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2 years ago

Plzzzzz helppppp plzzzzzzzzzzzzzzzzzzzzzzz

Physics

2 answers:

QveST [7]2 years ago

8 0

Answer:

A. Refraction

Explanation:

Helen [10]2 years ago

6 0

Answer:

Explanation:

ap3x

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A tennis ball is hit into the air with a racket. When is the balls kinetic energy the greatest?

madam [21]

Answer:

Kinetic energy is maximum when the player hits the ball.

Explanation:

Kinetic energy $=\frac{1}{2} mv^2$ , where m is the mass and v is the velocity.

So kinetic energy is proportional to square of velocity.

Velocity is maximum when the player hits the ball.

So kinetic energy is maximum when the player hits the ball.

3 0

3 years ago

Arlecino [84]

i think the data is not complete but that's according to me

7 0

3 years ago

A catcher "gives" with a baseball when catching it. If the baseball exerts a force of 437 N on the glove, so that the glove is d

sergiy2304 [10]

437x9
is ur answer. I'm not sure tho hope it helps

5 0

4 years ago

You are walking along a small country road one foggy morning and come to an intersection. While you are crossing, you hear an am

emmainna [20.7K]

Answer:

64.945 miles per hour

Explanation:

Since the frequency of sound heard is higher than actual frequency, the ambulance is moving towards you!

The frequency of sound waves as heard from a distance for a sound wave coming towards one at v₀ m/s and whose real frequency is f₀ is given by

+f = f₀/[1 - (v₀/v)]

+f = frequency of sound as heard from the distance away = 8.61 KHz

f₀ = real frequency of sound = 7.87 KHz

v₀ = velocity at which the sound source is moving towards the reference point = ?

v = velocity of sound waves = 343 m/s

8.61 = 7.87/(1 - (v₀/v))

1 - (v₀/343) = 0.9141

v₀/343 = 1 - 0.9141 = 0.0859

v₀ = 343 × 0.0859 = 29.48 m/s = 64.945 miles per hour

7 0

3 years ago

Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at

PSYCHO15rus [73]

Answer:

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

$\mu= \frac{m_1m_2}{m_1+m_2}$

now, since both the masses have mass m

therefore,

$\mu= \frac{m^2}{2m}$

= m/2

The final K.E. of the particles is

$KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2$

Distance between two particles is d and the gravitational potential energy between them is given by

$PE_{Gravitational}= \frac{Gmm}{d}$

By law of conservation of energy we have

$KE_{initial}+KE_{final}= PE_{gravitaional}$

Now plugging the values we get

$0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}$

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

$=\sqrt{\frac{Gm}{d} }$

This the required relation between G,m and d

5 0

3 years ago