Answer:
this pdf should help you out
Explanation:
Answer:
50 lb
Explanation:
Given,
The weight of astronaut's life support backpack on Earth (w) = 300 lb
Acceleration due to gravity on Earth (g) = 9.8 m/s²
Acceleration due to gravity on Moon = g'

We know that weight of an object on Earth is,


Similarly, weight on Moon will be




Thus the astronaut's life support backpack will weigh 50 lb on Moon.
Answer:
a) 
b) 
Explanation:
a)
Given:
amount of heat transfer occurred,
initial temperature of car, 
final temperature of car, 
We know that the change in entropy is given by:

(heat is transferred into the system of car)

b)
amount of heat transfer form the system of house,
initial temperature of house, 
final temperature of house, 



1) Assuming an adult person has an average mass of m=80 kg, and assuming it takes about 30 seconds to climb 5 meters of stairs, the energy used by the person is

So the power output is

And since the estimate we made is very rough, we can say that the power output of the person is comparable to the power output of the light bulb of 100 W.
2) Based on the results we found in the previous part of the exercise, since the power output of the person is comparable to the power output of 1 light bulb of 100 W, we can say that the person could have kept burning only one 100-W light bulb during the climb.
V = IR
By completing the equation, i found that the total power equation is : 4.8,
Which means that it's not exceed the power rating.
So i believe the answer would be : The string will remain lit
hope this helps