The S.T.O.P strategy is useful for situations involving negative peer pressure
1 answer:
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Answer:
a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m
Explanation:
This is an exercise in kinematics, the relationship of position and time is indicated
x = 5 t³ - 9t² -24 t - 8
a) ask when the velocity is zero
speed is defined by
v =
let's perform the derivative
v = 15 t² - 18t - 24
0 = 15 t² - 18t - 24
let's solve the quadratic equation
t =
t1 = -0.8 s
t2 = 2.0 s
the time has to be positive therefore the correct answer is t = 2.0 s
b) the position and distance traveled for a = 0
acceleration is defined by
a = dv / dt
a = 30 t - 18
a = 0
30 t = 18
t = 18/30
t = 0.6 s
we substitute this time in the expression of the position
x = 5 0.6³ - 9 0.6² - 24 0.6 - 8
x = 1.08 - 3.24 - 14.4 - 8
x = -24.56 m
we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s
the position for t = 0
x₀ = -8 m
the position for t = 0.6 s
x_f = - 24.56 m
the distance
ΔX = x_f - x₀
Δx = | -24.56 -(-8) |
Δx = 16.56 m
Answer:
= 285 Joules
Explanation:
a) answer can be found out in attachment
(b) The temperature for the isothermal compression is the same as the temp at the end of the isobaric expansion. Since pressure is held constant but volume doubles, we use the ideal gas law:
p V = nR T to see that the temperature also doubles.
.So... temp for isothermal compression = 355×2 = 710 K
.(c) The max pressure occurs at the top point. At this point, the volume is back to the original value but the temperature is twice the original value. So the pressure at this point is twice the original, or
max pressure = 2×240000 Pa = 480000 Pa = 4.80 x 10^5 Pa
(d) total work done by the piston = workdone during isothermal compression - work done during expansion =
= nRT ln(V initial / V final)-p (V initial - V final)
= nRT ln(2) - nR(T final - T initial)
= 0.250× 8.314 ×710×ln(2)-0.250×8.314× (710 - 355)
= 285 Joules
