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Elena-2011 [213]
3 years ago
13

Listed following is a set of statements describing individual stars or characteristics of stars. Match these to the appropriate

object category.A)red giant or supergiantB)main sequence starsC)white dwarfs1. The majority of stars in our galaxy2. very cool but very luminous3. not much larger in radius than Earth4. very hot but very dim5. the Sun, for example6. found in the upper right of the H-R diagram7. a very hot and very luminous star
Physics
1 answer:
Oksana_A [137]3 years ago
6 0

Answer:

Red giant or super giant → very cool but very luminous

                                      → found in the upper right of the H-R diagram.

Main sequence →The majority of stars in our galaxy

                        → Sun, for example

                        → a very hot and very luminous star

White dwarfs → very hot but very dim

                     → not much larger in radius than earth  

Explanation:

Giant:

When the stars run out of their fuel that is hydrogen for the nuclear fusion reactions then they convert into Giant stars.That's why they are very cool. Giant stars have the larger radius and luminosity then the main sequence stars.

Main Sequence:

Stars are called main sequence stars when their core temperature reaches up to 10 million kelvin and their start the nuclear fusion reactions of hydrogen into helium in the core of the star. That is why they are very hot and luminous. For example sun is known as to be in the stage of main sequence as the nuclear fusion reactions are happening in its core.

White dwarfs:

When the stars run out of their fuel then they shed the outer layer planetary nebula, the remaining core part that left behind is called as white dwarf. It's the most dense part as the most of the mass is concentrated in this part.

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Answer:

16200 J

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t = Time the battery is on = 60 hours

I = Current = 15\times 10^{-3}\ A

Average voltage

V=\dfrac{6+4}{2}=5\ V

Energy is given by

E=V\times I\times t

\\\Rightarrow E=5\times 15\times 10^{-3}\times 60\times 3600

\\\Rightarrow E=16200\ J

The energy delivered in the given time is 16200 J

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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:

<em>a) 6738.27 J</em>

<em>b) 61.908 J</em>

<em>c)  </em>\frac{4492.18}{v_{car} ^{2} }

<em></em>

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} = <em>6738.27 J</em>

<em></em>

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} = <em>61.908 J</em>

<em></em>

<em></em>

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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