Answer:
-58.876 kJ
Explanation:
m = mass of air = 1 kg
T₁ = Initial temperature = 15°C
T₂ = Final temperature = 97°C
Cp = Specific heat at constant pressure = 1.005 kJ/kgk
Cv = Specific heat at constant volume = 0.718 kJ/kgk
W = Work done
Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)
ΔU = Change in internal energy
Q = W+ΔU
⇒Q = W+mCvΔT
⇒0 = W+mCvΔT
⇒W = -mCvΔT
⇒Q = -1×0.718×(97-15)
⇒Q = -58.716 kJ
Answer:

Explanation:
It is given that,
Mass of the golf ball, m = 46 g = 0.046 kg
Terminal speed of the ball, v = 44 m/s
The drag force, 
Where, C is the drag coefficient. At terminal speed, the weight of the ball is balanced by the drag force.




Hence, this is the required solution.
3 times 6= 18. The average speed is 19 mph.
hope this helps!
I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J.
At point A, all this energy has converted into kinetic energy, which is:

And since K=7.35 J, we can find the velocity, v:
Answer:
304.89m
Explanation:
Given
acceleration a = 2.52m/s²
final speed v = 39.2m/s
initial speed = 0m/s (car accelerates from rest)
Using the equation of motion below to get the distance of Doc brown from Marty;
v² = u²+2as
substitute the given parameters
39.2² = 0²+2(2.52)s
1536.64 = 0+5.04s
divide both sides by 5.04
1536.64/5.04 = 5.04s/5.04
rearrange the equation
5.04s/5.04 = 1536.64/5.04
s = 304.89m
Hence He and Marty must stand at 304.89m to allow the car to accelerate from rest to a speed of 39.2 m/s?