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worty [1.4K]
3 years ago
10

A physical pendulum consists of a uniform solid disk (of radius R 2.35 cm) supported in a vertical plane by a pivot located a di

stance d 1.75 cm from the center of the disk.The disk is dis- placed by a small angle and released. What is the period of the resulting simple harmonic motion?
Physics
1 answer:
lutik1710 [3]3 years ago
7 0

Answer:

T = 0.3658

Explanation:

The expression to use to calculate the period is the following:

T = 2π √I/Mgd (1)

Where:

I: moment of Innertia of pendulum

g: gravity acceleration (9.81 m/s²)

d: distance of the pivot

M: mass of the disk.

Before we do anything, we will find first the moment of Innertia of the pendulum.

This can be calculated with the following expression:

I = 1/2 MR² + Md² (2)

At the moment we don't have the mass of the disk, but we don't need it, we will express I in function of M, and then, it will be canceled with the M of expression (1). Calculating M we have (Remember that the units of radius and distance should be in meter):

I = 1/2 M(0.0235)² + M(0.0175)²

I = (2.76x10^-4)M + (3.06x10^-4)M

I = (5.82x10^-4)M (3)

Now, we will replace this value in equation (1):

T = 2π √(5.82x10^-4)M / (9.81)*(0.0175)M ---> Here M cancels out

T = 2π √(5.82x10^-4) / (9.81)*(0.0175)

T = 2π * 0.0582

T = 0.3658 s

This is the period of the pendulum

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How much heat is required to convert 5.53 g of ice at -12.0 ∘C to water at 24.0 ∘C? (The heat capacity of ice is 2.09 J/(g⋅∘C),
fredd [130]

Answer:

2.55 × 10³ J =2.55 kJ

Explanation:

Specific heat capacity of ice =  37.8 J / mol °C

Specific heat capacity of water = 76.0 J/ mol °C

Ice at -12 °C is converted to ice at 0 °C by absorbing heat Q₁

Ice at 0°C melts to water at 0 °C. Let Heat  absorbed during this phase change be Q₂ .

Let heat  absorbed to raise the temperature of water from 0 C to 24°C be Q₃ .

Total heat = Q = Q₁ + Q₂ + Q₃

Q₁ = (37.8 j/mol C )(5.53 g /18.01532 g/ mol )( 0-(-12)) = 139.23749 j

Q₂ =(5.53 g/18.01532 g H₂O / mol ) (6.02 x10³ j) = 1847.905 j

Q₃ = (76 j/mol C) ( (5.53 g/18.01532 g H₂O / mol )(24-0) = 559.8968 j

Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j

= 2547.039 j = 2.55 × 10³ J =2.55 kJ

5 0
4 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'

7 0
3 years ago
An ac series circuit has an impedance of 60 Ohm and
Iteru [2.4K]

Answer:

Power factor of the AC series circuit is cos\phi=0.5

Explanation:

It is given that,

Impedance of the AC series circuit, Z = 60 ohms

Resistance of the AC series circuit, R = 30 ohms

We need to find the power factor of the circuit. It is given by :

cos\phi=\dfrac{R}{Z}

cos\phi=\dfrac{30}{60}

cos\phi=\dfrac{1}{2}

cos\phi=0.5

So, the power factor of the ac series circuit is cos\phi=0.5. Hence, this is the required solution.

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Shkiper50 [21]
<span>speed = wavelength x frequency
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3 years ago
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cricket20 [7]

Answer:

it's C (this is not a guess)

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