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harkovskaia [24]

3 years ago

13

A moving walkway has a speed of .4 m/s to the east. A stationary observer sees a man walking on the walkway with a velocity of 3

.0 m/s to the east. What's the mans velocity relative to the moving walkway?

1 answer:

SVEN [57.7K]3 years ago

8 0

Explanation:

Vnet = 0.4m/s - 3.0m/s = - 2.6m/s

so the man walks 2.6m/s towards the west

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A racecar drives at a constant speed down a straight track. The car is in _?_

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Which of the following terms is given to a pair of stars that appear to change position in the sky, indicating that they are orb

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Answer:

option A

Explanation:

The correct answer is option A

The binary star system is the system in which two stars are continuously orbiting each other,

In the eclipsing binary system, two stars revolve about there center of mass and in this system one one-star eclipse another star.

Spectroscopic binary stars are found from the observation of radial velocity and the brighter member of such binary can be seen to have continuously changed the wavelength and periodic velocity.

When the pairs of stars appear to change position in the sky then it is known as visual binary.

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Calculate the gravitational potential energy of 3kg book sitting on shelf 4m high

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2 years ago

Read 2 more answers

The special theory of relativity deals with space and time under what special condition?

sergij07 [2.7K]

<h2>Answer: Invariance of the speed of light in vacuum </h2>

Special relativity was proposed on 1905 by Einstein, who developed his theory based on the following two postulates:

<em>1. The laws of physics are the same in all inertial systems. There is no preferential system. </em>

<em>2. The speed of light in vacuum has the same value for all inertial systems. </em>

<em></em>

Focusing on the first postulate, it can be affirmed that any measurement on a body is made with reference to the system in which it is being measured.

In addition, it deals with the <u>dilation of time</u> stating that <u>time passes at different rates in regions of different gravitational potential</u>. That is, the greater the local distortion of space-time due to gravity, the slower the time passes.

On the other hand, following what relativity establishes, bodies within a gravitational field follow a curved space path.

7 0

2 years ago

Graphing: Michelle climbs a tree and drops the toy car once she has reached the top. Please create an Energy-Time graph to show

Vlada [557]

Answer:

Refer to the attachment for the graph. The shape of both functions should resemble part of a parabola. Assumption: air resistance on the car is negligible.

Explanation:

The toy car started with a large amount of (gravitational) potential energy (PE) when it is at the top of the tree. Since it wasn't moving (as it was within Michelle's grip,) its kinetic energy (KE) would be equal to zero.
As the car falls to the ground, its PE converts to KE.
When the car was about to reach the ground, its PE is almost zero, while its KE is at its maximum.

<h3>PE of the car over time</h3>

The size of gravitational PE depends on both the mass and the height of the object. In this case, assume that the mass of the car stayed the same, PE should be proportional to the height of the car.

Assume that air resistance on the car is negligible. The height $h$ of the car at time $t$ could be found with the equation:

$\displaystyle h = -\frac{1}{2}\, g\, t^2 + h_0 \, (\text{Initial height})$ ,

where

$g \approx \rm 9.81\; m \cdot s^{-2}$ near the surface of the earth, and
$h_0$ is the initial height of the car.

On the other hand, $\displaystyle \text{GPE} = m \, g \, h = -\frac{m \cdot g^2}{2}\, t^2 + \underbrace{m \cdot g \cdot h_0}_{\text{Initial GPE}}$ .

In other words, plotting the gravitational PE of the car against time would give a parabola. Since $\displaystyle -\frac{m \cdot g^2}{2} < 0$ (the quadratic coefficient is smaller than zero,) the parabola should open downwards. Besides, since at $t = 0$ the initial GPE is positive, the $y$ -intercept of this parabola should also be positive.

<h3>KE of the car over time</h3>

Assume that the air resistance on the car is negligible. The mechanical energy (ME) of the toy car should conserve (stay the same.) The mechanical energy of an object is the sum of its PE and KE. The PE of the toy car has already been found as a function of time. Therefore, simply subtract the expression of PE from mechanical energy to find an expression for KE.

To find the value of mechanical energy, consider the PE of the toy car before it was dropped. Since initially KE was equal to zero, the mechanical energy of the toy car would be equal to its initial PE. That's $m \cdot g \cdot h_0$ . If there's no air resistance, the value of ME would stay at

Subtract PE from ME to obtain an expression for KE:

$\begin{aligned} \text{KE} &= \text{ME} - \text{PE} \cr &= m \cdot g \cdot h_0 - \left(-\frac{m \cdot g^2}{2}\, t^2 + m \cdot g \cdot h_0\right)\cr &= \frac{m \cdot g^2}{2}\, t^2\end{aligned}$ .

That's also a parabola when plotted against $t$ . Note that since the quadratic coefficient $\displaystyle \frac{m \cdot g^2}{2}$ is positive, the parabola shall open upwards.

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3 years ago