Answer:
temperature on left side is 1.48 times the temperature on right
Explanation:
GIVEN DATA:

T1 = 525 K
T2 = 275 K
We know that


n and v remain same at both side. so we have

..............1
let final pressure is P and temp 

..................2
similarly
.............3
divide 2 equation by 3rd equation
![\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}](https://tex.z-dn.net/?f=%5Cfrac%7B21%7D%7B11%7D%5E%7B-2%2F3%7D%20%5Cfrac%7B21%7D%7B11%7D%5E%7B5%2F3%7D%20%3D%20%5B%5Cfrac%7BT_1%20%7Bf%7D%7D%7BT_2%20%7Bf%7D%7D%5D%5E%7B5%2F3%7D)

thus, temperature on left side is 1.48 times the temperature on right
Answer:
40mph
Explanation:
1st leg DATA:
time = 3 hrs ; speed = r mph ; distance = 3r miles
------------
2nd leg DATA:
speed = r mph ; distance = 12 miles
--------------------------------
3r + 12 = 132
3r = 120
rate = 40 mph
Answer: The answer is D
Explanation: i had the same question and i just guessed and got it first try
Answer:
copper will have more change in temperature as compare with aluminum
Explanation:
Hot piece of copper is made in contact with cold piece of aluminium
So here thermal energy transfer will take place from copper to aluminium
so by energy conservation we can say that heat given by copper is same as the heat absorbed by aluminium.
now we have

here we know that
= specific heat capacity of copper
= specific heat capacity of aluminum
given that specific heat capacity of aluminium is more than double that of copper
so we can say

so here if the mass of copper and aluminium is same then

so temperature change of copper is twice the temperature change of aluminium
So copper will have more change in temperature as compare with aluminum
Answer:
please give me brainlist and follow
Explanation:
The measuring sensitivity of liquid-in-glass thermometers increases with the amount of liquid in the thermometer. The more liquid there is, the more liquid will expand and rise in the glass tube. For this reason, liquid thermometers have a reservoir to increase the amount of liquid in the thermometer.