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IRINA_888 [86]
2 years ago
13

Complete the sentence below using one of the following

Physics
1 answer:
astra-53 [7]2 years ago
3 0

Answer:

equilibrium

Explanation:

acceleration will remain constant

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If you are holding a book and you let go of it, why does it fall? Explain in terms of forces, potential energy, and kinetic ener
inessss [21]

Answer:

Let the mass of the book be "m", acceleration due to gravity be "g", velocity be "v" and height be "h".

Now if we are holding a book at a certain height (h), <em><u>the potential energy will be maximum which is equal to mass× acceleration due to gravity× height (= mgh)</u>.</em>

(Remember: kinetic energy =0)

Now we consider that the book is dropped, in this case a force will act downward towards the centre of the earth, <em><u>Force= mass× acceleration due to gravity (F=mg)</u></em>. It is equal to the weight of the book.

While the book is falling, the potential energy stored in the book converts into kinetic energy and strikes the floor with <em><u>the maximum kinetic energy= (1/2)×mass×velocity² (=1/2mv²)</u>.</em>

(Remember: kinetic energy=0)

Due to this process the whole energy is conserved.

As the potential energy decreases kinetic energy increases.

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2 years ago
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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Explanation:

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The main physical characteristics of matter are mass, volume, weight, density, odor, and color. These are the characteristics that help us to see matter, feel matter, and taste matter.

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Please help<br> question #70
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can i get the question so that i can answer your question

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