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3 years ago

Which is a valid inertial frame of reference?

Physics

1 answer:

kow [346]3 years ago

6 0

The correct answer to the question will be C. a hot air balloon moving at 30° east of north with no net force

EXPLANATION:

Before coming into any conclusion, first we have to understand an inertial frame of reference.

An inertial frame of reference is a frame of reference which is an unaccelerated frame. In this frame of reference, Newton's laws are valid.

A falling rock can not be considered as an inertial frame as it is accelerating. When a body falls, the forces that act on the body are gravity and air resistance. The gravity acts in vertical downward direction while air resistance acts in vertical upward direction. There is a net force in the downward direction which means it is experiencing an acceleration.

The frictionless spinning merry-go-round can not be considered as an inertial frame of reference. It is so because a rotating frame can not be considered as an inertial frame. It is non inertial in nature.

A hot air balloon moving 30 degree east of north with no net force can be considered as an inertial frame as it is not accelerating.

A space shuttle whose boosters just ignited for take off can not be also inertial frame as it is an accelerated frame.

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Answer:

104 IMA

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3 years ago

What energy comes from a rining bell

Harrizon [31]

Answer: Sound Energy

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3 years ago

A beam of light in air is incident at an angle of 30º to the surface of a rectangular block of clear plastic (n = 1.46). The lig

Aneli [31]

Answer:

θ = 30°

Explanation:

Firts, the angle when the beam of light passes through the block cam be calculated using Snell Law:

$n_{1}sin(\theta_{1}) = n_{2}sin(\theeta_{2})$

<u>Where</u>:

n₁: is the index of refraction of the incident medium (air) = 1

θ₁: is the incident angle = 30°

n₂: is the medium 2 (plastic) = 1.46

θ₂: is the transmission angle

Hence, θ₂ is:

$sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1*sin(30)}{1.46} = 0.34 \rightarrow \theta_{2} = 20.03 ^{\circ}$

Now, when the beam of light re-emerges from the opposite side, we have:

n₁: is the index of refraction of the incident medium (plastic) = 1.46

θ₁: is the incident angle = 20.03°

n₂: is the medium 2 (air) = 1

θ₂: is the transmission angle

Hence, the angle to the normal to that surface (θ₂) is:

$sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1.46*sin(20.03)}{1} = 0.50 \rightarrow \theta_{2} = 30 ^{\circ}$

Therefore, we have that the beam of light will come out at the same angle of when it went in, since, it goes from air and enters to a plastic medium and then enters again in this medium to go out to air again. This was proved using the Snell Law.

I hope it helps you!

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3 years ago

Which of the following scenarios will cause the current to stop flowing within a circuit?

statuscvo [17]

A. The switch within the closed circuit is opened.

If the switch is opened current will stop flowing, as the current will meet a gap, that is not connected by a conductor.

8 0

3 years ago

Read 2 more answers

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Nikitich [7]

Answer:

<em>The correct option is C) spreads out over a large area on the object. </em>

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