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3 years ago

Why is light-year not related to time? (I just need scientific evidence why it isn't related to time)

Physics

1 answer:

valkas [14]3 years ago

6 0

Answer:

A light year is a measure of distance as opposed to a time measurement. This is because a light year is the distance light can travel in the time of a year.

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A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

7 0

3 years ago

What is it called when there’s an atom that has fewer neutrons than protons and more electrons than protons??l

HACTEHA [7]

Answer: The answer is an atom.

Explanation: This is because an atom has fewer neutrons than protons and more electrons than protons

6 0

3 years ago

Which of the following takes place in the combustion chamber of a gas turbine engine?A. A glow plug is used to add enough heat t

erica [24]

Answer:

Fuel oil is mixed with a proper portion of compressed air

Explanation:

A gas turbine has three main part, which are

combustion chamber
air compressor
power turbine

The combustion chamber is responsible for mixing fuel with a proper portion of compressed air.

The air compressor supplies air in sufficient quantity to satisfy the requirements of the combustion chamber

The power turbine produces the power to drive the air compressor.

6 0

3 years ago

Is a paper clip a conducted or insulator

Leni [432]

<span>So we want to know is a paper clip a conductor or an insulator. A conductor is a material that doesn't resist very much to the flow of electric current. An insulator is totally oposite of a conductor, it gives a lot of resistane to the flow of electric current. Metals are mostly conductors while rubber, plastics are insulators. Since paper clips are mostly made out of metals, they are a conductor. </span>

5 0

3 years ago

Suppose a moving car has 2000 J of kinetic energy. If the carʹs speed doubles, how much kinetic energy would it then have?

Bond [772]

Answer:

option A

Explanation:

given,

Kinetic energy of the car = 2000 J

speed of the car is doubled

we know,

$KE_1 = \dfrac{1}{2}mv^2$