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3 years ago

If you are asked to determine an object’s speed, what information must you have?

Physics

1 answer:

o-na [289]3 years ago

3 0

Speed = Distance/Time. So you are required to know A. Distance and period of time

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Plz help me answer this is my Final Exam!!!

Ostrovityanka [42]

It's answer number four

8 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:

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

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

$q_{1} = q / 2$ .

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

- What are (a) the x component, (b) the y component, and (c) the z component of r = a - b +c if a= 7.8 + 6.6 - 7.1 , b= -2.9 + 7

frez [133]

Answer:

Explanation:

a= 7.8i + 6.6j - 7.1k

b= -2.9 i+ 7.4 j+ 3.9k , and

c = 7.6i + 8.8j + 2.2k

r = a - b +c

=7.8i + 6.6j - 7.1k - ( -2.9i + 7.4j+ 3.9k )+ ( 7.6i + 8.8j + 2.2k)

= 7.8i + 6.6j - 7.1k +2.9i - 7.4j- 3.9k )+ 7.6i + 8.8j + 2.2k

= 18.3 i +18.3 j - k

the angle between r and the positive z axis.

cosθ = 18.3 / √18.3² +18.3² +1

the angle between r and the positive z axis.

= 18.3 / 25.75

cos θ= .71

45 degree

6 0

3 years ago

The distance between earth and sun is 15000000km. Light takes 499 seconds to reach earth from sun. Calculate the speed of light

iVinArrow [24]

To solve the problem we must know about the relationship between Speed, distance, and Time.

<h3>What is the relationship between Speed, distance, and Time?</h3>

We know that sped, distance, and time all are in a relationship to each other. this relationship can be given as,

$\rm{Speed = \dfrac{Distance}{Time}$

The speed of the light is 30,060.12 km/sec.

Given to us

The distance between the earth and the sun is 15000000km
Light takes 499 seconds to reach earth from the sun.

We know that speed can be described as,

$\rm{Speed = \dfrac{Distance}{Time}$

Therefore,

<h3>What is the speed of the light?</h3>

$\text{Speed of light} = \dfrac{\text{Distance between the earth and the sun}}{\text{Time taken by the light to travel the distance}}$

Substitute the value,

$\text{Speed of light} = \dfrac{15,000,000\ km}{499\ seconds}$

$\text{Speed of light} = 30,060.12\ km/sec$

Hence, the speed of the light is 30,060.12 km/sec.

Learn more about Speed, distance, and Time:

brainly.com/question/15100898

7 0

2 years ago

Read 2 more answers

When a wave is acted upon by an external damping force what happens to the energy of the wave

Nimfa-mama [501]

Answer:

A-the energy of the wave decreases gradually

Explanation:

when a wave is acted upon by an external damping force the energy of the wave decreases gradually.

The energy degrades into the form of heat which is considered to be of less value and use. The reason is because it disperses and spreads more widely.

So therefore it end up as heat with a little sound but that is close to none because that too disperses into heat i.e. decreased form of energy.

4 0

3 years ago