A train is moving with a speed of 50 miles per hour. A man standing in the rear of a train car throws a ball toward the front of
1 answer:
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Answer:
solution:
dT/dx =T2-T1/L
&
q_x = -k*(dT/dx)
<u>Case (1) </u>
dT/dx= (-20-50)/0.35==> -280 K/m
q_x =-50*(-280)*10^3==>14 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (3)
q_x =-50*(160)*10^3==>-8 kW
T2=T1+dT/dx*L=70+160*0.25==> 110° C
Case (4)
q_x =-50*(-80)*10^3==>4 kW
T1=T2-dT/dx*L=40+80*0.25==> 60° C
Case (5)
q_x =-50*(200)*10^3==>-10 kW
T1=T2-dT/dx*L=30-200*0.25==> -20° C
note:
all graph are attached
Answer:4.34 miles
Explanation:
first Elevation =
After 1 minute Elevation changes to
Ditsance travelled in 1 minute ==10 mile
Now
tan59=
H=xtan59
tan19=
H=
Equating H
we get
1.319x=10tan19
x=2.61 miles
H==4.34 miles
Answer:
Explanation:
We have,
The surface temperature of the star is 60,000 K
It is required to find the wavelength of a star that radiated greatest amount of energy. Wein's displacement law gives the relation between wavelength and temperature such that :
Here,
= wavelength
So, the wavelength of the star is .