Answer:
i believe there is something wrong in your question
Explanation:
Answer:
rise the air temperature is 0.179241 K
Explanation:
Given data
mass = 20000 kg
velocity = 18.5 m/s
long = 65 m
wide = 20 m
height = 12 m
density of the air = 1.20 kg/ m³
specific heat = 1020 J/(kg*K)
to find out
how much does the air temperature in the station rise
solution
we know here Energy lost by the train that is calculated by
loss in the kinetic energy that is = 1/2 m v²
loss in the kinetic energy = 0.5 × 20000 ×18.5²
loss in the kinetic energy is 3422500 J
and
this energy is used here to rise the air temperature that is KE / ( specific hat × mass )
so here
air volume = 65 ×20×12
air volume = 15600 m³
air mass = ρ × V = 1.2 × 15600
air mass = 18720 kg
so
rise the air temperature = 3422500 / ( 1020 × 18720)
rise the air temperature is 0.179241 K
Answer:
Half as much as it was
Explanation:
According to Charles law
PV = k
P is the pressure
V is the volume
If PV= k, then P1V1= P2V2
P1 and P2 are the initial and final pressure respectively
V1 and V2 are the initial and final volume respectively
Given
V1 =11.2litres
P1 = 0.860atm
If the volume is doubled, V2 = 2(11.2) = 22.4litres
Get P2
P2 = P1V1/V2
P2 = 0.860×11.2/22.4
P2 = 0.860/2
P2 = P1/2 (since P1 =0.860)
P2 = 1/2 P1
Hence the pressure becomes halved if the volume is doubled
Answer:
T = 0.00889 N*m
Explanation:
Given the initial speed
Vo = 9.0rev/s.
V = 65 rev/10s
V = 6.5 rev/s.
V = Vo + a*t
Solve to acceleration knowing the initial velocity and the velocity at 10 s
6.5 rev/s - 9 rev/s = a*10s
a = -0.25 rev/s^2.
Now the solve the time at stop time so V=0
V = Vo + a*t
0 = 9.0 - 0.25 rev/s *t,
t = 36 s The Stopping time.
36s - 10s = 26s
The torque can be find using the acceleration using the equation
T = I*a
I = 1/2*m*r^2
I = 1/2*0.725kg*(0.315m)^2= 0.0359kg*m^2
T = 0.0359kg*m^2*-0.25rev/s^2
T = 0.00889 N*m