"increasing the distance through which the force is applied"
Answer:
a 15.22 m/s
b 45.65 m
Explanation:
Using the same formula,
x = vt, where
x is now 45.65, and
t is 3 s, then
45.65 = 3v
v = 45.65/3
v = 15.22 m/s
See the attachment for the part b. We used the distance gotten in part B, to find question A
Answer:
uk = 0.25
Explanation:
Given:-
- An object comes to stop with acceleration, a = -2.45 m/s^2
Find:-
What is the coefficient of kinetic friction between the object and the floor?
Solution:-
- Assuming the object has mass (m) that slides over a rough surface with coefficient of kinetic friction (uk). There is only Frictional force (Ff) acting in the horizontal axis on the object opposing the motion (-x direction).
- We will apply equilibrium equation on the object in vertical direction.
N - m*g = 0
N = m*g
Where, N : Contact force exerted by the surface on the floor
g : Gravitational acceleration constant = 9.81 m/s^2
- Now apply Newton's second law of motion in the horizontal ( x-direction ):
- Ff = m*a
- The frictional force is related to contact force (N) by the following expression:
Ff = uk*N
- Substitute the 1st and 3rd expressions in the 2nd equation:
uk*m*g = -m*a
uk = a / g
- Plug in the values and solve for uk:
uk = - (-2.45) / 9.81
uk = 0.25
Answer:
The average emf induced in the coil is 175 mV
Explanation:
Given;
number of turns of the coil, N = 1060 turns
diameter of the coil, d = 20.0 cm = 0.2 m
magnitude of the magnetic field, B = 5.25 x 10⁻⁵ T
duration of change in field, t = 10 ms = 10 x 10⁻³ s
The average emf induced in the coil is given by;

where;
A is the area of the coil
A = πr²
r is the radius of the coil = 0.2 /2 = 0.1 m
A = π(0.1)² = 0.03142 m²

Therefore, the average emf induced in the coil is 175 mV