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2 years ago

14

An object moving north with an initial velocity of 14 m/s accelerates 5 m/s2 for 20 seconds. What is the final velocity of the o

bject?
39 m/s
90 m/s
114 m/s
414 m/s

2 answers:

ruslelena [56]2 years ago

6 0

Here is the answer ! Hope it helps.

lisabon 2012 [21]2 years ago

5 0

Answer:

option C

Explanation:

Final velocity of the object is 114 m / s. Hence, final velocity of the object is 114 m / s.

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Which of the following is NOT a factor in efficiency?<span><span>A.metabolism</span><span>B.type of movement</span><span>C.muscle efficiency</span><span>D.digestion</span></span>

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3 years ago

The sun is 1.5 × 108 km from Earth. The index of refraction for water is 1.349. How much longer would it take light from the sun

tankabanditka [31]

Answer:

175s

Explanation:

time it takes sunlight to reach the earth in vacuum

C=light speed=299792458m/s

X=1.5x10^8km=1.5x10^11m

c=X/t

T1=X/c

T1=1.5X10^11/299792458=500.34s

time it takes sunlight to reach the earth in water:

First we calculate the speed of light in water taking into account the refractive index

Cw=299792458m/s/1.349=222233104.5m/s

T2=1.5x10^11/222233104.5m/s=675s

additional time it would take for the light to reach the earth

ΔT=T2-T1=675-500=175s

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2 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:

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

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Answer:

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Explanation:

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3 years ago

The diagram shows Ned’s movement as he left his house and traveled to different places throughout the day. Which reference point

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The pet store would be the reference point because it is where he started and it will not move. Hope this helped.

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2 years ago

Read 2 more answers