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drek231 [11]
3 years ago
5

An observer measures the length (L), width (w), and height (h) of a box while stationary relative to the box. The observer then

travels at near light speeds parallel to the length (L) of the box. If the observer measures the width (w) of the box while moving, the measured value will now be:________
a. equal to w.
b. less than w.
c. greater than w.
d. information is needed. You need to know the density of the box.
Physics
1 answer:
Elenna [48]3 years ago
8 0

Answer:

b. less than w.

Explanation:

In this question, the application of length contraction is what helps us come to our conclusion. When an object moves very fast (relative to the observer), the length of the object seems to be smaller than it actually is (again, for the observer).

This is supported by the length contraction equation below:

L = L_0\sqrt{1-\frac{v^2}{c^2} }

Here, L is the observed length

L_0 is the original length of the object

v is the relative speed between the object and the observer

and c is the speed of light

Using this equation, we can see that as the speed between the object and the observer is increased to be close to that of light, the square root in the equation gives us values less than 1.0

This effectively decreases the length that is observed.

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cluponka [151]

Answer:

1) The distance further it takes Peter to arrive at the Coffee shop than Mia is 1.24 km

2) Mia's average speed is 6.00 km/hour

Peter's average speed is 8.48 km/hour

4) Mia's average velocity = Peter's average velocity = 6.00 km/hour

Explanation:

The given information from the diagram are;

The distance Peter jogs from school to the flower shop = 2.00 km

The distance Peter jogs from the Flower shop to the Coffee shop = 2.24 km.

The distance Mia walks from school directly to the Coffee shop = 3.00 km

The time it takes both Peter and Mia to arrive at the coffee shop = 30 minutes = 0.5 hour

1) The total distance Peter travels from school to the Coffee shop = 2.00 km + 2.24 km = 4.24 km

The distance Mia travels from school to the Coffee shop = 3.00 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 4.24 km - 3.00 km = 1.24 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 1.24 km

2) Average \ speed = \dfrac{Total \ distance \ traveled}{Total \ time \ taken \  in \ the \ journey}

Therefore, \ Mia's \ average \ speed = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Mia's average speed = 6.00 km/hour

Peter's \ average \ speed = \dfrac{4.24 \ km}{0.5 \ hour}= 8.48 \ km/hour

Peter's average speed = 8.48 km/hour

4) Average \ velocicty = \dfrac{Displacement }{Time  \ taken}

The displacement from the School to the Coffee shop is 3.00 km for both Mia and Peter

The time it takes both Peter and Mia to arrive at the Coffee shop from the school is 30 minutes = 0.5 hour

Therefore, \ Mia's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Mia's average velocity = 6.00 km/hour

Peter's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Therefore, Peter's average velocity is also = 6.00 km/hour

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Answer:

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Explanation:

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Answer:

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Explanation:

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