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erica [24]
3 years ago
15

A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 340 N is appl

ied. The 100 kg block sits on a horizontal frictionless surface, but there is friction between the two blocks.
Physics
1 answer:
Taya2010 [7]3 years ago
3 0

Answer:

The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

Explanation:

Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.

Given that,

Mass of block = 60 kg

Acceleration = 2.0 m/s²

Mass = 100 kg

Horizontal force = 340 N

Let the frictional force be f.

We need to calculate the frictional force

Using balance equation

F-f=ma

Put the value into the formula

340-f=60\times2.0

f=340-60\times2.0

f=220\ N

We need to calculate the coefficient of friction

Using formula of friction force

f= \mu mg

\mu=\dfrac{f}{mg}

\mu=\dfrac{220}{60\times9.8}

\mu =0.37

We need to calculate the acceleration of the 100 kg block

Using formula of newton's law

F = ma

a=\dfrac{F}{m}

a=\dfrac{220}{100}

a=2.2\ m/s^2

Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

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slavikrds [6]

Answer:

Option C. 30°C.

Explanation:

The following data were obtained from the question:

Mass of water (Mw) = 0.5 Kg

Specific heat capacity of water (Cw) = 4.18 KJ/Kg°C

Initial temperature of water (Tw1) =

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Change in temperature (ΔT) = T2 – Tw1 = T2 – 22°C

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Specific heat capacity of copper (Cc) = 0.386 KJ/kg°C

Initial temperature of copper (Tc1) = 115°C

Change in temperature (ΔT) = T2 – Tc1 = T2 – 115°C

Final temperature (T2) =..?

Note: Both the water and the piece copper will have the same final temperature and the heat will be zero since the water will cool the piece of copper.

Thus, we can determine the final temperature of the water and copper as follow:

Q = MwCwΔT + McCcΔT

0 = 0.5 x 4.18 x (T2 – 22) + 0.5 x 0.386 x (T2 – 115)

0 = 2.09 (T2 – 22) + 0.193 (T2 – 115)

0 = 2.09T2 – 45.98 + 0.193T2 – 22.195

Collect like terms

2.09T2 + 0.193T2 = 45.98 + 22.195

2.283T2 = 68.175

Divide both side by the coefficient of T2 i.e 2.283

T2 = 68.175/2.283

T2 = 29.8 ≈ 30°C

Therefore, the final temperature of water and copper is 30°C.

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Answer:

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