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2 years ago

Fill in the blanks correctly to describe the transfer of energy as a ball falls down and bounces back up.

Physics

1 answer:

Vedmedyk [2.9K]2 years ago

6 0

Explanation:

Here, we have four options. We need to fill the blanks to describe the transfer of energy as a ball falls down and bounces back up. The options are as :

Gravitational potential energy
elastic potential energy
kinetic energy
thermal energy

The ball is dropped and the Gravitational potential energy is converted into kinetic energy as it falls. Gravitational potential energy is associated with the height above ground surface and as the ball comes to ground the potential energy is zero and the is converted to kinetic energy.

When the ball hits the ground and comes to a momentary stop, some of the ball's energy is converted into elastic potential energy as the ball is distorted by the impact.

The impact also transfers thermal energy to the ground. As the ball moves back upwards, kinetic energy is transformed into gravitational potential energy.

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A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = 61.908 J

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = $\frac{1}{2}mv_{car} ^{2}$

where m is the mass of the car

$v_{car}$ is the velocity of the car

Equating the energy

2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

mass of the car m = $\frac{4492.18}{v_{car} ^{2} }$

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2 years ago

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Rashid [163]

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2 years ago

Read 2 more answers

A slender uniform rod 100.00 cm long is used as a meter stick. Two parallel axes that are perpendicular to the rod are considere

Nataliya [291]

Answer:

The correct answer is D $I_{30}$ / $I_{50}$ = 1.5

Explanation:

In this exercise the moment of inertia equation should be used

I = ∫ r² dm

In addition to the parallel axis theorem

I = $I_{cm}$ + M D²

Where $I_{cm}$ is the moment of the center of mass, M is the total mass of the body and D the distance from this point to the axis of interest

Let's apply these relationships to our problem, the center of mass of a uniform rod coincides with its geometric center, in this case the rod is 1 m long, so the center of mass is in

L = 100.00 cm (1m / 100 cm) = 1.0000 m

$x_{cm}$ = 50 cm = 0.50 m

Let's calculate the moment of inertia for this point, suppose the rod is on the x-axis and use the concept of linear density

λ = M / L = dm / dx

dm = λ dx

Let's replace in the moment of inertia equation

I = ∫ x² ( λ dx)

We integrate

I = λ x³ / 3

We evaluate between the lower limits x = -L/2 to the upper limit x = L/2

I = λ/3 [(L/2)³ - (-L/2)³] = lam/3 [L³/8 - (-L³ / 8)]

I = λ/3 L³/4

I = 1/12 λ L³

Let's replace the linear density with its value

I = 1/12 (M/L) L³

I = 1/12 M L²

Let's calculate with the given values

I = 1/12 M 1²

I = 1/12 M

This point is the center of mass of the rod

Icm = I = 1/12 M = 8.333 10-2 M

Now let's use the parallel axis theorem to calculate the moment of resection of the new axis, which is 0.30 m from one end, in this case the distance is

D = $x_{cm}$ - x

D = 0.50 - 0.30

D = 0.20 m

Let's calculate

$I_{30}$ = $I_{cm}$ + M D²

$I_{30}$ = 1/12 M + M 0.202

$I_{30}$ = M (1/12 + 0.04)

$I_{30}$ = M 0.123

To find the relationship between the two moments of inertia, divide the quantities

$I_{30}$ / $I_{50}$ = M 0.123 / (M 8.3 10-2)

$I_{30}$ / $I_{50}$ = 1.48

The correct answer is d 1.5

6 0

2 years ago

if you apply a Force of F1 to area A1 on one side of a hydraulic jack, and the second side of the jack has an area that is twice

Firlakuza [10]

Answer:

$2F_{1}$

Explanation:

F₁ = Force on one side of the jack

A₁ = Area of cross-section of one side of the jack

F₂ = Force on second side of the jack

A₂ = Area of cross-section of second side of the jack = 2 A₁

Using pascal's law

$\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{A_{_{2}}}$

$\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{2A_{_{1}}}$

$F_{1}= \frac{F_{2}}{2}\\$

$F_{2}= 2F_{1}$

7 0

2 years ago

11–8 Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0

Greeley [361]

Answer:

Explanation:

position of centre of mass of door from surface of water

= 10 + 1.1 / 2

= 10.55 m

Pressure on centre of mass

atmospheric pressure + pressure due to water column

10 ⁵ + hdg

= 10⁵ + 10.55 x 1000 x 9.8

= 2.0339 x 10⁵ Pa

the net force acting on the door (normal to its surface)

= pressure at the centre x area of the door

= .9 x 1.1 x 2.0339 x 10⁵

= 2.01356 x 10⁵ N

pressure centre will be at 10.55 m below the surface.

When the car is filled with air or it is filled with water , in both the cases pressure centre will lie at the centre of the car .

7 0

2 years ago