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RoseWind [281]
2 years ago
10

A rotating turntable (rt=4.50 m) is rotating at a constant rate. At the edge of the turntable is a mass (m = 3.00 kg) on the end

of a 5.00 m - long string (L). If theta = 40.0o,determine:
Physics
1 answer:
navik [9.2K]2 years ago
4 0

The magnitude of the tension in the string is 9.31 N.

<h3>Tension in the string</h3>

The magnitude of the tension in the string in the horizontal circle formed by the turn table is calculated as follows;

T = Fcos(θ)

where;

  • F is centripetal force
  • θ is inclination of the string

T = \frac{mv^2}{r} \times cos(\theta)\\\\T = \frac{3 \times 4.5^2}{5} \times cos(40)\\\\T = 9.31 \ N

Thus, the magnitude of the tension in the string is 9.31 N.

The complete question is below:

A rotating turntable (rt=4.50 m) is rotating at a constant rate. At the edge of the turntable is a mass (m = 3.00 kg) on the end of a 5.00 m - long string (L). If theta = 40.0o,determine the magnitude of the tension in the string.

Learn more about tension in horizontal circle here: brainly.com/question/12803719

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The steam boiler in a power plant depends on the fuel that it is using, but a coal-fired power plant with modern technology its efficiency is about 40%

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In order to better understand the concept of efficiency it is as if we pay 100 dollars of gasoline for our weekly use, but of that 100 dollars the car only uses 10 dollars to do that activity the rest of the money the 90 dollars were lost because of the inefficiencies of the vehicle.

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3 years ago
(b) The distance of mass from mass A if there is no gravitational force acted on C
shepuryov [24]

Answer:

(a) The force, acting on object 'C' is approximately 2.66972 × 10⁻¹⁰ Newtons

(b) The distance of 'C' from 'A', in the direction particle 'B' if there is no  meters gravitational force acting on 'C' is appromimately 0.829 meters or 1.877 meters

Explanation:

The given parameters are;

The mass of particle, A, m₁ = 2 kg

The mass of particle, B, m₂ = 0.3 kg

The mass of particle, C, m₃ = 0.05 kg

The distance between particle 'A' and particle 'B', r₁ = 0.15 m

The distance between particle 'B' and particle 'C', r₂ = 0.05 m

(a) The gravitational force, 'F', is given as follows;

F =G \times \dfrac{m_{1} \times m_{2}}{r^{2}}

Where;

F = The force between the two masses

G = The gravitation constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m₁ = The mass of object 1

m₂ = The mass of object 2

If 'C' is placed at 0.05 m from 'B', we have;

F₂₃ =  6.67430 × 10⁻¹¹ × 0.05 × 0.3/(0.05²) ≈ 4.00458 × 10⁻¹⁰

The gravitational force between force between particle 'B' and particle 'C', F₂₃ = 4.00458 × 10⁻¹⁰ N (towards the right)

F₁₃ =  6.67430 × 10⁻¹¹ × 0.05 × 2/(0.1²) ≈ × 10⁻¹⁰

The gravitational force between force between particle 'A' and particle 'B', F₁₃ = 6.6743 × 10⁻¹⁰ N (towards the left)

The force, 'F', acting on object 'C' = F₁₃ - F₂₃

F = (6.6743 - 4.00458) × 10⁻¹⁰ = 2.66972 × 10⁻¹⁰ N

The force, acting on object 'C' ≈ 2.66972 × 10⁻¹⁰ N

(b), When there is no gravitational force acting on 'C', let the distance of 'C' from 'A' = x

We have;

F₂₃ = F₁₂

F_{23} =G \times \dfrac{m_{1} \times m_{2}}{r_1^{2}} = F_{13} =G \times \dfrac{m_{1} \times m_{3}}{r_2^{2}}

By plugging in the values and removing like terms, we get;

\dfrac{0.3 \times 0.05}{(1.15 - x)^{2}}  = \dfrac{2 \times 0.05}{x^2}

(1.15 - x)² × 2 × 0.05 = 0.3 × 0.05 × x²

0.1·x² - 0.23·x + 1.3225 = 0.015·x²

0.1·x² - 0.23·x + 1.3225 - 0.015·x² = 0

0.085·x² - 0.23·x + 0.13225= 0

x = (0.23± √((-0.23)² - 4 × 0.085 × ( 0.13225)))/(2 × 0.085))

x ≈ 0.829, or x ≈ 1.877

Therefore, the distance of 'C' from 'A', if there is no gravitational force acting on 'C', x ≈ 0.829 m, or x = 1.877 m, in the direction of 'B'

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3 years ago
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