If you mean the tree, evergreen trees can exploded if theres extreme stress on the trunk
Answer:
the extension recorded by the student would be smaller than the actual extension of the spring
Answer:
Explanation:
When a body is held against a vertical wall , to keep them in balanced position , normal force is applied on their surface . this force creates normal reaction which acts against the normal force and it is equal to the normal force as per newton's third law . Ultimately friction force is created which is proportional to normal force and it acts in vertically upward direction . It prevents the body from falling down .
Hence normal force = reaction force .
From second law also net force is zero , so if normal force is N and reaction force is R
R - N = mass x acceleration = mass x 0 = 0
R = N .
Ranking normal force from highest to smallest
150 N , 130 N , 120 N
B )
Frictional force is equal to the weight of the body because the body is held at rest .
Ranking of frictional force form largest to smallest
7 kg , 5 kg , 3 kg , 1 kg .
Here frictional force is irrespective of the normal force acting on the body because frictional force adjusts itself so that it becomes equal to weight in all cases here because it always balances the weight of the body .
Answer:
The coefficient of kinetic friction between the crate and the floor can be calculated using the formula μ = Ff / N, where Ff is the frictional force, N is the normal force, and μ is the coefficient of kinetic friction.
In this case, the normal force is equal to the weight of the crate, which is 24 kg * 9.8 m/s2 = 235.2 N. The frictional force can be calculated using the formula Ff = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.
If we substitute the values for N and Ff into the formula for the coefficient of kinetic friction, we get:μ = 53 N / 235.2 N = 0.225
Therefore, the coefficient of kinetic friction between the crate and the floor is 0.225.
Answer:
A) 31 kJ
B) 1.92 KJ
C) 40 , 2.48
Explanation:
weight of person ( m ) = 79 kg
height of jump ( h ) = 0.510 m
Compression of joint material ( d ) = 1.30 cm ≈ 0.013 m
A) calculate the force
Fd = mgh
F = mgh / d
W = mg
F(net) = W + F = mg ( 1 + 
= 79 * 9.81 ( 1 + (0.51 / 0.013) )
= 774.99 ( 40.231 ) ≈ 31 KJ
B) calculate the force when the stopping distance = 0.345 m
d = 0.345 m
Fd = mgh hence F = mgh / d
F(net) = W + F = mg ( 1 +
= 79 * 9.81 ( 1 + (0.51 / 0.345) )
= 774.99 ( 2.478 ) = 1.92 KJ
C) Ratio of force in part a with weight of person
= 31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40
Ratio of force in part b with weight of person
= 1920 / 774.99 = 2.48
