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ella [17]
3 years ago
7

A rocket passing the earth with high speed will look shorter for the stationary observer measuring the size on earth than travel

er in rocket. O True O False
Physics
1 answer:
neonofarm [45]3 years ago
5 0

Answer:

True

Explanation:

This can be explained by the special theory of relativity for length contraction.

Length contraction is observed in the direction of motion of an object when an object moves with speed closer to the speed of light.

The length of the rocket in this case, appears shorter to the observer on earth in the stationary reference frame which is improper frame whereas the traveler in the rocket is in the same inertial frame which is proper for the rocket's size measurement.

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If the escalator pulls you up a slope of 30.0o​ ​ and moves with a velocity of 3.10m/s. What is the vertical component of your v
ioda

Answer:

1.55\ \text{m/s}

2.68\ \text{m/s}

Explanation:

v = Velocity of the elevator = 3.1 m/s

\theta = Angle of the slope = 30^{\circ}

Vertical component is given by

v_y=v\sin\theta\\\Rightarrow v_y=3.1\sin30^{\circ}\\\Rightarrow v_y=1.55\ \text{m/s}

The vertical component of the velocity is 1.55\ \text{m/s}.

Horizontal component is given by

v_x=v\cos\theta\\\Rightarrow v_x=3.1\times \cos30^{\circ}\\\Rightarrow v_x=2.68\ \text{m/s}

The horizontal component of the velocity is 2.68\ \text{m/s}.

8 0
2 years ago
A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglect
zaharov [31]

Answer:

X=X_o+\dfrac{1}{2}gt^2

Explanation:

Given that

Length = L

At initial over hanging length = Xo

Lets take the length =X after time t

The velocity of length will become V

Now by energy conservation

\dfrac{1}{2}mV^2=mg(X-X_o)

So

V=\sqrt{2g(X-X_o)}

We know that

\dfrac{dX}{dt}=V

\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}

\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX

At t= 0 ,X=Xo

So we can say that

X=X_o+\dfrac{1}{2}gt^2

So the length of cable after time t

X=X_o+\dfrac{1}{2}gt^2

6 0
3 years ago
Ben rushin is waiting at a stoplight. when it finally turns green, ben accelerated from rest at a rate of a 6.00 m/s2 for a time
jasenka [17]

In the 4.10 seconds that elapsed, Ben reaches a velocity of

v_f=v_0+at\implies v_f=0\,\dfrac{\mathrm m}{\mathrm s}+\left(6.00\,\dfrac{\mathrm m}{\mathrm s^2}\right)(4.10\,\mathrm s)

\implies v_f=24.6\,\dfrac{\mathrm m}{\mathrm s}

In this time, his displacement \Delta x satisfies

{v_f}^2-{v_0}^2=2a\Delta x\implies\left(24.6\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(0\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(6.00\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x

\implies\Delta x=50.4\,\mathrm m

4 0
3 years ago
A uniform magnetic field B=5.210 T is perpendicular to the plane of the paper (into the page). The current-carrying wire of valu
ehidna [41]

Answer:5.21 N

Explanation:

Given

B=5.210 T

I=2 A

L=0.5 m

Given Wire is perpendicular to Magnetic field

\theta =90^{\circ}

F=IL\times B

F=BIL sin\theta

F=5.210\cdot 2\cdot 0.5 sin(90)

F=5.210 N

as  1 Tesla =1 N/A/m

4 0
3 years ago
Sound waves that enter the ear canal are directed to the ____, causing it to vibrate.
andre [41]

According to the research, the correct option the eardrum. Sound waves that enter the ear canal are directed to the <u>eardrum</u>, causing it to vibrate.

<h3>What is the eardrum?</h3>

It is the membrane found in the middle ear of vertebrate animals, separating this sector from the external auditory canal.

When sound waves enter through the external auditory canal, the eardrum vibrates, transmitting its movement to the middle ear through a series of bones and in this way, the pressure change is transformed into a mechanical movement.

Therefore, we can conclude that according to the research, the correct option is the eardrum. Sound waves that enter the ear canal are directed to the <u>eardrum</u>, causing it to vibrate.

Learn more about the eardrum here: brainly.com/question/12770491

#SPJ1

4 0
1 year ago
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