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Andrei [34K]
3 years ago
9

A child on a playground swing is swinging back and forth (one complete oscillation) once every four seconds, as seen by her fath

er standing next to the swing. At the same time, a spaceship hurtles by at a speed close to the speed of light. According to special relativity (and ignoring the Doppler effect for this question), a person on the spaceship finds that the time for one full swing is:
A - less than 4 seconds when the spaceship is approaching the swing and longer than 4 seconds when it is moving away.
B - less than 4 seconds.
C - more than 4 seconds.
D - equal to 4 seconds.
Physics
1 answer:
Sati [7]3 years ago
4 0

Answer:

A

Explanation:

Solution:-

- According to the law of relativity the relative speed between two moving objects is inversely proportional to the the time taken.

- Ignoring Doppler Effect.

- So if the relative speeds of two objects in motion i.e ( swing and spaceship) are positive then the time frame of reference for both object relative to other other decreases. So in other words if spaceship approaches the swing i.e relative velocity is positive then the time period of oscillation observed would be less than actual i.e less than 4 seconds.

- Similarly, if spaceship moves away from the swing i.e relative velocity is negative then the time period of oscillation observed would be more than actual i.e more than 4 seconds.

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Dvinal [7]

Answer:

734.16 kg m/s^{2}

Explanation:

The problem is asking for the Force of pushing off the ground.

  • The formula of Force is: F = mass x acceleration

Given = <em>Mass</em>: 600 newtons (N)

             <em>Acceleration</em>: 12 m/s^{2}

We have to convert the mass into kg first. Remember that <u>1 kg is equal to 9.80665 newtons.</u>

Let x be the<em> mass in newtons</em>.

Let's convert: \frac{1 kg}{9.80665 N} x \frac{x}{600 N} = \frac{600}{9.80665} = 61.18 kg

Phil's weight is 61.18 kg

Let's go back to finding the force.

F = m x a

F = 61.18 kg x 12 m/s^{2}

F = 734.16 kg m/s^{2}

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3 years ago
A technician attaches one lead of a digital voltmeter to the ground terminal of the TP sensor and the other meter lead to the ne
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Neither A or B

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The 37.3mv is not the signal voltage

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Calculate the magnitude of the force between two 3.60 mC point challenges 9.3 cm apart
vladimir1956 [14]

Q = magnitude of charge on each of the two point charge = 3.60 mC = 3.60 x 10⁻³ C

r = distance between the two point charges = 9.3 cm = 0.093 m

k = constant = 9 x 10⁹ Nm²/C²

F = magnitude of the force between the two point charges = ?

according to coulomb's law , force between two charges is given as

F = k Q²/r²

inserting the values

F = (9 x 10⁹) (3.60 x 10⁻³)²/(0.093)²

F = 1.35 x 10⁷ N

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Gallium iodide is also found in a dimer form Ga₂I₆.

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