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SashulF [63]
3 years ago
6

5.

Physics
1 answer:
Vera_Pavlovna [14]3 years ago
4 0
Its a 10:1 ratio so every 1 hour, the speed goes up 10 miles.
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How much power is needed to move a 2000 kg mass from the bottom of the 150 m talk great pyramid in Egypt up a ramp if the ramp h
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How Heavy? More than 2,300,000 limestone and granite blocks were pushed, pulled, and dragged into place on the Great Pyramid. The average weight of a block is about 2.3 metric tons (2.5 tons).

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A ball is thrown off the top of a building and lands on the ground below.
natita [175]

Answer:

Mass and velocity.

Explanation:

Kinetic energy <u>is the energy that an object has due to its movement</u>, mathematically it is represented as follows:

K=\frac{1}{2} mv^2

where m is the mass of the object, and v is its velocity at a given point in time.

So we can see that to find the kinetic energy just before the ball hits the gound, we need the quantities:

  • mass of the ball
  • velocity of the ball before it hits the ground

With the knowledge of these two quantities the kinetic energy of the ball  before touching the gound can be determined.

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3 years ago
The CIA investigated hypnosis as a possible tool for interrogating prisoners. Why did they decide it was unsuitable for that
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D. Hypnosis can make the subjects talk, but they talk only about their childhoods.

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7 0
3 years ago
Read 2 more answers
A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
alisha [4.7K]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

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