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svlad2 [7]
3 years ago
6

A golf ball bounces down a flight of steel stairs, striking several steps on the way down, but never hitting the edge of a step.

The ball starts at the top step with a vertical velocity component of zero. If all the collisions with the stairs are elastic, and if the vertical height of the staircase is 2.63 m, determine the bounce height when the ball reaches the bottom of the stairs. Neglect air resistance.
Physics
1 answer:
frez [133]3 years ago
3 0

Answer:

2.63m

Explanation:

In order to use this expression, we must calculate the values of Velocity (V0) and Velocity(Ve). The initial velocity has only a horizontal component V(0) = V(0x), the final velocity also has a horizontal component since the ball it's at a point of trajectory, V(e) = V(ef). No forces act in the horizontal direction so the momentum of the ball in this direction is conserved, hence, V(0) = V(e)

Therefore, h(e) = h(0) = 2.63m

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I’ll give brainliest!! please help and answer correctly! plsss answer quick
Rashid [163]

Answer: The motion of the object will remain the same

Explanation:

6 0
3 years ago
A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a
Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C =  50 - x cm

We know that ,electric field is given as

E=K\dfrac{q}{r^2}

K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

3 0
3 years ago
Someone please help me
Lapatulllka [165]
False... I hope that helps ;)
6 0
3 years ago
The distance recorded for riding a motorcycle on its rear wheel without stopping is more than 320 km! Suppose the rider in this
antiseptic1488 [7]

Answer:

<h3>14.97m/s</h3>

Explanation:

Given

Initial velocity of the car u = 8m/s

Distance travelled by the rider S = 40m

Acceleration a = 2m/s²

Required

rider's velocity after the acceleration v

Using the equation of motion

v² = u²+2as

v² = 8²+2(2)(40)

v² = 64+160

v² = 224

v = √224

v = 14.97m/s

Hence the rider's velocity after the acceleration is 14.97m/s

5 0
3 years ago
A force of 100N is applied to move an object a horizontal distance of 20m to the right. The work done by this force on the objec
horsena [70]
WORKDONE = FORCE * DISPLACEMENT
W=F*S
HERE, THE FORCE = 100N AND DISTANCE = 20M
WORKDONE = 100*20
WORKDONE=2000
ITS S.I UNIT IS JOULE OR J
SO, 2000J
5 0
3 years ago
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