Answer:
1) ΔE = -800 J, 2) ΔE = 200 J, 3) ΔE = -200 J, 4) ΔE = 800 J
Explanation:
For this exercise let's use the first law of thermodynamics
ΔE = Q + W
Where we will apply this expression to several cases
1) output Q = 500 J and does work of W = 300 J.
The two processes involve an energy output
ΔE = - 500 - 300
ΔE = -800 J
Therefore, the internal energy of the gas decreases by this amount,
2) enter Q = 500J and do work of W = 300 J
Positive heat is added and does negative work
ΔE = 500 - 300
ΔE = 200 J
3) they leave Q = 500 J and do (work on the gas) W = 300J
negative heat and positive work come out
ΔE = -500 + 300
ΔE = -200 J
4) heat enters Q = 500 J, work on gas W = 300 J
both positive
ΔE = 500 + 300
ΔE = 800 J
Yes. A pencil sitting on the desk in potential energy. <span />
Two thermometers, calibrated in celsius and fahrenheit respectively, are put into a liquid. the reading on the fahrenheit scale is four times the reading on the celsius scale. the temperature of the liquid is:
Answer
given,
speed of the basketball player, v_x = 4.75 m/s
height of the jump = ?
a) initial vertical velocity = 0 m/s
final vertical velocity = ?
height, h = 0.75 m
using equation of motion
v² = u² + 2 g h
v² = 0² + 2 x 9.8 x 0.75
v² = 14.7
v = 3.83 m/s
b) let horizontal distance= x m
maximum height at time , t s
time taken to reach maximum height

t = 0.39 s
horizontal distance
= v_x × t
= 4.75 × 0.39
= 1.85 m
horizontal distance is equal to 1.85 m