It’s either 0.05 or 20. Assuming that the coefficient friction is a damping factor, I feel like 0.05 would be correct m
Answer: The velocity at different marked time points are given as
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
Explanation:
The slope of the tangent of the curve indicates the instantaneous velocity. So if the slope of the tangent is positive, that Is, the tangent makes a positive angle (above the horizontal axis) with the horizontal
axis, then the velocity at this point is positive, and if the slope of the tangent is negative, that is the tangent makes a negative angle with the horizontal axis (below the horizontal axis), then the velocity at this point is negative.
When the tangent of the line is parallel to the horizontal axis, the velocity is 0.
From the position-time graph attached, the sign on the instantaneous velocity for each time marked on the graph is given below
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
QED!
Answer:
(a) 0.345 T
(b) 0.389 T
Solution:
As per the question:
Hall emf, 
Magnetic Field, B = 0.10 T
Hall emf, 
Now,
Drift velocity, 

Now, the expression for the electric field is given by:
(1)
And

Thus eqn (1) becomes
where
d = distance
(2)
(a) When 

(b) When 

Answer:
Explanation:
Let h be the height .
initial velocity in first case u = 0
final velocity v = 6 m /s
acceleration due to gravity g = 9.8 m /s²
v² = u² + 2 g h
6² = 0 + 2 x 9.8 x h
h = 1.837 m .
For second case u = 3 m /s
v² = u² + 2 gh
= 3² + 2 x 1.837 x 9.8
= 9 + 36
= 45 m
v = 6.7 m /s
To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.
PART A) We will begin by finding the two net distances.

And the distance 'd' is



Through the free-body diagram the tension components are given by


Here we can watch that,

Dividing both expression we have that,

Replacing the values,


PART B) Using the vertical component we can find the tension,



