Answer:
i answered this question not that long ago....it's D
Explanation:
Answer:
The frictional force is 
Explanation:
From the question we are told that
The coefficient of kinetic force is μk = 0.35
The normal force felt by the puck is 
Generally the frictional force that acts on the puck is mathematically represented as

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1. No they aren’t because they all belong to different sports and are used differently
Answer:
F = 156.3 N
Explanation:
Let's start with the top block, apply Newton's second law
F - fr = 0
F = fr
fr = 52.1 N
Now we can work with the bottom block
In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal
we apply Newton's second law
Y axis
N - W₁ -W₂ = 0
N = W₁ + W₂
as the two blocks are identical
N = 2W
X axis
F - fr₁ - fr₂ = 0
F = fr₁ + fr₂
indicates that the lower block is moving below block 1, therefore the upper friction force is
fr₁ = 52.1 N
fr₁ = μ N
a
s the normal in the lower block of twice the friction force is
fr₂ = μ 2N
fr₂ = 2 μ N
fr₂ = 2 fr₁
we substitute
F = fr₁ + 2 fr₁
F = 3 fr₁
F = 3 52.1
F = 156.3 N
Answer:
The value is 
Explanation:
From the question we are told that
The mass of the ice cube is 
The temperature of the ice cube is
The mass of the copper cube is 
The final temperature of both substance is 
Generally form the law of thermal energy conservation,
The heat lost by the copper cube = heat gained by the ice cube
Generally the heat lost by the copper cube is mathematically represented as
![Q = m_c * c_c * [T_c - T_f ]](https://tex.z-dn.net/?f=Q%20%3D%20%20m_c%20%20%2A%20%20c_c%20%2A%20%20%5BT_c%20%20-%20%20T_f%20%5D)
The specific heat of copper is 
Generally the heat gained by the ice cube is mathematically represented as

Here L is the latent heat of fusion of the ice with value 
So

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So
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