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3 years ago

With out inertia, how would an object that is experiencing a centripetal force behave?

Physics

1 answer:

densk [106]3 years ago

4 0

<h2>The body will move towards the center .</h2>

Explanation:

When the object moves in a circle , the centripetal force is applied towards the center .

Due to inertia of motion , equal and opposite force is developed , acting outwards from center of circle .

This force is called centrifugal force , which balances the centripetal force and the body remains in a circle .

If there is no inertia , this outward force will be zero . thus the body will move towards the center due to its one force called centripetal force .

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Infrared radiation has frequencies from 3.0×1011 to 3.0×1014 Hz, whereas the frequency region for radio wave radiation is 3.0×10

expeople1 [14]

Answer:

1. The speed of infrared radiation is greater than radio wave radiation

2. The wavelength of infrared radiation is less than the radio wave radiation

Explanation:

Given:

The radio wave radiation N₁ is 3x10⁵ to 3x10⁷ Hz and the infrared radiation N₂ is 3x10¹¹ to 3x10¹⁴ Hz. The speed is:

$\frac{velocity_{2} }{velocity_{1} } =\frac{3x10^{11} }{3x10^{5} } to\frac{3x10^{14} }{3x10^{7} } =1x10^{6} to1x10^{7}$

1. The speed of infrared radiation is greater than radio wave radiation

2. The wavelength is:

λ = C/N

where more C, less λ

Then, the wavelength of infrared radiation is less than the radio wave radiation

4 0

3 years ago

The Moon and Earth rotate about their common center of mass, which is located about 4700 km from the center of Earth. (This is 1

Nesterboy [21]

Answer:

a) $a_c=3.41x10^{-5} \frac{m}{s^2}$

b) $a_c=3.34x10^{-5}\frac{m}{s^2}$

If we analyze the two values obtained for the centripetal acceleration we see that are similar. Based on this we can say that the centripetal force would be similar to the gravitational force between the Earth and Moon.

Explanation:

1) Notation and important concepts

Centripetal acceleration is defined as "The acceleration experienced while in uniform circular motion. It always points toward the center of rotation and is perpendicular to the linear velocity."

Angular frequency is defined as "(ω), or radial frequency, measures angular displacement per unit time and the units are usually degrees (or radians) per second. "

T = 27.3 d represent the time required by the Earth around an specific point given in the problem

G= universal constant $6.673x10^{-11}\frac{Nm^2}{kg^2}$

M= represent the mass of the Moon=7.35x10^{22}kg

2) Part a

$a=\frac{GM}{r^2}$ (1)

The Earth-Moon Distance Is $3.84x10^8 km$ (average Value) and both rotates at a point located about 4700 km from the center of Earth, so the radius for this case would be the difference between these two values

$r=(3.84x10^8 km)-(4.7x10^6)=3.793x10^8 m$

Since we have the radius now we can replace into equation (1)

$a=\frac{(6.673x10^{-11}\frac{Nm^2}{kg^2})(7.35x10^22 kg)}{(3.793x10^8 m)^2}=3.41x10^{-5} \frac{m}{s^2}$

And the acceleration due to the Moon's gravity would be $3.41x10^{-5} \frac{m}{s^2}$ at the point required.

3)Part b

For this case we can find the centripetal acceleration from this formula:

$a_c =r \omega^2$ (2)

But on this case we don't have the angular frequency so we can find it with this formula

$\omega =\frac{2\pi}{T}$ (3)

But since the period is on days we need to convert that into seconds

$27.3dx\frac{86400s}{1d}=2358720sec$

Replacing the value into equation (3) we got:

$\omega =\frac{2\pi}{2358720}=2.664x10^{-6}\frac{rad}{s}$

Now we can find the centripetal acceleration with the equation (2), the new radius on this case since our reference is the Earth and the point is located 4700km=4700000m from the center of Earth then the new value for the radius would be r=4700000m

$a_c =r \omega^2 =(4700000m)(2.664x10^{-6}\frac{rad}{s})^2=3.34x10^{-5}\frac{m}{s^2}$