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Artemon [7]
2 years ago
9

While watching a recent science fiction movie, one Klingon spaceship blows up a Droid spaceship with a laser gun. The Klingon cr

ew watches out the port window and covers their ears to muffle the noise from the explosion. Scientifically, this scene is inaccurate because A) light can not reflect off objects in the vacuum of space. B) they would not be able to see anything outside the window. Eliminate C) lasers can not be transmitted through the vacuum of space. D) the sound of the explosion would not be transmitted back to their ship.
Physics
2 answers:
svetlana [45]2 years ago
8 0
The answer is D.  Sound cannot travel in vacuums. Hope this helps homie
elena-14-01-66 [18.8K]2 years ago
3 0
D is the correct answer
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Which of the following is not true about centripetal force? Check all that
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A, B Y E

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acid and bases

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Kepler’s third law can be used to derive the relation between the orbital period, P (measured in days), and the semimajor axis,
NikAS [45]
Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant

Given:
P = 687 days
A = 1.52 AU

Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³

Answer:  1.3439 x 10⁵ (days²/AU³)

8 0
2 years ago
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A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve
Gemiola [76]

Answer:

(a)   v = 5.42m/s

(b)   vo = 4.64m/s

(c)   a = 2874.28m/s^2

(d)   Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

v=\sqrt{2gh}        (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m

THe compression of the ball when it strikes the floor is 5.11*10^-3m

4 0
3 years ago
An astronaut is moving in space when a big explosion occurs about 50 metres BEHIND him. How will the astronaut come to know abou
kakasveta [241]
The matter from the explosion can reach him, hitting him. He should be able to feel that.
6 0
2 years ago
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