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Vika [28.1K]
3 years ago
12

A fireman is sliding down a fire pole. As he speeds up, he tightens his grip on the pole, thus increasing the vertical frictiona

l force that the pole exerts on the fireman. When the force on his hands equals his weight, what happens to the fireman?
A. The fireman comes to a stop.
B. The fireman descends with slower and slower speed.
C. The fireman descends with a smaller but non-zero acceleration.
D. The fireman continues to descend, but with constant speed.
E. The acceleration of the fireman is now upward.
Physics
1 answer:
Gnesinka [82]3 years ago
6 0

Answer:

The fireman continues to descend but with constant speed.

Explanation:  

According to Newton first law of motion, The object which is rest continues to remain at rest or object moving with constant speed move with constant velocity when acted upon by an unbalanced force.

When the net force acting on the object becomes zero the object moves with constant speed.

Hence when the frictional force acting on the fireman hands equals to his body weight , the net force acting on the fireman becomes zero and thus fireman continues to descend with constant speed.

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Answer:

This is how I figured it out:

  1. 215.5 rounded to one significant figure is 200
  2. 101.02555 rounded to one significant figure is 100.
  3. 200 + 100 = 300.

Hope this helps!

Explanation:

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Answer:

Explanation:

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m2v_0+2mv_0=mv_1+2mv_2 \quad (1/m) \quad 4v_0=v_1+2v_2\\

Kinetic energy conservation

\displaystyle \frac{1}{2}m(2v_0)^2+\frac{1}{2}2mv_0^2=\frac{1}{2}mv_1^2+\frac{1}{2}2mv_2^2 \quad (1/m) \quad 6v_0^2=v_1^2+2v_2^2

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Answer:

1353.38 Watt

Explanation:

T₁ = Initial temperature of the house = 35°C

T₂ = Final temperature of the house = 20°C

Δt = Time taken to cool the house = 38 min = 38×60 = 2280 s

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Cv = Specific heat at constant volume = 0.72 kJ/kgK

Cp = Specific heat at constant pressure = 1.0 kJ/kgK

Heat removed

q = mCvΔT

⇒q = 800×720×(35-20)

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Average rate of hear removal

Q=\frac{q}{\Delta t}\\\Rightarrow Q=\frac{8640000}{2280}\\\Rightarrow Q=3789.47\ W

COP=\frac{Q}{W}\\\Rightarrow W=\frac{Q}{COP}\\\Rightarrow W=\frac{3789.47}{2.8}\\\Rightarrow W=1353.38\ W

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