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Over [174]
2 years ago
14

A powerful missile reaches a speed of 5 kilometers per second in 10 seconds after its launch. What is the average acceleration o

f the missile during this period?
a) 0.2 meters/second^2
b) 0.4 centimeters/second^2
c) 0.5 meters/second^2
d) 0.5 kilometers/second^2
Physics
2 answers:
Aloiza [94]2 years ago
7 0
Average acceleration<span> is the change in velocity divided by an elapsed time.  So, we can calculate it as follows:

acceleration = 5 - 0 / 10 - 0 = 0.5 kilometers/second^2

Therefore the correct answer is option D.

Hope this answers the question. Have a nice day.</span>
AveGali [126]2 years ago
4 0

Average acceleration  =  (change in speed)  /  (time for the change)

The missile's change in speed is  (5,000 - 0)  =  5,000 m/s

Average acceleration  =  (5,000 m/s)  /  (10 sec)

                                 =      500 m/s²       (about  51 Gs)

Inconveniently, this isn't one of the choices on the list. Is there something wrong, either with the choices or with my solution ?

No. Relax. Everything is OK.

500 meters is the same thing as 0.5 kilometer.  So my answer can also be written as  0.5 km/s² .  That doesn't change anything, and it IS one of the listed choices.

The average acceleration is <em>0.5 km/sec²  (d)</em> .

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Answer:

 E = k q / a²   (1.3535) (- i ^ + j ^)

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Explanation:

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        E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃

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      E₁₃ = k q / d²

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       cos 45 = E₁₃ₓ / E₁₃

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       E_{13y} = E₁₃ sin 45

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X axis

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      Eₓ = - k q / a² (1 + cos 45/2)

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      Eₓ = - k q / a²   (1 + 0.707 / 2)

      Eₓ = - k q / a²    (1.3535)

Y axis  

      E_y = E₁₄ + E_{13y}

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       E_y = k q / a²       (1.3535)

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In modulus and angle form, let's use Pythagoras' theorem for the angle

       E = √ (Eₓ² + E_y²)

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        tan θ = E_y / Eₓ

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           θ‘= 90 + 45

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