Answer:
Explanation:
Given data
length=100mm
Diameter=5mm
Thermal conductivity=5 W/m.K
Power=50 W
Temperature=25°C
The temperature of heater surface follows from the rate equation written as:
Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium
Substitute the given values
![S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B2%5Cpi%20%280.1m%29%7D%7Bln%5B%5Cfrac%7B4%2A0.1m%7D%7B0.005m%7D%20%5D%7D%5C%5C%20S%3D0.143m)
The temperature of heater is then:
The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.
Answer:
Explanation:
Call the bike on the right A
Call the bike on the left B
The car begins it's time when it passes A
4 minutes later, it passes B.
But B has moved in 4 minutes and that is the key to the problem.
How far has B moved.
t = 4 minutes = 4/60 hours = 1/15 of an hour.
d = ?
rate = 30 km / hr
d = r * t
d = 30 km/hr * 1/15 hours = 2 km
The distance between the bikes is 5 km.
So the car has traveled 5 - 2 = 3 km
d = 3 km
r = ?
t = 4 minutes = 1/15 hour
r = d/t = 3/(1/15)= 3 / 0.066666666 = 45 km/hr.
a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J
b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J
Applying Law of Energy conservation :
K 1+U 1
=K 2+U 2
⇒K 1− r 1GmM
=K 2− r 2 GmM
where M=5.0×10 23kg,r1
=> R=3.0×10 6m and m=10kg
(a) If K 1
=5.0×10 7J and r 2
=4.0×10 6 m, then the above equation leads to
K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J
(b) In this case, we require K 2
=0 and r2
=8.0×10 6m, and solve for K 1:K 1
=K 2 +GmM (r 11− r 21)=6.9×10 7 J
Learn more about Kinetic energy on:
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