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2 years ago

When I drop a bouncy ball it bounces back up to me. What force causes it to bounce?

2 answers:

6 0

I do not think that there is a "force" that causes it to bounce. it's the rubber (or whatever material) makes up the ball.

SVEN [57.7K]2 years ago

6 0

Gravity pulls the ball down to the centre of the Earth but the ground resists the ball in the opposite direction. If there was no resistance the ball would phase through the ground. This is Newton's Second Law. However since the ball is made of rubber (just an assumption), the force of the floor resisting the ball is higher than the force of gravity, hence the ball then flies back up.

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"The White Shark" allows riders to start from rest on a tube and then slide down a 44 meter slide. It takes the rider 6.2 second

Leni [432]

<span>Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. Calculations of such is straightforward, if we are given the final velocity, the initial velocity and the total time interval. However, we are not given these values. We are only left by using the kinematic equation expressed as:

d = v0t + at^2/2

We cancel the term with v0 since it is initially at rest,

d = at^2/2
44 = a(6.2)^2/2
a = 2.3 m/s^2

</span>

6 0

2 years ago

Who is the first president of China ?

Kitty [74]

The President is Xi Jinping

6 0

3 years ago

Read 2 more answers

The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how

liraira [26]

Answer:

$x = 0.198456 \ m$

$h = 1.3061 \ m$

$v = 5.06 \ m/s$

$d = 4.0273 \ m$

Explanation:

Considering the first question

From the question we are told that

The spring constant is $k = 32.50 N/m$

The potential energy is $PE = 0.640 \ J$

Generally the potential energy stored in spring is mathematically represented as $PE = \frac{1}{2} * k * x^2$

=> $0.640= \frac{1}{2} * 32.50 * x^2$

=> $x = \sqrt{0.03938}$

=> $x = 0.198456 \ m$

Considering the second question

From the question we are told that

The mass of the dart is m = 0.050 kg

Generally from the law of energy conservation

$PE = mgh$

=> $0.640 = 0.050 * 9.8 * h$

=> $h = 1.3061 \ m$

Considering the third question

The height at which the dart was fired horizontally is $H = 3.90\ m$

Generally from the law of energy conservation

$PE = KE$

Here KE is kinetic energy of the dart which is mathematical represented as

$KE = \frac{1}{2} * mv^2$

=> $0.640 = \frac{1}{2} * 0.050 * v^2$

=> $v^2 = 25.6$

=> $v = 5.06 \ m/s$

Considering the fourth question

Generally the total time of flight of the dart is mathematically represented as

$t = \frac{ 2 * H }{g}$

=> $t = \frac{ 2 * 3.90 }{9.8 }$

=> $t = 0.7959 \ s$

Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as

$d = v * t$

=> $d = 5.06 * 0.7959$

=> $d = 4.0273 \ m$

5 0

2 years ago

Planet a and planet b are in circular orbits around a distant star. planet a is 7.8 times farther from the star than is planet

gogolik [260]

To find the ratio of planetary speeds Va/Vb we need the orbital velocity formula:

V=√({G*M}/R), where G is the gravitational constant, M is the mass of the distant star and R is the distance of the planet from the star it is orbiting.

So Va/Vb=[√( {G*M}/Ra) ] / [√( {G*M}/Rb) ], in our case Ra = 7.8*Rb

Va/Vb=[ √( {G*M}/{7.8*Rb} ) ] / [√( {G*M}/Rb )], we put everything under one square root by the rule: (√a) / (√b) = √(a/b)

Va/Vb=√ [ { (G*M)/(7.8*Rb) } / { (G*M)/(Rb) } ], when we cancel out G, M and Rb we get:

Va/Vb=√(1/7.8)/(1/1)=√(1/7.8)=0.358 so the ratio of Va/Vb = 0.358.

6 0

3 years ago

Identify the energy transformation is image below

-BARSIC- [3]

Answer:

chemical to mechanical

hope it helps have a nice day

3 0

2 years ago