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erica [24]
2 years ago
12

A plane starting from rest accelerates at 3m/s2 for 25s. Calculate the increase in velocity after:

Physics
1 answer:
horsena [70]2 years ago
5 0
  • Initial velocity=u=0m/s
  • Acceleration=a=3m/s^2

Time not needed now

Case-1:-

  • Time=1s
  • Final velocity=v

\\ \rm\hookrightarrow v=u+at

\\ \rm\hookrightarrow v=0+3(1)

\\ \rm\hookrightarrow v=3m/s

Case-2:-

  • Time=3s

\\ \rm\hookrightarrow v=0+3(3)

\\ \rm\hookrightarrow v=9m/s

Case-3:-

  • Time=25s

\\ \rm\hookrightarrow v=0+3(25)

\\ \rm\hookrightarrow v=75m/s

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In collision of the steel ball and the steel plate, the collision is an inelastic collision and there is loss in the kinetic energy.

<h3>What are collisions?</h3>

Collisions occur when two objects that are moving in the same directions or in different direction meet each other and collide.

There are two types of collisions:

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In the collision of the steel ball and the steel plate, there is loss in the kinetic energy of the steel ball which is converted to sound energy.

In conclusion, the collision of the steel and steel plate is an inelastic collision.

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1 year ago
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A baseball is hit that just goes over a wall that is 45.4m high. If the baseball is traveling at 46.2 m/s at an angle of 32.7° b
mario62 [17]

Answer:

54.9 m/s at 44.9 degrees

Explanation:

If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.

Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s

Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s

SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.

Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use

Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2

We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m

If we take the first derivative of the equation of position, we get the equation for speed

V(t) = Vy0 + a * t

We know that being t2 the moment the ball goes over the wall

V(t2) = -25 m/s

Y(t2) = 45.4 m

So:

45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2

-25 = Vy0 + a * t2

Then:

Vy0 = -25 - a * t2

So:

45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2

0 = -44.4 - 25 * t2 - 1/2 * a * t2^2

a = -9.81 m/s^2

0 = -44.4 - 25 * t2 + 4.9 * t2^2

Solving this quadratic equation we get:

t1 = -1.39 s

t2 = 6.5 s

Since we are looking for a positive value we disregard t1.

Now we can obtain Vy0:

Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s

Since horizontal speed is constant Vx0 = 38.9 m/s

By Pythagoras theorem we obtain the value of the initial speed:

V0 = \sqrt{Vx0^2 + Vy0^2} = \sqrt{38.9^2 + 38.76^2} = 54.9 m/s

The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90

a = atan(Vy0/Vx0) = 44.9 degrees

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3 years ago
I really need Brainliest i am desprate
kondor19780726 [428]

Hy tikki! I've asked some questions, so of you find the questions as easy, then answer it. I'll surely mark you as brainliest :)

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Solid and liquids are much more denser than gas because their molecules are close to each other and with that the molecules of them can't move that freely unlike the gas molecules. Also, because of being near to each other the molecules of solid and liquids became heavy making them dense.
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