Answer: Qc * (1 - Th/Tc)
Explanation:
Maximum Work done = Qh * η
Wmax = Qh * (1 - Tc/Th)
For reversible cycle of engine: Qc/Tc + Qh/Th = 0
Qh = -(Qc*Th)/Tc
Substituting Qh back into equation for work:
Wmax = -(Qc*Th)/Tc * (1 - Tc/Th)
Wmax = -(Qc*Th)/Tc + Qc
Wmax = Qc * (1 - Th/Tc)
Answer:
Explanation:
Speed of car =22.5miles/hr
U=22.5miles/hour
Applied brake and come to rest
Final velocity, =0
t, =2sec
Given that,
Speed=distance /time
Then,
Distance, =speed, ×time
Converting mile/hour to m/s
Given that
Use: 1 mile= 1600 m, 1 h= 3600s
22.5miles/hour × 1600m/mile × 1hour/3600s
Therefore, 22.5mile/hour=10m/s
Using speed =10m/s
Distance =speed ×time
Distance=10×2
Distance, =20m
The distance travelled before coming to rest is 20m.
Answer with Explanation:
We are given that
a.Length of segment,l=20 m
Magnetic force ,F=
Substitute the values
Hence, the magnetic force exert by each segment on the other=0.0119 N
b.We know that when current carrying in the wires are in same direction then the force will attract to each other.
Hence, the force will be attractive.
Answer:
0.733J/g°C
Explanation:
Using the formula
Q=mcΔθ
Q=38.5J, c=? , m=17.5g , Δθ=3°C
c= Q/(mΔθ)
c=0.733J/g°C