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Bumek [7]
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 0.68 of its initial angular speed of 260 rev/min. Assuming a constant angular acceleration, find its angular speed (rev/min) at the end of the second minute.
Physics
1 answer:
Soloha48 [4]3 years ago
6 0

Answer:

ω₂ = 93.6 rev / min

Explanation:

ω₀ = 260 rev / min

ω₁ = 0.68*ω₀ = 0.68*(260 rev / min) = 176.8 rev / min

ω₂ = ?

t₁ = 1 min

t₂ = 2 min

We can apply the equation:

ω₁ = ω₀ + α*t₁     ⇒    α = (ω₁ - ω₀) / t₁  

⇒   α = (176.8 rev / min - 260 rev / min) / 1 min = - 83.2 rev / min²

then we can use the same formula, knowing the angular acceleration:

ω₂ = ω₀ + α*t₂   ⇒   ω₂ = (260 rev / min) + (- 83.2 rev / min²)*(2 min)

⇒   ω₂ = 93.6 rev / min

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C. Why you must push harder to move a car farther.

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Mathematically, it is given by the formula;

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Hence, Newton's 2nd Law explains why you must push harder to move a car farther because of its mass. Thus, it is important to increase the force that the engine provides and decrease the mass of the car.

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Answer: moving object :)

Explanation:

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<span>Antimony I am pretty sure is one. </span>
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3 years ago
To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv
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Answer:

             E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

       E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element  and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

         λ = Q / L

If we derive from the length we have

        λ = dq/dx       ⇒    dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

        dE = k dq / x²2

        dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

        E = k \int\limits^{d+L}_d {\lambda/x^{2}} \, dx

We take out the constant magnitudes and perform the integral

        E = k λ (-1/x){(-1/x)}^{d+L} _{d}

   

Evaluating

        E = k λ [ 1/d  - 1/ (d+L)]

Using   λ = Q/L

        E = k Q/L [ 1/d  - 1/ (d+L)]

 

let's use a bit of arithmetic to simplify the expression

     [ 1/d  - 1/ (d+L)]   = L /[d(d+L)]

The final result is

     E = k Q / [d(d+L)]

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Answer:

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3 years ago
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