In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its sp
1 answer:
Answer:
the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
Explanation:
Given that;
weight of vehicle = 4000 lbs
we know that 1 kg = 2.20462
so
m = 4000 / 2.20462 = 1814.37 kg
Initial velocity = 60 mph = 26.8224 m/s
Final velocity = 30 mph = 13.4112 m/s
now we determine change in kinetic energy
Δk = m(
² -
² )
we substitute
Δk = ×1814.37( (26.8224)² - (13.4112)² )
Δk = × 1814.37 × 539.5808
Δk = 489500 Joules
we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule
so
Δk = 489500 / 3.6 × 10⁶
Δk = 0.13597 ≈ 0.136 kWh
Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
You might be interested in
Answer:
The average speed of the fielder is 5.24 m/s
Explanation:
The position vector of the ball after it was hit can be calculated using the following equation:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
r = position vector at time t.
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = launching angle.
y0 = initial vertical position
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
Please, see the attached figure for a graphical description of the problem.
When the ball is caught, its position vector will be (see r1 in the figure):
r1 = (r1x, 0.873 m)
Then, using the equation of the position vector written above:
r1x = x0 + v0 · t · cos α
0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²
Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:
r1x = v0 · t · cos α
0.873 m = v0 · t · sin α + 1/2 · g · t²
Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:
0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²
0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²
Solving the quadratic equation:
t = 0.03 s and t = 5.72 s.
It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.
With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:
r1x = v0 · t · cos α
r1x = 37.2 m/s · 5.72 s · cos 49.3°
r1x = 1.39 × 10² m
The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.
The average velocity is calculated as the traveled distance over time, then:
average velocity = treveled distance / elapsed time
average velocity = 30.0 m / 5.72 s = 5.24 m/s
