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timofeeve [1]
3 years ago
15

How would I calculate the X from a known velocity when launching a ball off a table?

Physics
2 answers:
Anon25 [30]3 years ago
4 0
Here are the steps you would need to follow:

#1). Define what 'the 'X is, and how it's related to the ball.
#2). Be clear on how 'the X' is related to the 'known velocity'.
#3). Identify how the 'known velocity' is related to the action of the ball when it's launched.

With this information in front of you, you'll have a much better chance
of answering the question.

With none of it in front of me, I have no chance at all.
mario62 [17]3 years ago
3 0
Use the equation:

X=ut+ \frac{1}{2} at^{2}
where u is the initial velocity, a is the acceleration downwards (in this case a=g=9.81m s^{-2})

Hope this helps :)
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Ruff, the 50 cm tall Labrador Retriever stands 3m from a plane mirror and looks at his image. What is Ruffs image position and h
GaryK [48]
Ruff's image is 50m behind the mirror surface and the image is also 3m tall.

This is because it is a plane mirror.
5 0
3 years ago
A world class sprinter can accelerate at 5.0 m/s squared, from the rest in the starting blocks to their top speed in 4.0 seconds
Airida [17]

Answer:

The distance covered by the sprinter, s = 40 m

Explanation:

Given data,  

The initial velocity of the sprinter, u = 0 m/s

The acceleration of the sprinter, a = 5 m/s²

The time period of acceleration of the sprinter, t = 4 s

Using the II equations of motion

                   s = ut + ½ at²

                       = 0 + ½ (5) (4)²

                       = 40 m

Hence, the distance covered by the sprinter, s = 40 m                    

4 0
2 years ago
Which letter on the diagram below represents the retina of the eye? A) A B) B C) C D) D
Lorico [155]

Answer:

THE ANSWER IS OPTION C.

4 0
3 years ago
Read 2 more answers
In a physics experiment, two equal-mass carts roll towards each other on a level, low-friction track. One cart rolls rightward a
xz_007 [3.2K]

Answer:

The correct answer is option a.

Explanation:

Conservation of momentum :

m_1u_1+m_2u_2=m_1v_1+m_1v_2

Where :

m_1, m_2 = masses of object collided

u_1,u_2 = initial velocity before collision

v_1,v_2 = final velocity after collision

We have :

Two equal-mass carts roll towards each other.

m_1=m_2=M

Initial velocity of m_1=u_1=2 m/s

Initial velocity of m_2=u_2=-1 m/s (opposite direction)

Final velocity of m_1=v_1=v (same direction )

Final velocity of m_2=v_2=v  (same direction)

M\times 2 m/s+M(-1 m/s)=Mv+Mv

1 m/s=2v

v = 0.5 m/s

rg135

The speed of the carts after their collision is 0.5 m/s.

7 0
3 years ago
If the cd rotates clockwise at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angu
Katen [24]

The solution for this problem is:

500 revolution per minute = 8.33rev /s = 2π*8.33 rad /s = 52.36 rad /s 

Angular velocity ω = 2π N

Angular acceleration α= (ω2 - ω1) /t

ω2 = 0

α = - ω1/t = -2π N /t

N = 500 rpm = 8.33 r p s.

α = -2π 8.33 /2.6 =- 20 rad/s^2

3 0
3 years ago
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