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miskamm [114]
3 years ago
15

A heavy flywheel rotating on its central axis is slowing down because of friction in its bearings. At the end of the first minut

e of slowing, its angular speed is 0.70 of its initial angular speed of 125 rev/min. Assuming constant frictional forces, find its angular speed at the end of the second minute.
Physics
1 answer:
jonny [76]3 years ago
6 0

Answer:

ωf' =5.16 rad/s

Explanation:

Given that

Initial angular speed ,ωi= 125 rev/min    ( 1 rev/min = 0.10472 rad/s)

ωi=13.08 rad/s

The angular speed after 1 min ,ωf= 0.7 x 125 = 87.5 rev/min

ωf=9.1 rad/s

Lets angular acceleration =α

We know that

ωf = ωi + α t

Now by putting the values in the above equation we get

9.1 = 13.08 + α x 60             ( after 1 min ,t= 60 s)

\alpha =\dfrac{9.1-13.08}{60}

α = -0.066 rad/s²

The angular speed after 2 min

ωf' = ωi + α t

ωf' =13.08 - 0.066 x 2 x 60 rad/s

ωf' =5.16 rad/s

Therefore the angular speed after 2 min will be 5.16 rad/s.

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Answer:

B. 14.4 N

Rotational speed (Angular Velocity) = 2

The Radius of the circle = 1.2 m

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= 14.4 N

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Mega rayquaza and zygarde 100% form vs lunala, solgaleo, and red's mega charizard
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Explanation:

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3 years ago
what is the energy (in eV units) carried by one photon violet light that has a wavelength of 4.5e-7?
DaniilM [7]
The energy of a photon is given by
E=hf
where h is the Planck constant and f is the photon frequency.

We can find the photon's frequency by using the following relationship:
f= \frac{c}{\lambda}
where c is the speed of light and \lambda is the photon's wavelength. By plugging numbers into the equation, we find
f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{4.5 \cdot 10^{-7} m}=6.67 \cdot 10^{14}Hz

And so now we can find the photon energy
E=hf=(6.6 \cdot 10^{-34} Js)(6.67 \cdot 10^{14}Hz )=4.4 \cdot 10^{-19} J

We know that 1 Joule corresponds to
1 J = 1.6 \cdot 10^{-19} eV
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E= \frac{4.4 \cdot 10^{-19} J }{1.6 \cdot 10^{-19} J/eV}=2.75 eV
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4 years ago
Which of the following keys moves the insertion point to the beginning of data in a cell? (Points : 2)
Korvikt [17]
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6 0
3 years ago
Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are cha
lbvjy [14]

Answer:

Electric flux;

Φ = 30.095 × 10⁴ N.m²/C

Explanation:

We are given;

Charge on plate; q = 17 µC = 17 × 10^(-6) C

Area of the plates; A_p = 180 cm² = 180 × 10^(-4) m²

Angle between the normal of the area and electric field; θ = 4°

Radius;r = 3 cm = 3 × 10^(-2) m = 0.03 m

Permittivity of free space;ε_o = 8.85 × 10^(-12) C²/N.m²

The charge density on the plate is given by the formula;

σ = q/A_p

Thus;

σ = (17 × 10^(-6))/(180 × 10^(-4))

σ = 0.944 × 10^(-3) C/m²

Also, the electric field is given by the formula;

E = σ/ε_o

E = (0.944 × 10^(-3))/(8.85 × 10^(-12))

E = 1.067 × 10^(8) N/C

Now, the formula for electric flux for uniform electric field is given as;

Φ = EAcos θ

Where A = πr² = π × 0.03² = 9π × 10^(-4) m²

Thus;

Φ = 1.067 × 10^(8) × 9π × 10^(-4) × cos 4

Φ = 30.095 × 10⁴ N.m²/C

3 0
3 years ago
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