Answer:
Acceleration of the car,
Explanation:
Initial sped of the car, u = 0
Distance covered by the car, d = 100 m
Time taken, t = 4.5 s
Let a is the acceleration of the car. It can be calculated using second equation of kinematics as :
So, the acceleration of the car is . Hence, this is the required solution.
Answer:
Q=1005 J
t= 0.67 sec
Explanation:
Lets take condition of room is 1 atm and 25°C.
Heat capacity ,c = 21 J /K.mol
If we assume that air is ideal gas that
P V = n R T
V= 107250 L
At STP number of moles given as
V=22.4 L at S.T.P.
n=4787.94 moles
n= 4.784 Kmoles
So heat required to raise 10°C temperature
Q = n x c x ΔT
Q = 4.78794 x 21 x 10
Q=1004.64 J
Time t
t= Q/P
P= 1.5 KW
t = 1.004.64 /1.5
t= 0.66 sec
Answer:
-40 kJ
80 kJ
Explanation:
Work is equal to the area under the pressure vs volume graph.
W = ∫ᵥ₁ᵛ² P dV
2.27) Pressure and volume are linearly related. When we graph P vs V, the area under the line is a trapezoid. So the work is:
W = ½ (P₁ + P₂) (V₂ − V₁)
W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)
W = -40 kJ
2.29) Pressure and volume are inversely proportional:
pV = k
The initial pressure and volume are 500 kPa and 0.1 m³. So the constant is:
(500) (0.1) = k
k = 50
The final pressure is 100 kPa. So the final volume is:
(100) V = 50
V = 0.5
The work is therefore:
W = ∫ᵥ₁ᵛ² P dV
W = ∫₀₁⁰⁵ (50/V) dV
W = 50 ln(V) |₀₁⁰⁵
W = 50 (ln 0.5 − ln 0.1)
W ≈ 80 kJ
