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tangare [24]
3 years ago
14

A support crossbeam on a high speed train is made from a titanium rod that has a length of 0.650m when measured by an observer o

n the train. The beam is positioned such that the observer on the train measures an angle of 39degrees between the crossbeam and the floor of the train. (Figure 1) At what speed u would the train have to be traveling in order for an observer standing outside the train to measure the angle as 83degrees ? In the figure, S indicates the reference frame of the observer standing outside the train and S' indicates the reference frame of the observer on the train.
The beam has two dimensions: a height that is perpendicular to the floor of the train and a width that is parallel to the floor of the train. The angle between the beam and the floor appears different as measured by observers on the moving train and on the platform for which of the following reasons?

A. The crossbeam is moving relative to the observer on the platform so the height appears contracted.
B. The crossbeam is moving relative to the observer on the platform so the width appears contracted.
C. The crossbeam is moving relative to the observer on the train so the width appears contracted.
D. The crossbeam is moving relative to the observer on the train so the height appears contracted.
Physics
1 answer:
Vilka [71]3 years ago
3 0

Answer:

A) The crossbeam is moving relative to the observer on the platform so the height appears contracted.

Explanation:

The observer on the train and the beam are in the same reference frame. That means observer on the train will measure the proper length of the beam not the contracted length . the observer is outside and the plank is in the moving system,it will appear to be moving.

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Explanation:

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Convert 500ml to how many liters
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1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.
NikAS [45]

Answer:

1.

109.6 cm ,  - 1.74 , real

2.

1.5

Explanation:

1.

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{40} = \frac{1}{63} + \frac{1}{d}

d = 109.6 cm

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{-109.6}{63}

m = - 1.74

The image is real

2

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}

a = 10

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{- (- (a +5))}{a}

m = \frac{(5 + 10)}{10}

m = 1.5

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. You are in an airplane flying over the ocean. How could you most accurately determine your distance from the water?
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Explanation:

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