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2 years ago

How can a large force result in a relatively small power?

Physics

1 answer:

Sav [38]2 years ago

8 0

Answer:

hey mate here is your answer

So if an object has a very small velocity (not moving very far over time, even though a large force may be applied to it, the Power will remain small. ... Stepping on the gas, or "speeding up" the car, is applying a force which will increase velocity and increase power.

please mark me as a brainliest

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When you turn up the volume on a radio, which of the following change: velocity of sound, intensity, pitch, amplitude, frequency

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3 years ago

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A child rides her bike at a rate of 12.0 km/hr down the street. A squirrel suddenly runs in front of her so she applies the brak

Sedbober [7]

By v = u - at
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I hope this helped!

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3 years ago

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A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ .

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

3 0

2 years ago

Please help!

den301095 [7]

Hello there,

It takes 300 newtons of force and a distance of 20 meters for a moving cart to come to a stop. How much kinetic energy did this cart have?

Answer: 6000

8 0

3 years ago

Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.

Korolek [52]

Answer:

If the acceleration is constant, the movements equations are:

a(t) = A.

for the velocity we can integrate over time:

v(t) = A*t + v0

where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:

$\int\limits^{10}_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)$

Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.

4 0

3 years ago