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larisa86 [58]
4 years ago
14

Find α, the angular acceleration of the wheel, which results from F⃗ pulling the string to the left. Use the standard convention

that counterclockwise angular accelerations are positive. Express the angular acceleration, α, in terms of F, r, m, and k (but not Iw).
Physics
1 answer:
kap26 [50]4 years ago
8 0

Answer:

α = F/(k×m×r)

Explanation:

When the wheel is pulled to turn in a counterclockwise direction, the wheel will have a moment of inertia given by Iw = k×m×r²

Where k = the radius of gyration of the wheel which is a dimensionless quantity less than one.

m = the mass of the wheel

r = the radius of the wheel

First and foremost, we relate the torque (τ) about the axle of the wheel to the force (F) applied on the wheel and we have that τ = r × F

We then relate the torque on the wheel to the angular acceleration (α), we have that τ = Iw × α, where Iw is the moment of inertia of the wheel as explained above

Substituting for torque τ and moment of inertia I into the above equation we have that

r × F = k×m×r² × α

solving for α we have that

α = r × F /(k×m×r²)

Therefore

α = F/(k×m×r)

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Two trains, each having a speed of 33 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h fli
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66 km

Explanation:

Given that:

The speed of the two trains = 33 km/h

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The distance apart between the two trains =  60 km

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We know that:

speed  of the train = distance traveled × time

Making the time t the subject of the formula:

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time = 30 km / 33 km/h

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Thus, the bird flying at a given speed of 60 km/h in a time of 0.909 / hr will cover a total distance of :

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distance (d) = \dfrac{60 \ km/hr}{0.909 \ /hr}

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3 years ago
What is the focal length of concave mirror that magnifies , by a factor of +3.2 , an object that is placed 30cm from the mirror?
guapka [62]

Answer:

23 cm

Explanation:

The formula for magnification is;

Magnification = image distance / object distance

use values as;

3.2 = v / 30   where v is image distance

v =30*3.2

v=96 cm

The relationship of the focal length with image distance and object distance is expressed as;

\frac{1}{u} +\frac{1}{v} =\frac{1}{f}

where f is the focal length and u is object distance

use values in the equation as;

\frac{1}{30} +\frac{1}{96 }  =\frac{1}{f}

\frac{1}{f} =\frac{7}{160}

f=160/7

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f= 23 cm ----------nearest a cm

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Answer:

The entropy change of the Universe that occurs is 19.346 J/K

Explanation:

Given;

temperature of the sun, T_s = 5,300 K

temperature of the Earth, T_E = 293 K

radiation energy transferred by the sun to the earth, E = 6000 J

The sun loses Q of heat and therefore decreases its entropy by the amount

\delta S_{sun} = \frac{-Q}{T_s}

The earth gains Q of heat and therefore increases its entropy by the amount

\delta S_{Earth} = \frac{-Q}{T_E}

The total entropy change is:

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Therefore, the entropy change of the Universe that occurs is 19.346 J/K

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3 years ago
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