A car has a momentum of 20,000 kg • m/s. What would the car’s momentum be if its velocity doubles?
2 answers:
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Answer:
(a). Check attachment.
(b). 280.305 J.
(c). 31.81 kpa; 38.26K.
(d). 24.05K.
(e). 24.05k; 40kpa.
(f). -138.6J.
Explanation:
(a). Kindly check the attached picture for the diagram showing the four process.
1 - 2 = adiabatic expansion process.
2 - 3 = Isochoric process.
3 - 4 = isothermal process.
4 - 1 = isochoric process.
(b). Recall that the process from 1 to is an adiabatic expansion process.
NB: b = 5/3 for a monoatomic gas.
Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].
= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.
Thus, the workdone = 280.305 J.
(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.
T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.
(d). The process 2 - 3 is an Isochoric process, then;
T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.
(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.
The pressure can be determine as below;
P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.
(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J
Answer:
Explanation:
We shall take the help of vector form of displacement . Taking east as i and north as j
4.0m N = 4 j
7.5 m E = 7.5 i
6.8 m S = - 6.8 j
3.7 m E, = 3.7 i
3.6 m S = - 3.6 j
5.3 m W = - 5.3 i
3.7 m N, = 3.7 j
5.6 m W = - 5.6 i
4.4 m S = - 4.4 j
4.9 m W = - 4.9 i
Total displacement = 4j +7.5 i -6.8j+3.7i-3.6j-5.3i+3.7j-5.6i-4.4j-4.9i
= -4.6 i -7.1 j
magnitude of displacement =
= 8.46 m
Direction
Tanθ = 7.1/ 4.6
θ = 57⁰ south of west .
distance walked = 4+7.5 +6.8+3.7+3.6+5.3+3.7+5.6+4.4+4.9
= 49.5 m