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3 years ago

I need help on 54 plz

Physics

2 answers:

Nutka1998 [239]3 years ago

4 0

Acceleration = (change in speed) / (time for the change)

Change in speed = (ending speed) - (starting speed)

Change in speed = (20 m/s) - (10 m/s)

Change in speed = 10 m/s

Acceleration = (10 m/s) / (5 sec)

Acceleration = 2 m/s² forward

dalvyx [7]3 years ago

3 0

Answer:

a = 2 m/s²

Explanation:

average acceleration = change of velocity / change of time

a = Δv/Δt = (20 - 10) / 5 = 10/5 = 2

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3 years ago

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A skier leaves the horizontal end of a ramp with a velocity of 25.0 m/s and lands 70.0 m from the base of the ramp. How high is

Valentin [98]

Answer:

The end of the ramp is 38.416 m high

Explanation:

Horizontal Motion

When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.

The maximum horizontal distance traveled by the object can be calculated as follows:

$\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}$

If the maximum horizontal distance is known, we can solve the above equation for h:

$\displaystyle h=\frac {d^2g}{2v^2}$

The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:

$\displaystyle h=\frac {70^2\cdot 9.8}{2\cdot 25^2}$

$\displaystyle h=\frac {4900\cdot 9.8}{2\cdot 625}$

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2 years ago

An archer defending a castle is on a 15.5 m high wall. He shoots an arrow straight down at 22.8 m/s. How much time does it take

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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

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

Now plugging the values we get

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

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

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3 years ago

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