Answer:
Heat required to melt 1 lb of ice is 151.469 KJ
Explanation:
We have given mass of ice = 1 lb
We know that 1 lb = 0.4535 kg
Latent heat of fusion for ice =334 KJ/kg
Amount if heat for fusion of ice is given by
, here m is mass of ice and L is latent heat of fusion
So heat 
So heat required to melt 1 lb of ice is equal to 151.469 KJ
Question three is C and question 4 is b
Answer:
The answer to your question is: Ep = 14715 Joules
Explanation:
Data
mass = 50 kg
Potential energy = ?
height = 30 m
g = 9.81 m/s²
Equation
Ep = mgh
Substitution
Ep = (50)(9.81)(30)
Ep = 14715 Joules
Answer:
1:04-1:10 hours
Explanation:
You'll need a <em>Recreational dive planner</em> table, I annexed a copy, now you'll follow the next steps:
- In the first part of your table, you'll look for the distance row (in feet) of your first dive, for this specific exercise you'll find 60, once you locate it you'll go down that column until you reach the time you'll dive, in this case, 45 (minutes) or the closest value (47).
- You'll check and keep the letter in that 47 row (S) for future use.
- Now you have to go to the second part of your table and look for the distance column, in feet, of your second dive. We find 60 and then going right in the blue row, we'll look for the time (35) or its closest value (36).
- Finally, we have to check the letter for 36 minutes (F) and we'll make it met with the letter S in the first portion of your tables. This will give us an interval of time, 1:04-1:10 in this case.
I hope you find this information useful and interesting! Good luck!
