v2
<v2> = (Eoq / m) 2 <cos2 t>
The average of cos2 t = ½, substitute and calculate
K = ½ m (Eoq / m)² ½
K = ¼ Eo² q² / m
