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2 years ago

Describe the mechanical energy and the thermal energy of a bouncing basketball

Physics

1 answer:

miskamm [114]2 years ago

8 0

Answer:

when the ball is bounced gravitationally energy is released (more info below)

Explanation:

When a force is applied down to the ball, the gravitational potential energy is converted into kinetic energy before it hits the earth. Cinetic energy is converted into vibration, thermal, and static electrical energy, before the momentum of the ball pushes the ball back up into the hands of the people.

hope this helped!

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A one-dimensional conservative force is given by the function F(x) = (2.00 N/m) x +(1.00 N/m3) x3. What is the change in potenti

topjm [15]

Answer:

Explanation:

Given that

F(x) = (2.00 N/m) x +(1.00 N/m3) x3.

Which can be written as

F(x) = 2x +x³ N/m

Potential energy from x=1 to x=2

P.E is given as

P.E=∫F(x)dx

P.E=∫2x+x³ dx. From x=1 to x=2

P.E= 2x²/2 + x⁴/4 from x=1 to x=2

P.E= x² + x⁴/4 from x=1 to x=2

P.E=(2²+2⁴/4)-(1² +1⁴/4)

P.E=(4+4)-(1+1/4)

P.E, =8-5/4

P.E=27/4

P.E=6.75J

Since it is moving from lower position to upper position, then P.E should be negative

P.E = - 6.75J

8 0

2 years ago

A boy throws a water ballon such that it hits his sister standing 10m away. The boy threw the water balloon at an angle of 35 de

Tresset [83]

Answer: 7.734 m/s

Explanation:

We have the following data:

$\theta=35\°$ The angle at which the water ballon was thrown

$x=10 m$ The horizontal distance of the water ballon

$g=-9.8 m/s^{2}$ The acceleration due gravity

We need to find the initial velocity $V_{o}$ at which the water ballon was thrown, and we can find it by the following equation:

$x=V_{o}cos \theta T$ (1)

Where $T=2t$ is the total time the water ballon is on air

On the other hand, when we talk about parabolic motion (as in this situation) the water ballon reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time $T$ it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

$V=V_{o}+gt=0$ (2)

Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (3)

Remembering $T=2t$ :

$T=2\frac{-V_{o}}{g}$ (4)

Substituting (4) in (1):

$x=V_{o}cos \theta (2\frac{-V_{o}}{g})$ (5)

Isolating $V_{o}$ :

$V_{o}=\sqrt{\frac{x g}{-2 cos \theta}}$ (6)

$V_{o}=\sqrt{\frac{(10 m)(9.8 m/s^{2})}{-2 cos(35\°)}}$ (7)

Finally:

$V_{o}=7.734 m/s$

4 0

2 years ago

A probe is launched from earth and lands on mars the gravitational acceleration on mars is 3.7 m/s. Which statement best compare

Ugo [173]

There are no correct statements among the choices you gave us.

When the probe arrives on the surface of Mars, its mass is the same
as it was on Earth, and at every place along the way.

But its weight is (3.7/9.8) = 37.8% of what it weighed on Earth.

8 0

2 years ago

Read 2 more answers

3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag

aleksandr82 [10.1K]

Answer:

(a) $W=-19.25J$

(b) $W=-52.8J$

Explanation:

Hello.

(a) In this case, since the initial volume is 18.5 dm³ and the final volume is 21 dm³ (18.5 +2.5), we can compute the work at constant pressure as shown below:

$W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J$