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2 years ago

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The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. Which of the velocity components (n

eglecting air resistance) remains constant throughout the flight?

2 answers:

DochEvi [55]2 years ago

6 0

Answer:

Horizontal component of velocity will remains conserved which is v = 12 m/s

Explanation:

Two components of velocity of the projectile is given as

$\vec v = 20 cos53 \hat i + 20 sin53 \hat j$

$\vec v = 12\hat i + 16\hat j$

now we know that the acceleration is only due to gravity as we are considering the force is only vertically downwards due to gravity and there is no air resistance during whole motion.

So we will have acceleration as

$\vec a = 0\hat i - g\hat j$

now we can say that here we do not have any acceleration along horizontal direction so this component of velocity will remain conserved.

So the constant component of velocity is along horizontal component and it is given by v = 12 m/s

Gekata [30.6K]2 years ago

5 0

Neglecting air resistance, the horizontal component remains constant. The angle doesn't matter.

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From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

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Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy is mathematically given as

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For more information on this visit

brainly.com/question/23379286?referrer=searchResults

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