Answer is a.)
Given: Mass m = 70 Kg Height h = 10 m; g = 10 m/s²
Required. Gravitational Potential Energy or P.E
Formula: P.E = mgh
= (70 Kg)(10 m/s²)( 10 m)
P.E = 7,000 Kg.m²/s² or
P.E = 7,000 J
Answer:
Ts=51.83C
Explanation:
First we calculate the surface area of the cylinder, neglecting the top and bottom covers as indicated by the question
Cilinder Area= A=πDL
L=200mm=0.2m
D=20mm=0.02m
A=π(0.02m)(0.2m)=0.012566m^2
we use the equation for heat transfer by convection
q=ha(Ts-T)
q= heat=2Kw=2000W
A=Area=0.012566m^2
Ts=surface temperature
T=water temperature=20C
Solving for ts
Ts=q/(ha)+T
Ts=2000/(5000*0.012566m^2)+20=51.83C
Taro's error is when he stated that the total energy of the ball and the club system is increasing. This is not true. The total energy of the system is not increasing. According to the first law of thermodynamics, <span>total energy of a system is always constant; energy can be transformed from one form to another however it cannot be created or destroyed. </span><span>Energy is conserved. </span>So, for this problem the total energy of the system should remain constant at all times.
Say the initial point is (0,0)
The final point is
x = 200 + 135*cos(30) = 200 + 135*sqrt(3)/2 = 316.91 ft
y = 135*sin(30) = 135/2 = 67.5 ft
Resultant vector = (316.91, 67.5) - (0,0) = 316.91, 67.5) ft
Answer:
Explanation:
Given that:
Absolute temperature of the body, 
- emissivity of the body,
<u>Using Stefan Boltzmann Law of thermal radiation:</u>
where:
(Stefan Boltzmann constant)
Now putting the respective values:
