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3 years ago

Differences Between light year and astronomical unit in two points .

1 answer:

5 0

One astronomical unit (AU) is the average distance from the Earth to the Sun, which is about 93-million miles or about 8.5 light-minutes.

A light-year is the distance light travels in one year in a vacuum, which is about 6-trillion miles. One light-year is about 64,500 times longer than one AU.

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If an object can make 10 revolutions in a minute, what's is period

lidiya [134]

Answer:

Period = 6 seconds.

Explanation:

Given the following data;

Number of oscillation, n = 10

Time = 1 minute to seconds = 60 seconds

To find the period;

A period is the number of revolutions in a minute.

$Period = \frac {time}{number \; of \; oscillations}$

Substituting into the equation, we have;

$Period = \frac {60}{10}$

Period = 6 seconds.

8 0

3 years ago

A Venturi tube may be used as the inlet to an automobile carburetor. If an inlet pipe with a diameter of 2.0 cm diameter narrows

valkas [14]

Answer:

I think it's 81

4 0

4 years ago

Read 2 more answers

A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneou

Roman55 [17]

Answer:

The solid sphere will reach the bottom first.

Explanation:

In order to develop this problem and give it a correct solution, it is necessary to collect the concepts related to energy conservation. To apply this concept, we first highlight the importance of conserving energy so we will match the final and initial energies. Once this value has been obtained, we will concentrate on finding the speed, and solving what is related to the Inertia.

In this way we know that,

$\Delta KE = - \Delta PE$

$KE_t + KE_r = mgh$

We know as well that the lineal and angular energy are given by,

$KE_r = \frac{1}{2}I\omega^2$

And the tangential kinetic energy as

$KE_t = \frac{1}{2} mv^2$

Where $\omega = \frac{v}{R}$

Replacing

$\frac{1}{2}mv^2 + \frac{1}{2}I\frac{v}{R} = mgh$

Re-arrange for v,

$v=\sqrt{\frac{2mgh}{m+I/R^2}}$

We have here three different objects: solid cylinder, hollow pipe and solid sphere. We need the moment inertia of this objects and replace in the previous equation found, then,

For hollow pipe:

$I_{hp}=mR^2$

$v_{hp}=\sqrt{\frac{2mgh}{m+(mR^2)/R^2}}$

$v_{hp}=\sqrt{\frac{2mgh}{m+m)}$

$v_{hp}=\sqrt{gh}$

For solid cylinder:

$I_{sc}=\frac{1}{2}mR^2$

$v_{sc}=\sqrt{\frac{2mgh}{m+(1/2mR^2)/R^2}}$

$v_{sc}=\sqrt{\frac{2mgh}{m+1/2m}}$

$v_{sc}=\sqrt{\frac{3}{4}gh}$

For solid sphere,

$I_{ss}=\frac{2}{5}mR^2$

$v_{ss}=\sqrt{\frac{2mgh}{m+(2/5mR^2)/R^2}}$

$v_{ss}=\sqrt{\frac{2mgh}{m+2/5m}}$

$v_{ss}=\sqrt{\frac{10}{7}gh}$

Then comparing the speed of the three objects we have:

$v_{hp}$

$\sqrt{gh}$

3 0

4 years ago

There is an electric field in the region between the two plates. The magnitude of this electric field is ed. This imposes anothe

krok68 [10]

Answer:

it is essential that the charge on the plates are of the same magnitude, but in the opposite direction

Explanation:

The configuration of parallel plates is called a capacitor and is widely used to create constant electric fields inside.

To obtain this field it is essential that the charge on the plates are of the same magnitude, but in the opposite direction

This is so that the fields created by each plate can be added inside and subtracted from the outside of the plates

5 0

3 years ago

What behavior of light proves that it is a wave? Hint: will this ONLY happen because light is a wave, or could it also be true o

kiruha [24]

It’s red shift and blue shift because This only occurs when the frequency of the wave is made longer or shorter due to the movement of the source relative to the observer

4 0

3 years ago