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Inessa [10]
1 year ago
14

At the circus, the Human Cannonball is

Physics
1 answer:
svlad2 [7]1 year ago
6 0

The correct answer is  1.4285714.

In physics, velocity is characterised as a vector measurement of the motion's direction and speed. To be more precise, the rate of change in an object's position relative to a frame of reference and time is another way to describe velocity. The definition of velocity simply states the rate of motion of an object in a specific direction. It determines how quickly or slowly something is going.

Velocity = distance/ time  

Thus time = distance/velocity

Here velocity = 350m/s

diatnce = 500 m

time = 500/350

time = 1.42857142857

t= 200m /350m/s  = 1.4285714

To learn more about velocity refer the link:

brainly.com/question/18084516

#SPJ9

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a

Explanation:

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Lighting a match results in light and heat. Since energy was not created by the match, what transformation took place?
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Which describes how a simple machine can make work easier ?
sveta [45]
A simple machine can make work easier by reduce the amount of energy needed to perform a task, therefore, B. <span>it magnifies the potential energy so that the kinetic energy is greater</span> is the correct answer.
7 0
3 years ago
What effect does the vertical acceleration have on the horizontal velocity of the projectile?
KengaRu [80]
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None, if air resistance is ignored.

Explanation:
At any instant, the projectile has vertical and horizontal components of velocity.
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3 0
3 years ago
A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve
Gemiola [76]

Answer:

(a)   v = 5.42m/s

(b)   vo = 4.64m/s

(c)   a = 2874.28m/s^2

(d)   Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

v=\sqrt{2gh}        (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m

THe compression of the ball when it strikes the floor is 5.11*10^-3m

4 0
3 years ago
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