4 years ago

A runner runs 4875 ft in 6.85 minutes. what is the runnerâs average speed in miles per hour?

Physics

1 answer:

scoray [572]4 years ago

7 0

About 8.09 miles is the average speed mph

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An object is placed 60 cm in front of a convex lens with a focal length of 20 cm. Where does the image form?

nata0808 [166]

We are given with the focal length f equal to 20 cm and distance of the object do equal to 60 centimeters. We can compute for the distance of the image di from the lens equation
1/f = 1/do + 1/di
1/di = 1/f - 1/do = (do - f) / do f
di = do f / (do - f)
Substituting the values to the derived equation, we can now calculate for the image distance di:
di = do f / (do - f)
di = 60*20 / (60 - 20) = 30 cm
Therefore, the convex lens will produce an image at 30 centimeters.

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3 years ago

What needs to be done to balance this

navik [9.2K]

Answer:

I'm not sure, but "Add coefficient 2 to O2...". You didn't really write out the equation properly.

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3 years ago

A box is pushed 40 m by a mover. The amount of work done was 2,240 J. How much force was exerted on the box?

yawa3891 [41]

2240/40=56
56 is your answer

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4 years ago

Identify the traces on the oscilloscope screen and explain what happens to the pitch of the sound in both A and B. Mention your

jolli1 [7]

Answer:

the assessment is due at 11 you only have 1 hour do it fasst

Explanation:

the assessment is due at 11 you only have 1 hour do it fasst

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2 years ago

A futuristic train rushes past you at an incredible, relativistic speed such that it appears to only be 80% as long as you know

Fed [463]

Answer:

v = 0.6c = 1.8 x 10⁸ m/s

Explanation:

From Einstein's theory of relativity, we know that the length of an object contracts while traveling at a speed relative to speed of light. The contraction is according to following equation:

$L = L_{0}\sqrt{1 - \frac{v^2}{c^2}}$

where,

L = Relative Length

L₀ = Rest Length

According to given:

L = 0.8 L₀

c = speed of light = 3 x 10⁸ m/s

v = relative speed of train = ?

Therefore,

$0.8L_{0} = L_{0}\sqrt{1 - \frac{v^2}{c^2} } \\\\0.8 = \sqrt{1 - \frac{v^2}{c^2} }$

squaring on both sides:

$0.64 = 1-\frac{v^2}{c^2}\\\\\frac{v^2}{c^2} = 1-0.64 \\v^2= 0.36c^2\\$

taking square root on both sides:

3 0

3 years ago