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3 years ago

How is a flywheel constructed to maximize its rotational inertia?

Physics

1 answer:

BigorU [14]3 years ago

6 0

Answer:

It is constructed with a high mass and a high raidus.

Explanation:

The rotational inertia I for every object is calculated as:

cMR^2 = I

where c is a constant, M is the mass of the object and R the radius of the object.

So, for a flywheel, the rotational inertia is calculated as:

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

Then, for constructed a flywheel with the maximun rotational inertia we have to set the maximum mass and the maximun radius.

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What is the temperature 32 degrees Fahrenheit in degrees Celsius? A. 0°C B. 20°C C. 10°C D. –10°C

omeli [17]

The answer would 0. The reasoning of this is because freezing point in celsius is always 0 degrees but in fahrenheit the freezing point is 32 degrees.

8 0

3 years ago

A projectile is shot on level ground with a

erastova [34]

Answer:

time rising = 34 / 9.8 = 3.47 sec

total time in air = 2 * 3.47 sec = 6.94 sec

(time rising must equal time falling)

R = 17 m/s * 6.94 s = 118 m

Can also use range formula

R = v^2 sin (2 theta) / g

tan theta = 34 / 17 = 2

theta = 63.4 deg

2 theta = 126.9 deg

sin 126.9 = .8

v^2 = 17^2 + 34^2 = 1445 m^2/s^2

R = 1445 * .8 / 9.8 = 118 m agreeing with answer found above

5 0

3 years ago

Urgente!!!!! <br> Necesito ayuda con esto!!!!

ahrayia [7]

media.discordapp.net/attachments/782414373888458783/826224189828366377/video0.mp4

5 0

3 years ago

Can anyone answer this fast pls

nekit [7.7K]

I believe the answer would be 4.5. because it wouldnt be c or d. and 2 seems too small.

3 0

3 years ago

Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p

boyakko [2]

Answer:

The number of free electrons per cubic meter is $7.61\times 10^{28}\ m^{-3}$

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, $\sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}$

Density of metal, $\rho=10.5\ g/cm^3$

Atomic weight, A = 107.87 g/mol

Let n is the number of free electrons per cubic meter such that,

$n=1.3\ N$

$n=1.3(\dfrac{\rho N_A}{A})$

Where

$\rho$ is the density of silver atom

$N_A$ is the Avogadro number

A is the atomic weight of silver

$n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})$

$n=7.61\times 10^{22}\ cm^{-3}$

$n=7.61\times 10^{28}\ m^{-3}$

Hence, this is the required solution.

6 0

3 years ago