<span>Heat transfer between two substances is affected by specific heat and the "Temperature difference between them"
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Answer:

Explanation:
Using kinematics equations:

Use
due to condition of distance traveled.
Solving second equation for time, there are two solutions. t=0 and

Use the expression in the first equation to have

Using trigonometric identities, you have the answer of the distance.
By doing the ratio for two different angles, you have the second answer. Due to sine function properties, the distances can be the same to complementary angles. Example, for 20° and 70°, the distance is the same.
Explanation:
Assuming the wall is frictionless, there are four forces acting on the ladder.
Weight pulling down at the center of the ladder (mg).
Reaction force pushing to the left at the wall (Rw).
Reaction force pushing up at the foot of the ladder (Rf).
Friction force pushing to the right at the foot of the ladder (Ff).
(a) Calculate the reaction force at the wall.
Take the sum of the moments about the foot of the ladder.
∑τ = Iα
Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0
Rw (3.0 sin 60°) = mg (1.5 cos 60°)
Rw = mg / (2 tan 60°)
Rw = (10 kg) (9.8 m/s²) / (2√3)
Rw = 28 N
(b) State the friction at the foot of the ladder.
Take the sum of the forces in the x direction.
∑F = ma
Ff − Rw = 0
Ff = Rw
Ff = 28 N
(c) State the reaction at the foot of the ladder.
Take the sum of the forces in the y direction.
∑F = ma
Rf − mg = 0
Rf = mg
Rf = 98 N
Answer:
the answer is B.
Explanation:
The claim is correct because Student Y can apply a force that is greater in magnitude than the frictional forces that are exerted on the student-student-skateboard system