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3 years ago

10

The average speed of an object which moves 100 m in 5 seconds is?

1 answer:

CaHeK987 [17]3 years ago

3 0

Answer:

A car

Explanation:

A car can travel 100 m in 5 seconds

Hope this helps!

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How does the temperature of water affect the speed of the sound waves?

nikitadnepr [17]

Answer:

Temperature is also a condition that affects the speed of sound. Heat, like sound, is a form of kinetic energy. Molecules at higher temperatures have more energy, thus they can vibrate faster. Since the molecules vibrate faster, sound waves can travel more quickly.

5 0

3 years ago

The electric motor of a model train accelerates the train from rest to 0.540 m/s in 27.0 ms. The total mass of the train is 610

Gnesinka [82]

Answer:

The value is $P =3.294 \ W$

Explanation:

From the question we are that

The velocity $v = 0.540 \ m/s$

The time taken is $t = 27.0 ms = 27.0 *10^{-3} \ s$

The total mass of the train is $m = 610 \ g = 0.610 \ kg$

Generally the average power delivered is mathematically represented as

$P =\frac{KE }{t}$

$P =\frac{ \frac{1}{2} * m * v^2 }{t}$

=> $P =\frac{ \frac{1}{2} * 0.610 * 0.540 ^2 }{ 27.0 *10^{-3}}$

=> $P =3.294 \ W$

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2 years ago

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Which type of experiment involves changing only one variable at a time

BlackZzzverrR [31]

Controlled Experiment

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3 years ago

Read 2 more answers

What is the difference between a 4x100 and a 4x400 relay

qaws [65]

A 4x100 relay is where 4 people run 100 meters and a 4x400 relay is where 4 people run 400 meters

6 0

3 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

3 years ago