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Alecsey [184]
2 years ago
12

A phonograph record 0.15 m in its radius rotates 18 times per 90 seconds what is the frequency?

Physics
1 answer:
ioda2 years ago
6 0

Answer:

The frequency of the phonograph record is 0.2 Hz

Explanation:

The frequency of an object moving in uniform circular motion is the number of completed cycles the object makes in a specified time period

The given parameters of the phonograph record are;

The radius of the record = 0.15 m

The number of times the phonograph record rotates, n = 18 times

The time it takes the phonograph record to rotate the 18 times, t = 90 seconds

The frequency of the phonograph record, f = (The number of times the phonograph record rotates) ÷ (The time it takes the phonograph record to rotate the 18 times)

∴ The frequency of the phonograph record, f = n/t = 18/(90 s) = 0.2 Hz

The frequency of the phonograph record = 0.2 Hz.

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Answer:

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Explanation:

We must solve this problem in parts, first we calculate each force and then we apply Newton's law to add the forces.

Let's use Coulomb's law to calculate each force

         F = k \frac{q_1q_2}{r_{12}^2}

particles 1 and 2

q₁ = 8.0 10⁻⁶ C, q₂ = 3.5 10⁻⁶ C x₁₂ = 0.10 m

         F₁₂ = 9 10⁹ 8.0 3.5 10⁻¹² / 0.1²

         F₁₂ = 2.59 10¹ N

Since the two charges are of the same sign, this force is repulsive and is directed towards the positive side of the x axis.

particles 2 and 3

q₂ = 3.6 10⁻⁶ C, q₃ = 2.5 10⁻⁶ C, x₂₃ = 0.15 m

we calculate

        F₂₃ = 9 10⁹ 3.5 2.5 10⁻¹²/ 0.15²

        F₂₃ = 3.5 N

as the charge is of different sign, the force is attractive, therefore it is directed to the right of the load 2

Now we add the forces as vectors

        F_total = ∑ F = F₁₂ + F₂₃

        F_total = 25.2 +3.5

        F_total = 29.4 N

directed to the right of particle 2

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A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along
storchak [24]

Answer:

v = \frac{kQ}{a}  

Explanation:

We define the linear density of charge as:

\lambda = \frac{Q}{L}

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The potential created at the center by an differential element of charge is:

dv = \frac{kdq}{r}

          where k is the coulomb's constant

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Thus.

v = \int_{}^{}\frac{kdq}{a}  

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• How much work is<br>required to lift a 2kg<br>object 2m high?<br>​
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The amount of Energy to lift an object is (mass) * (acceleration due to gravity) * (height)

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Explanation:

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Marizza181 [45]

Answer:

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Explanation:

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