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vlada-n [284]
3 years ago
9

The Starship Enterprise speeds past an asteroid at 0.920c. If an observer on the asteroid sees 10.0 seconds pass on her watch, h

ow long would that time interval be if measured by an observer on the Enterprise?
Physics
1 answer:
lukranit [14]3 years ago
5 0

Answer:

25.52 seconds

Explanation:

Speed of space ship = 0.92 c = v

c = Speed of light

Time observed from asteroid = Δt = 10 seconds

Time dilation

\Delta t'=\frac{\Delta t}{\sqrt{1-\frac {v^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{10}{\sqrt{1-\frac {0.92^2c^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{10}{\sqrt{1-0.92^2}}\\\Rightarrow \Delta t'=25.52\ s

∴ Time interval be if measured by an observer on the Enterprise would be 25.52 seconds

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A car's velocity as a function of time is given by vx(t)=α+βt2, where α=3.00m/s and β=0.100m/s3.
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1) Analyze the equation

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That is a quadratic equation, so the graph is a parabola.

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i) vertex

ii) concavity

iii) y-intercepts

iv) x - intercepts

Also, for all graphs you need the domain and the range.

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4) Find the x-intercepts (Vₓ = 0)

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8) You can also build a table with several points in the domain

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t = 2; Vₓ = 3.00 + 0.100 (2)² = 3.40

t = 3; Vₓ = 3.00 + 0.100 (3)² = 3.90

t = 4; Vₓ = 3.00 + 0.100 (4)² = 4.60

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3 years ago
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Explanation:

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