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3 years ago

An example of objects the move in revolution

1 answer:

8 0

An example of objects that move in revolution is a ball on a string end or a planet that is moving around a star.

<u>Explanation:</u>

Rotation is the movement of an object in circular path. When the rotation's axis does not pass through the object, it is revolution. The axis of rotation may also be out of the object completely. This results in making the object to revolve around the rotation's axis. An object going in an elliptical path is revolution.It is the total time taken to complete one orbit.

Some of the examples for an object revolving are the following:

A ball on the string end.
A planet that revolves around a star. An orbit is a term that is used to describe this type of motion.

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1. True or False: All sound happens through vibration.<br> A True<br> B False

Genrish500 [490]

Answer:

True

Explanation:

Sound is all about vibrations.

The source of a sound vibrates, bumping into nearby air molecules which in turn bump into their neighbor's, and so forth.

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3 years ago

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Suppose the Earth's magnetic field at the equator has magnitude 0.00005 T and a northerly direction at all points. How fast must

sasho [114]

Answer:

Velocity will be $v=1.291\times 10^8m/sec$

Explanation:

We have given magnetic field B = 0.00005 T

Mass m = 238 U

We know that $1u=1.66\times 10^{-27}kg$

So 238 U $=238\times 1.66\times 10^{-27}=395.08\times 10^{-27}kg$

Radius $=R+1.44=6378+1.44=6379.44KM$

We know that magnetic force is given by

$F=qvB$ which is equal to the centripetal force

So $qvB=\frac{mv^2}{r}$

$1.6\times 10^{-19}\times v\times 0.00005=\frac{395.08\times 10^{-27}v^2}{6379.44}$

$v=1.291\times 10^8m/sec$

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3 years ago

The elements carbon, nitrogen, and oxygen are'll part of the same _____ on the periodic table?

andreev551 [17]

C, N and O all belong to the same period, in which it's 2nd Period.

4 0

4 years ago

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A 40kg-skier starts at the top of a 12-meter high slope. At the bottom, she is traveling 10 m/s. How much energy does she lose t

KIM [24]

Energy at top = U = mgh = 40 * 9.8 * 12 = 4704 J

Energy at bottom = 1/2 mv² = 1/2 * 40 * 10² = 4000 / 2 = 2000 J

Energy Lost = Final - Initial = 4704 - 2000 = 2704 J

In short, Your Answer would be Option D

Hope this helps!

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4 years ago

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Ablock is released from rest al height d= 40 cm and slides down a frictionless ramp and onto a first plateau, which has length d

Levart [38]

Here is the complete part of the question

A block is released from rest at height d= 40 cm and slides down a frictionless ramp and onto a first plateau, which has length d and where the coefficient of kinetic friction is 0.50. If the block is still moving, it then slides down a second frictionless ramp through height d/2 and onto a lower plateau, which has length d/2 and where the coefficient of kinetic friction is again 0.50. If the block is still moving, it then slides up a frictionless ramp until it (momentarily) stops. Where does the block stop?

Answer:

0.3 m

Explanation:

Given that :

the height h = 40 cm = 0.40 m

Coefficient of kinetic friction is $\mu$ =0.50

Using the Law of conservation of energy = $\frac{1}{2} m \mu_1^2 = mgd$

As the blocks slides down a frictionless ramp and onto a first rough plateau region.So kinetic energy is decreased to :

$\frac{1}{2}mu^2_2 = mgd - \mu_2_k mgd \\ \\ u^2_2 = 2gd - 2 \mu_kgd$

If the block is still moving, it then slides down a second frictionless ramp through an height h = d/2

Then , we can say that the gained kinetic energy is :

$\frac{1}{2} mu_2^2 = mg (\frac{d}{2})+\frac{1}{2}mu_2^2 \\ \\ \frac{1}{2} mu_2^2 = mg (\frac{d}{2})+ mgd - \mu_k mgd \\ \\ \frac{1}{2} mu_2^2 = 2g(\frac{d}{2})+2gd - 2 \mu_k gd$

Futhermore , it moves on the horizontal surface where the coefficient of friction causes some of the kinetic energy to disappear

So, the final value of kinetic energy at the end just before climbing is :

$\frac{1}{2}mv^2 = \frac{1}{2}m \mu_2 ^2 - \mu_k mg (\frac{d}{2}) \\ \\ \frac{1}{2}mv^2 = mg \frac{d}{2} + mgd - \mu_k mgd - \mu_k mg (\frac{d}{2}) \\ \\ v^2 = gd + 2gd - 2 \mu_kgd -\mu_kgd$

$= 3gd - 3 \mu_k gd \\ \\ = 3[g- \mu_kg ]d$

Let represent H to be the height above the lower plateau when it momentarily stops; From the law of conservation of energy :

$\frac{1}{2}mv^2 = mgH \\ \\ \frac{3}{2}[g-\mu_kg]d = gH \\ \\ H = \frac{3}{2}[1-\mu_k]d \\ \\ = \frac{3}{2}[1-(0.50)](0.40 \ m) \\ \\ =0.3 m$

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3 years ago