The formula d = t² + 5 t expresses a car's distance (in feet) from a stop sign, d, in terms of the number of seconds t since it
1 answer:
Answer:
The car's average speed is 16 feet per second.
Explanation:
The expression for the average speed is as follows;
Given expression of distance:
Put t= 4 s.
d(4)= 36 ft
Put t= 7 s in the given expression.
d(7)= 84 ft
Calculate the car's average speed over the interval of time from t = 4 seconds to t = 7 seconds.
A.S= 16 feet per second
Therefore, the car's average speed is 16 feet per second.
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- Force=12N
- Work=9J
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
