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Sophie [7]
3 years ago
9

The crew of an enemy spacecraft attempts to escape from your spacecraft by moving away from you at 0.259 of the speed of light.

but all is not lost! you launch a space torpedo toward the foe at 0.349 of the speed of light with respect to you. at what speed in kilometers per second does the enemy crew observe the torpedo approaching its spacecraft?
Physics
1 answer:
Vladimir79 [104]3 years ago
4 0
If you do not have to use relative physics but classic physics, this is how you solve it:

Speed of light = c = 3 * 10^5 km/s

Speed of your foe respect to you: 0.259c

Speed of the torpedo respect to you: 0.349c

Speed of the torpedo respect your foe: 0.349c - 0.259c = 0.09c

Conversion to km/s = 0.09 * 3.0 * 10^5 km/s = 27000 km/s

Note that this solution, using classic physics do not take into account time and space dilation.

Answer: 27000 km/s
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105 m/s

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5 0
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If the back of the truck is 1.3 m above the ground and the ramp is inclined at 22 ∘ , how much time do the workers have to get t
solong [7]
Refer to the diagram shown below.

Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.

The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m

The component of the piano's weight along the ramp is
mg sin(22°)
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ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²

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t = 1.375 s

Answer: 1.375 s

3 0
2 years ago
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