Answer:
1.9976 m
Explanation:
An iron bar whose initial length is 2m has its temperature increased from 20 ° C to 120 ° C. What is the final length of the bar? (consider that the coefficient of expansion of iron) is 1.2 x 10-5 1 / ° C (3 points)
Given that,
Initial length of an iron bar, 
The temperature increased from 20°C to 120°C
The coefficient of expansion of iron is 
We need to find the final length of the bar.
The linear expansion equation is given by :

Where
is the change in length

Also, 
L is final length

So, the final length of the bar is 1.9976 m
Answer:
It's due to the distance from either ends of strings origin...
Explanation:
As we know that waves behave moving in a flow from one side to another side and this gives a prospective of motion. Suppose a wave is pinched from the near one end of a guitar then due to the distortion created by the point of tie of strings the wave super imposes and moves with a velocity v and produces a wave frequency f. as we the pinching go down to the center the wave stabilizes itself to a stationary origin right at the center and the frequency then changes accordingly as moving down on the string.
Answer:
The power output of this engine is 
The the maximum (Carnot) efficiency is 
The actual efficiency of this engine is 
Explanation:
From the question we are told that
The temperature of the hot reservoir is 
The temperature of the cold reservoir is 
The energy absorbed from the hot reservoir is 
The energy exhausts into cold reservoir is 
The power output is mathematically represented as

Where t is the time taken which we will assume to be 1 hour = 3600 s
W is the workdone which is mathematically represented as

substituting values

So


The Carnot efficiency is mathematically represented as



The actual efficiency is mathematically represented as

substituting values


Answer:
Heat required = mass× latent heat Q = 0.15 × 871 ×