Answer:
L= 0.4 kgm²/s
Explanation:
The angular momentum of a hallow spinning sphere is
L = Iω
L = (0.02kgm²) × (20rad/s)
L= 0.4kgm²/s
Answer:
-11.11 degree Celsius
Explanation:
The change was 44 degree fanhereit
To 56 degree fanhereit
Therefore the temperature range can be calculated as follows
56-44
= 12 degree fanhereit to Celsius
= 12-32×5/9
= -20×5/9
= 100/9
= -11.11 degree Celsius
Answer:
The block will not move.
Explanation:
We'll begin by calculating the frictional force. This can be obtained as follow:
Coefficient of friction (µ) = 0.6
Mass of block (m) = 3 Kg
Acceleration due to gravity (g) = 10 m/s²
Normal reaction (R) = mg = 3 × 10 = 30 N
Frictional force (Fբ) =?
Fբ = µR
Fբ = 0.6 × 30
Fբ = 18 N
From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.
Answer:
a. 20m/s
b.50N
c. Turkey has a larger mass than the ball. Neglible final acceleration and therefore remains stationery.
Explanation:
a. Given the force as 50N, times as 0.2seconds and the weight of the ball as 0.5 kg, it's final velocity can be calculated as:

Hence, the velocity of the ball after the kick is 20m/s
b.The force felt by the turkey:
#Applying Newton's 3rd Law of motion, opposite and equal reaction:
-The turkey felt a force of 50N but in the opposite direction to the same force felt by the ball.
c. Using the law of momentum conservation:
-Due to ther external forces exerted on the turkey, it remains stationery.
-The turkey has a larger mass than the ball. It will therefore have a negligible acceleration if any and thus remains stationery.
-Momentum is not conserved due to these external forces.
Answer:
(a) Magnitude of static friction force is 109 N
(b) Minimum possible value of static friction is 0.356
Solution:
As per the question;
Horizontal force exerted by the girl, F = 109 N
Mass of the crate, m = 31.2 kg
Now,
(a) To calculate the magnitude of static friction force:
Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:
F = f = 109 N
(b) The maximum possible force of friction between the floor and the crate is given by:

where
N = Normal reaction = mg
Thus

For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.



