Let the rise in temperature be 
The expansion in length due to change in temperature is given by the expression lαΔt , where l is the length, α is the coefficient of linear expansion, Δt is the change in temperature.
Here l = 93 m, α = 
, and Δt =
So expansion in length = 93**5 = 0.007905 m = 
So order of magnitude in change in length = -3
<span>when it returns to its original level after encountering air resistance, its kinetic energy is
decreased.
In fact, part of the energy has been dissipated due to the air resistance.
The mechanical energy of the ball as it starts the motion is:
</span>
<span>where K is the kinetic energy, and where there is no potential energy since we use the initial height of the ball as reference level.
If there is no air resistance, this total energy is conserved, therefore when the ball returns to its original height, the kinetic energy will still be 100 J. However, because of the presence of the air resistance, the total mechanical energy is not conserved, and part of the total energy of the ball has been dissipated through the air. Therefore, when the ball returns to its original level, the kinetic energy will be less than 100 J.</span>
Answer:
Explanation:
a )
Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .
v = u + a t
20 = 0 + 2 t
t = 20 /2 = 10 s .
Total time = 10 s + 20 s + 5 s = 35 s .
b) Average velocity = Total distance travelled / total time
Distance travelled in first 10 s
S₁ = ut + 1/2 a t²
= 0 + .5 x 2 x 10²
= 100 m
Distance travelled in next 20 s
S₂= 20s x 20 m/s = 400 m
Distance travelled in last 5 s .
deceleration in last 5 s
v = u + at
0 = 20 m/s + a x 5
a = - 4 m/s²
v² = u² - 2 a s
0 = (20 m/s)² - 2 x 4 m/s² x s
s = 50 m
S₃ = 50 m
Total distance = S₁ + S₂ + S₃
= 100 m + 400 m + 50 m
= 550 m .
Average velocity = 550 m / 35 s
= 15.71 m /s .
it would be C laminated soda lime glass
Answer:
change in internal energy 3.62*10^5 J kg^{-1}
change in enthalapy 5.07*10^5 J kg^{-1}
change in entropy 382.79 J kg^{-1} K^{-1}
Explanation:
adiabatic constant 
specific heat is given as 
gas constant =287 J⋅kg−1⋅K−1
specific heat at constant volume
change in internal energy 
change in enthalapy 
change in entropy
