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3 years ago

Starting from rest, a person runs with a constant acceleration, traveling 40 meters in 10 seconds. What is their final velocity?

Physics

1 answer:

Assoli18 [71]3 years ago

6 0

Answer:

Final velocity v = 8.944 m/sec

Explanation:

We have given distance S = 40 meters

Time t = 10 sec

As it starts from rest so initial velocity u = 0

From second equation of motion $s=ut+\frac{1}{2}at^2$

$40=0\times 10+\frac{1}{2}a10^2$

$a=0.8944m/sec^2$

Now from first equation of motion $v=u+at$ , here v is final velocity, u is initial velocity, a is acceleration and t is time

So $v=u+at=0+0.8944\times 10=8.944m/sec$

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horsena [70]

Answer:

Explanation:

refractive index of ember = sin of angle of incidence / sin of angle of refraction

= sin 35 / sin24

= .5735 / .4067

= 1.41

This is refractive index of ember with respect to water

refractive index of ember with respect to water

= wμe = μe / μw

μe = wμe x μw

= 1.33 x 1.41

= 1.87

refractive index of ember with respect to air = 1.87 .

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3 years ago

A measure of randomness or disorder

vodka [1.7K]

It is simply called Entropy.

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3 years ago

Read 2 more answers

a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien

Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

$\Sigma F=ma_{c}=m\frac{v^{2}}{R}$

Where:

m is the mass of the car
v is the speed
R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

$F_{f}=m\frac{v^{2}}{R}$

$F_{f}=2100\frac{18^{2}}{83}$

$F_{f}=8197.60 \: N$

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

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3 years ago

Learning Task No. 5 Identify the word or words being described by each statement. Choose y box below. 1. It is the process of ch

Rashid [163]

Identifying the word that is described by each statements in the question is listed below:

Evaporation
Transpiration
Hydrosphere
Water Cycle
Condensation

<h3>Meaning of Water </h3>

Water can be simply defined as any fluid or substance that is odurless, colorless and tasteless.

Water is a very important factor and substance that sustains life in humans and plants on earth.

Water tends to undergo different changes and processes in plant, animals and even as a substance on its own

In conclusion the above listed processes are all related to water.

Learn more about Water: brainly.com/question/1313076

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1 year ago

At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex

telo118 [61]

Answer:

346 * 10⁶ m

Explanation:

The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let $F_{e}$ be the force of gravity exerted by the earth

and let $F_{m}$ be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d, is given by:

$F = \frac{Gm_{1} m_{2} }{d^{2} }$

Mass of the earth, $m_{e} = 5.97 * 10^{24} kg$

Mass of the moon, $m_{m} = 7.348 * 10^{22} kg$

Mass of the satellite, $m_{s} = ?$

$F_{e} = \frac{G*5.97 * 10^{24} M }{d^{2} }$ ...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

$F_{m} = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }$ ............................(2)

Equating equations (1) and (2)

$\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }$

$(5.97 * 10^{24})(14.78 * 10^{16} -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d + 5.97 * 10^{24}d^{2} = 7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\$

Factorising out $10^{24}$

$1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0$

Solving for d in the quadratic equation above:

d = 346 * 10⁶ m

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3 years ago