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
Answer:
a new moon is quite near the Sun in the sky
Explanation:
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
The attached file has a detailed solution of the given problem.
Answer:
B. The elastic portion of a straight-line, downward-sloping demand curve corresponds to the segment above the midpoint.
Explanation:
Elasticity measures the sensitivity of one variable to another. Specifically it is a figure that indicates the percentage variation that a variable will experience in response to a variation of another one percent.
The elasticity of demand measures the reaction of demand when one of the factors that affects it varies.
<u>Elasticity - Price of demand.</u>
easure the sensitivity of the quantity demanded to price variations. It indicates the percentage variation that the quantity demanded of a good will experience if its price rises by 1 percent.
<u>
Elastic Demand
</u>
The demand quantity is relatively sensitive to price variations, so the total expenditure on the product decreases when the price rises, the price elasticity takes value greater than -∞ but less than -1
Answer:0.061
Explanation:
Given

Temperature of soup 
heat capacity of soup 
Here Temperature of soup is constantly decreasing
suppose T is the temperature of soup at any instant
efficiency is given by



integrating From
to 


![W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}](https://tex.z-dn.net/?f=W%3Dc_v%5Cleft%20%5B%20T-T_C%5Cln%20T%5Cright%20%5D_%7BT_H%7D%5E%7BT_C%7D)
![W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]](https://tex.z-dn.net/?f=W%3Dc_v%5Cleft%20%5B%20%5Cleft%20%28%20T_C-T_H%5Cright%20%29-T_C%5Cleft%20%28%20%5Cln%20%5Cfrac%7BT_C%7D%7BT_H%7D%5Cright%20%29%5Cright%20%5D)
Now heat lost by soup is given by

Fraction of the total heat that is lost by the soup can be turned is given by

![=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bc_v%5Cleft%20%5B%20%5Cleft%20%28%20T_C-T_H%5Cright%20%29-T_C%5Cleft%20%28%20%5Cln%20%5Cfrac%7BT_C%7D%7BT_H%7D%5Cright%20%29%5Cright%20%5D%7D%7Bc_v%28T_C-T_H%29%7D)



