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QveST [7]
4 years ago
9

People hoping to travel to other worlds are faced with huge challenges. One of the biggest is the time required for a journey. T

he nearest star is 4.1×1016m 4.1 × 10 16 m away. Suppose you had a spacecraft that could accelerate at 1.5 g g for two thirds of a year, then continue at a constant speed. (This is far beyond what can be achieved with any known technology.)How long would it take you to reach the nearest star to earth?
Physics
1 answer:
Elena L [17]4 years ago
5 0

Answer:

It would take 8.22037 hrs away. Wouldn't it?

Explanation:

Because

    4.11016

    4.11016

            15

= 8.22037

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madreJ [45]
C. Negative force. The dog isn't going to learn that way.

6 0
3 years ago
supposed to vacuum cleaner uses 120 J of electrical energy. If 45 J used to pull air into a vacuum cleaner, how efficient is the
Ainat [17]

we know that

Power efficiency is defined as the ratio of the output power divided by the input power:

so

efficiency =\frac{ Power\  out}{Power\  in} *100

in this problem

Power\  out=45J\\ Power\  in=120J

Substitute in the formula above

efficiency =\frac{ 45}{120} *100

efficiency =37.5%

therefore

the answer is

the efficient is 37.5%

4 0
3 years ago
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For a plane mirror,how is the object s related to the distance s?
dlinn [17]
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Rays and wave fronts can generally be used to represent light when the light is interacting with objects that are much larger than the wavelength of light, which is about 500 nm. In particular, we'll use rays and wave fronts to analyze how light interacts with mirrors and lenses.

3 0
4 years ago
Which diagram BEST represents the light waves of a bright red brake light on the left side to a dim violet night light on the ri
Paraphin [41]

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6 0
3 years ago
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A baseball is thrown at an angle of 20° relative to the ground at a speed of 25 m/s if the ball was caught 50 m from the thrower
scoray [572]

Answer:

2.1 s

Explanation:

The motion of the ball is a projectile motion. We know that the horizontal range of the ball is

d = 50 m

And that the initial speed of the ball is

u=25 m/s

at an angle of

\theta=20^{\circ}

So, the horizontal speed of the ball (which is constant during the entire motion) is

u_x = u cos \theta = 25 \cdot cos 20^{\circ} = 23.5 m/s

And since the horizontal range is 50 m, the time taken for the ball to cover this distance was

t=\frac{d}{u_x}=\frac{50}{23.5}=2.1 s

which is the time the ball spent in air.

8 0
3 years ago
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