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3 years ago

IF it possible for an object to move for 10 seconds at a high speed and end up with an average velocity of zero? true or false

2 answers:

4 0

True, if you move something forward at 100 miles an hour but your on something moving backwards 100 miles an hour you up staying in the same location, aka zero velocity.

guapka [62]3 years ago

4 0

It's true and possible. Velocity over a period of time depends on the displacement during that time. Displacement is measured between the start point and end point. If, say, you're moving along a circular track, and you stop when you reach the point where you started from, your displacement is zero, so your average velocity for the trip around the circle would be zero.

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Which is an example of projectile motion?

stiv31 [10]

A person throwing a rock

5 0

3 years ago

Read 2 more answers

11. A cyclist accelerates from 0 m/s to 10 m/s in 3 seconds. What is his acceleration ? Is this acceleration higher than that of

Marat540 [252]

$a = \frac{v - u}{t}$

v = final velocity

u = initial velocity

t = time taken

the acceleration of the cyclist is

$\frac{10 - 0}{3} = 3.333333....$

approximately 3.33 m/s^2

the acceleration of the car is

$\frac{40 - 0 }{8} = 5.0$

5.0 m/s^2

$5.0 > 3.33 \\ so \: the \: answer \: is \: no$

6 0

3 years ago

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: