<span>internet tension = mass * acceleration internet tension = 23 – Friction tension = 14 * acceleration Friction tension = µ * 14 * 9.8 = µ * 137.2 23 – µ * 137.2 = 14 * acceleration Distance = undemanding speed * time undemanding speed = ½ * (preliminary speed + very final speed) Distance = ½ * (preliminary speed + very final speed) * time Distance = 8.a million m, preliminary speed = 0 m/s, very final speed = a million.8 m/s 8.a million = ½ * (0 + a million.8) * t Time = 8.a million ÷ 0.9 = 9 seconds Acceleration = (very final speed – preliminary speed) ÷ time Acceleration = (a million.8 – 0) ÷ 9 = 0.2 m/s^2 23 – µ * 137.2 = 14 * 0.2 resolve for µ</span>
Answer:
28 m/s^2
Explanation:
distance, s = 14 m
time, t = 2 - 1 = 1 s
initial velocity, u = 0 m/s
Let a be the acceleration.
Use third equation of motion


a = 28 m/s^2
Thus, the acceleration is 28 m/s^2.
Answer:
Terminal speed, v = 6901.07 m/s
Explanation:
It is given that,
Mass of the horizontal bar, m = 30 g = 0.03 kg
Length of the bar, l = 13 cm = 0.13 m
Magnetic field, 
Resistance, R = 1.2 ohms
We need to find the terminal speed oat which the bar falls. When terminal speed is reached,
Force of gravity = magnetic force
..................(1)
i is the current flowing
l is the length of the rod
Due to the motion in rods, an emf is induced in the coil which is given by :
, v is the speed of the bar


Equation (1) becomes,



v = 6901.07 m/s
So, the terminal speed at which the bar falls is 6901.07 m/s. Hence, this is the required solution.
Answer:
The electric field at origin is 3600 N/C
Solution:
As per the question:
Charge density of rod 1, 
Charge density of rod 2, 
Now,
To calculate the electric field at origin:
We know that the electric field due to a long rod is given by:

Also,
(1)
where
K = electrostatic constant = 
R = Distance
= linear charge density
Now,
In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.
At x = - 1 cm = - 0.01 m:
Using eqn (1):

(towards)
Now, at x = 1 cm = 0.01 m :
Using eqn (1):

(towards)
Now, the total field at the origin is the sum of both the fields:

Answer:
The answer is explained below.
Explanation:
All the point on the disk has same angular acceleration. Here, the point P is at the midway between the center and the rim of the disk and the point Q is at rim of the disk.
So, the distance of the point Q from the axis is twicee the distance of the point P from the axis.
<em>Rp - R</em>
<em>Rq - 2R</em>
The linear acceleration is
α2 - Rα
So, the linear acceleration of Q is twice as great as the linear acceleration of P.
The speed of the particle when it is in the circular motion depends on the radius of the particle.
In this case, the speed of point Q is twice the speed of point P.