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The boom of a fire truck raises a fireman (and his equipment – total weight 280 lbf) 60 ft into the air to fight a building fire. (a) Showing all your work and using unity conversion ratios, calculate the work done by the boom on the fireman in units of Btu. (b) If the useful power supplied by the boom to lift the fireman is 3.50 hp, estimate how long it takes to lift the fireman.

Respuesta :

Answer:

a)Work done by fireman=   2.15  Btu

b) Time t= 0.86 sec

Explanation:

Given that

Weight = 280 lbf

We know that 1 lbf = 4.44 N

so 280 lbf = 1245.5 N

Weight =1245.5 N

Height h = 60 ft

We know that

1 ft = 0.3048 m

So 60 ft = 18.28 m

 h =18.28 m

Power = 3.5 hp

We know that

1 hp =0.74 KW

So 3.5 hp = 2.61 KW

Power = 2.61 KJ/s

So the work done by fireman = Weight x h

Now by putting the values

Work done by fireman= 1245.5 x 18.28 J

Work done by fireman=   2267.74 J

Work done by fireman=   2.26774  KJ

We know that 1 Btu= 1.05 KJ

So   2.266 KJ = 2.15 Btu

Work done by fireman=   2.15  Btu

We know that ,rate of work is called power.

Power x time =  work

2.61 x t = 2.26

So t= 0.86 sec

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