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2 years ago

Can you please answer these ASAP!

Physics

1 answer:

SVEN [57.7K]2 years ago

4 0

Answer:

Let's explain this briefly.

Suppose that we have a piece of ice (this is, solid water) now we give energy to the piece of ice, so the temperature of the ice increases. There is a point where the piece of ice will start a change of phase, at this point the temperature of the ice stops increasing because all the energy we give to the ice is used in the change of phase.

Once we have a complete change of phase, the temperature can increase again, and now we will have liquid water.

If we keep increasing the temperature we will see this happen again, when we have the transition from liquid to gas.

(and a similar thing happen when we have a material in a given phase and we remove heat from the material).

In the images we can see the different changes of phase of water.

1) In the first image we can see the circle in a part where the temperature is constant, so the temperature does not change in this part, which means that there is a change of phase happening.

2) Here we have the circle in a diagonal line, so here the temperature is changing, meaning that we have an increase of temperature in this region.

3) Here we want to know what the x-axis represents, this should rerpesent the energy that is being given to the material (so in some parts we see that the temperature increases and in other parts we see that the material changes of phase)

Then here the correct option is heat over time.

4) The freezing point is the temperature in which the change of phase from liquid to solid happens (or solid to liquid).

In the graph we can see that this change of phase happens at the temperature T = -210°C

Then the correct option is -210°C (The last option)

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3 years ago

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3 years ago

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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

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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