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Elden [556K]
2 years ago
8

Iron has a mass of 7.87 g per cubic centimeter of volume, and the mass of an iron atom is 9.27 × 10-26 kg. If you simplify and t

reat each atom as a cube, (a) what is the average volume (in cubic meters) required for each iron atom and (b) what is the distance (in meters) between the centers of adjacent atoms?
Physics
1 answer:
Alecsey [184]2 years ago
4 0
Known values; mass per / volume and mass of each atom 

average volume=<span> volume per atom. </span>
<span>density*mass per atom = volume per atom </span>

<span>1cm^3 / 7.87 gram * 9.27 *10^-26 kg /atom* 1000 gram / kg=
</span>=<span>1.178 * 10^-22 cm^3 per atom </span>

<span>Convert that to mass </span>
<span>1 cm^3 * (1 m / 100 cm)^3 </span>
 average volume= <span>1.178 *10^-29 m^3 per  atom </span>

<span>distance: </span>

<span>Volume of a cube is L^3 </span>


<span>so L = (1.178*10^-29)^(1/3) </span>

<span>Half way through the cube is distance 1/2 L. </span>
<span>Second cube is 1/2 L </span>
<span>So distance between the center of 2 cubes = L </span>

=2.28 *10^-10 m 
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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:

<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} }

3 0
3 years ago
6th grade science !!!!!!!!!!!!!!!​
posledela

Answer:

71

Explanation:

5 0
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The curve below shows the percentage of population of aquatic species that die in response to doses of pollutant A:
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Based on the trend produced by the dose - response graph, it would be best to evacuate the residents in other to prevent the increasing percentage of deaths due to the rising level of pollutant A.

  • The curve shows that the pollutant level in mg/kg of pollutant A is still increasing, hence, groundwater monitoring alone won't be the best decision to make.

  • Since the pollutant level is still increasing, then the spill level still need effective monitoring.

  • Evacuation of residents seems to be the best decision that should be taken based on the information interpreted on the graph.

Therefore, Evacuating residents to prevent rising death percentage is required as the pollutant level is yet to subside.

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8 0
2 years ago
The latent heat of fusion for Aluminium is 3.97 x 105. How much energy would be required to melt 0.75 kg of it?
RoseWind [281]

Answer:

E = 2.9775\times10^5 J

Explanation:

Given:  The latent heat of fusion for Aluminum is L = 3.97\times10^5  J/Kg

mass to be malted m = 0.75 Kg

Energy require to melt E = mL

E = 3.97\times10^5\times0.75 = 2.9775\times10^5 J

Therefore, energy required to melt 0.75 Kg aluminum

E = 2.9775\times10^5 J

5 0
2 years ago
What is the shape of the orbit of satellites?
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All objects in orbit must follow the path of an ellipse (one of Keplers laws)
8 0
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