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Mars2501 [29]
3 years ago
10

A car with a mass of 1240kg starts from rest and accelerates to a speed of 19.4m/s in 10.2s. Assume that the force of resistance

remains constant at 376N during this time. What is the average power developed by the car's engine?
Physics
1 answer:
garri49 [273]3 years ago
6 0

Answer:

Power, P = 38459.16 W

Explanation:

Given that,

Mass of a car, m = 1240 kg

Initial speed, u = 0

Final speed, v = 19.4 m/s

Time, t = 10.2 s

The resistance force, R = 376 N

We need to find the average power developed by the car's engine. It is given by the formula as follows :

P=F\times v\\\\P=(ma-R)v\\\\P=(\dfrac{mv}{t}-R)v\\\\P=(\dfrac{1240\times 19.4}{10.2}-376)\times 19.4\\\\P=38459.16\ W

So, the required power is 38459.16 watts.

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It has been argued that power plants should make use of off-peak hours to generate mechanical energy and store it until it is ne
sdas [7]

Answer:

Explanation:

90 rpm = 90 / 60 rps

= 1.5 rps

= 1.5 x 2π rad /s

angular velocity of flywheel

ω = 3π rad /s

Let I be the moment of inertia of flywheel

kinetic energy = (1/2) I ω²

(1/2) I ω² = 10⁷ J

I = 2 x  10⁷ / ω²

=2 x  10⁷ / (3π)²

= 2.2538 x 10⁵ kg m²

Let radius of wheel be R

I = 1/2 M R² , M is mass of flywheel

= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .

1/2 πR⁴ x t x d = 2.2538 x 10⁵

R⁴ = 2 x 2.2538 x 10⁵ / πt d

= 4.5076 x 10⁵ / 3.14 x .1 x 7800

= 184

R= 3.683 m .

diameter = 7.366 m .

b ) centripetal accn required

= ω² R

= 9π² x 3.683

= 326.816 m /s²

3 0
2 years ago
Which of the following equations illustrates the law of conservation of
Andrej [43]
I think the answer is letter B
4 0
3 years ago
You and a partner sit on the floor and stretch out a coiled spring to a length of 7.2 meters. You shake the coil so you
vekshin1

Answer:

Approximately 5.9\; {\rm m\cdot s^{-1}} (assuming that the partner is holding the other end of the coil stationary.)

Explanation:

In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.

The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.

There would be a node between:

  • the antinode at the end of the spring that is being shaken, and
  • the antinode between the two ends of this spring.

Overall, the nodes and antinodes on this spring would be:

  • node at the end that is being held still,
  • antinode (as mentioned in the question),
  • node (inferred, not mentioned in the question), and
  • antinode at the end that is being shaken.

The distance between two adjacent nodes is equal to one-half (that is, (1/2)) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is, (1/4)) of the wavelength of the wave.

Thus, if the wavelength of the wave in this question is \lambda, the length of this spring would be:

\displaystyle \frac{1}{2}\, \lambda + \frac{1}{4}\, \lambda = \frac{3}{4}\, \lambda.

The question states that the length of this coiled spring is 7.2\; {\rm m}. In other words, (3/4) \, \lambda = 7.2\; {\rm m}. The wavelength of this wave would be (7.2\; {\rm m}) / (3/4) = 9.6\; {\rm m}.

The frequency f of this wave is the number of cycles in unit time:

\begin{aligned} f &= \frac{10}{16.3\; {\rm s}} \approx 0.613\; {\rm s^{-1}}\end{aligned}.

Hence, the speed v of this wave would be:

\begin{aligned} v &= \lambda\, f \\ &=9.6\; {\rm m} \times 0.613\; {\rm s^{-1}} \\ &\approx 5.9\; {\rm m \cdot s^{-1}}\end{aligned}.

3 0
2 years ago
Melvin is traveling south on I-95 at 29 m/s (65 mph) when a deer jumps into his path, 50 m ahead. a. If his reaction time is 0.1
aleksandr82 [10.1K]

Answer:

a. 5.22 meters

b. 2.9 seconds

c. No, Melvin does not hit the deer

Explanation:

The parameters with which Melvin is travelling are as follows;

The speed of Melvin's motion, u = 29 m/s

The distance from Melvin at which the deer jumps into the path = 50 m

a. Distance, d = Velocity, u × Time, t

The time it takes Melvin to react = 0.18 seconds

The distance, "d₁" Melvin travels before his foot hits the break = The velocity with which Melvin was traveling, "u" × The time duration it takes Melvin to hit the brakes, "t₁"

∴ d₁ = 29 m/s × 0.18 s = 5.22 m

The distance, Melvin travels before his foot hits the break = d₁ = 5.22 m

b. Melvin's acceleration after his foot hits the brakes, a = -10 m/s²

Therefore, we have;

The time it takes "t₂" it takes for him to come to a complete stop given as follows;

y = u + a × t₂

Where;

v = The final velocity after Melvin comes to a complete stop = 0 m/s

By substituting the known values, we have;

0 = 29 m/s + (-10 m/s²) × t₂ = 29 m/s - 10 m/s² × t₂

∴ 29 m/s = 10 m/s² × t₂

t₂ = (29 m/s)/(10 m/s²) = 2.9 s

The time it takes it takes for him to come to a complete stop = t₂ = 2.9 s

c. The distance, "d₂", Melvin reaches while accelerating (decelerating) at -10 m/s² to come to a complete stop is given as follows;

v² = u² + 2·a·d₂

Therefore, we have;

0² = (29 m/s)² + 2 × (-10 m/s) × d₂ = (29 m/s)² - 2 × 10 m/s × d₂

∴  (29 m/s)² = 2 × 10 m/s × d₂

d₂ = ((29 m/s)²)/(2 × 10 m/s²) = (841 m²/s²)/(20 m/s²) = 42.05 m

The distance, Melvin reaches while accelerating (decelerating) at -10 m/s² to come to a complete stop = d₂ = 42.05 m

Given that d₂ = 42.05 m < 50 m (The distance separating Melvin's initial location and the deer, Melvin does not hit the deer.

3 0
2 years ago
If the plotted points on a speed-time graph do not form a straight line, what do you know about the object's acceleration?
slega [8]
Remember, that while sped is constant, acceleration is not. Acceleration is when velicity changes. So the graph which shows the slop <span>of a velocity vs time describes acceleration.
</span>If we have the straight line on the graph it means that the slope is always the same whereas the <span>non-linear graphs has a variable slope that changes depending on your point in the graph.
</span>To conclude - if your graph is not a straight line it has variable acc at many points.<span>

</span>
6 0
2 years ago
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