Answer:
a is the answer to the question
Answer:
28 cm and 32 cm
Explanation:
1. The spring pendulum hangs vertically, oscillates harmonic with amplitude 2cm and angular frequency 20 rad/s. The natural length of
a spring is 30cm. What is the minimum and maximum length of the spring during the oscillation? Take g = 10m/s2.
As the amplitude is 2 cm and the natural length is 30 cm. So, it oscillates between 30 -2 = 28 cm to 30 + 2 = 32 cm.
So, the minimum length is 28 cm and the maximum length is 32 cm.
Answer: 62 μT
Explanation:
Given
Length of rod, l = 1.33 m
Velocity of rod, v = 3.19 m/s
Induced emf, e = 0.263*10^-3 V
Using Faraday's law, the induced emf of a rod can be gotten by the formula
e = blv where,
e = induced emf of the rod
b = magnetic field of the rod
l = length of the rod
v = velocity of the rod. On substituting, we have
0.263*10^-3 = b * 1.33 * 3.19
0.263*10^-3 = b * 4.2427
b = 0.263*10^-3 / 4.2427
b = 0.0000620 T
b = 62 μT
Thus, the strength of the magnetic field is 62 μT
a) 10 m/s
b) 25 m
Explanation:
a)
The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:
where
u is the initial velocity
v is the final velocity
a is the acceleration
t is the time
For the body in this problem:
u = 0 (the body starts from rest)
is the acceleration
t = 5 s is the time
So, the final velocity is
b)
In this second part, we want to calculate the distance travelled by the body.
We can do it by using another suvat equation:
where
u is the initial velocity
v is the final velocity
a is the acceleration
s is the distance travelled
Here we have
u = 0 (the body starts from rest)
is the acceleration
v = 10 m/s is the final velocity
Solving for s,
