An L-R-C series circuit is connected to a 120 Hz ac source that has Vrms = 82.0 V. The circuit has a resistance of 71.0 Ω and an impedance at this frequency of 107 Ω and an impedance at this frequency of 105Ω. What average power is delivered to the circuit by the source?

The researchers need to compare those who contracted the disease to those who did not.

Answer:

Explanation:

The moment of inertia of the disk I  = 1/2 m R² where R is radius of the disc and m is its mass .

putting the values

I = .5 x 95 x .38²

= 6.86 kg m²

n = 87 rpm = 87 / 60 rps

n = 1.45 rps

angular velocity ω = 2π n , n is frequency of rotation .

= 2 x 3.14 x 1.45

= 9.106 radian /s

frictional force = 16 x .2

= 3.2 N

torque created by frictional force = 3.2 x .38

= 1.216 N.m

angular acceleration = torque / moment of inertia

= - 3.2 / 6.86

α  = - 0.4665 rad /s²

b ) ω² = ω₀² +  2 α θ , where α is angular acceleration

0 = 9.106² - 2 x .4665 θ

θ = 88.87 radian

no of turns = 88.87 / 2π

= 14.15  turns

Answer:

A. The sound wave will reflect off Buildings and automobiles.

Explanation:

This is because the sound waves would more likely propagate through diffraction through buildings and transmission through the air. It is also more likely to be absorbed by buildings than for multiple reflections to occur off buildings and automobiles. In the process of reflection, these materials would absorb the sound energy thereby reducing its ability to reflect.

Answer:

 I = 7.96 W / m²

Explanation:

The light bulb emits a power of P = 100W, this power is distributed over the surface of a sphere, thus the emission is in all directions.

Intensity is defined by power per unit area

            I = P / A

The area of ​​a sphere is

         A = 4π r²

we substitute

         I = P / (4π r²)

in this case it tells us that the distance is r = 1 m

let's calculate

        I = 100 / (4π 1²)

        I = 7.96 W / m²

Answer:

Explanation:

We shall apply conservation of momentum law in vector form to solve the problem .

Initial momentum = 0

momentum of 12 g piece

= .012 x 37 i since it moves along x axis .

= .444 i

momentum of 22 g

= .022 x 34 j

= .748 j

Let momentum of third piece = p

total momentum

= p + .444 i + .748 j

so

applying conservation law of momentum

p + .444 i + .748 j  = 0

p = - .444 i -  .748 j  

magnitude of p

= √ ( .444² + .748² )

= .87 kg m /s

mass of third piece = 58 - ( 12 + 22 )

= 24 g = .024 kg

if v be its velocity

.024 v = .87

v = 36.25 m / s .

Answer:

No, a rigid body cannot experience any acceleration when the resultant force acting on the body is zero.

Explanation:

If the net force on a body is zero, then it means that all the forces acting on the body are balanced and cancel out one another. This sate of equilibrium can be static equilibrium (like that of a rigid body), or dynamic equilibrium (that of a car moving with constant velocity)

For a body under this type of equilibrium,

ΣF = 0   ...1

where ΣF is the resultant force (total effective force due to all the forces acting on the body)

For a body to accelerate, there must be a force acting on it. The acceleration of a body is proportional to the force applied, for a constant mass of the body. The relationship between the net force and mass is given as

ΣF = ma   ...2

where m is the mass of the body

a is the acceleration of the body

Substituting equation 2 into equation 1, we have

0 = ma

therefore,

a = 0

this means that if the resultant force acting on a rigid body is zero, then there won't be any force available to produce acceleration on the body.

Answer:

His angular velocity will increase.

Explanation:

According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.

The angular momentum of a system = [tex]I[/tex]'ω'

where

[tex]I[/tex]' is the initial rotational inertia

ω' is the initial angular velocity

the rotational inertia = [tex]mr'^{2}[/tex]

where m is the mass of the system

and r' is the initial radius of rotation

Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.

we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to [tex]I[/tex].

From

[tex]I[/tex]'ω' = [tex]I[/tex]ω

since [tex]I[/tex] is now reduced, ω will be greater than ω'

therefore, the angular velocity increases.

Answer:

The  charge on the dust particle is  [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Explanation:

From the question we are told that

    The length is  [tex]l = 2.0 \ m[/tex]

    The width is  [tex]w = 4.0 \ m[/tex]

   The charge is  [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]

    The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]

Generally the electric field on the carpet is mathematically represented as

           [tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]

Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

           [tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]

           [tex]E = -70621.5 \ N/C[/tex]

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        [tex]F__{E}} = F__{G}}[/tex]

=>     [tex]q_d * E = m * g[/tex]

=>      [tex]q_d = \frac{m * g}{E}[/tex]

=>      [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]

=>     [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Answer

76200meters

Explanation:

we know that 1km=1000meters

to convert km into meters we we divide km by meters

=76.2/1000

=76200meters

Answer:

I. light house 1 will be warmer during the day ii. light house 2 will be warmer at night.

Explanation:

Because the land conducts heat better than water the light house farthest away from the water will get hotter during as the ground will heat up faster than the water. But this also means that the ground will lose heat faster at night where the water won't making the light house closest to the water hotter at night.

Answer:

Explanation:

Using the Continuity equation

v X A = v' xA'

so if A is 1/2of A' then A velocity must be 2 times the A'

after-contraction v = 2 x 5.0m/s = 10m/s

Using the Bernoulli equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

, the "h" terms cancel

3.5 x 10^ 5Pa + ½ x 1000kg/m³x (5.0m/s)² = p₂ + ½ x 1000kg/m³ x (10m/s)²

p₂ = 342500pa

Answer:

a)   Δt = 24.96 s , b)  τ = 0.078 N m

Explanation:

This is a rotational kinematics exercise

        θ = w₀ t - ½ α t²

Let's reduce the magnitudes the SI system

       θ = 60 rev (2π rad / 1 rev) = 376.99 rad

       w₀ = 7.0 rot / s (2π rad / 1 rpt) = 43.98 rad / s

      α = (w₀ t - θ) 2 / t²

let's calculate the annular acceleration

      α = (43.98 10 - 376.99) 2/10²

      α = 1,258 rad / s²

Let's find the time it takes to reach zero angular velocity (w = 0)

        w = w₀ - alf t

         t = (w₀ - 0) / α

         t = 43.98 / 1.258

         t = 34.96 s

this is the total time, the time remaining is

         Δt = t-10

         Δt = 24.96 s

To find the braking torque, we use Newton's law for angular motion

        τ = I α

the moment of inertia of a circular ring is

       I = M r²

we substitute

         τ = M r² α

we calculate

        τ = 0.625  0.315²  1.258

        τ = 0.078 N m

The total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.

Given data:

The initial angular speed of wheel is, [tex]\omega = 7.0 \;\rm rps[/tex]   (rps means rotation per second).

The time interval is, t' = 10.0 s.

The number of rotations made by wheel is, n = 60.0.

The mass of bike wheel is, m = 0.625 kg.

The radius of wheel is, r = 0.315 m.

The problem is based on rotational kinematics. So, apply the second rotational equation of motion as,

[tex]\theta = \omega t-\dfrac{1}{2} \alpha t'^{2}[/tex]

Here, [tex]\theta[/tex] is the angular displacement, and its value is,

[tex]\theta =2\pi \times 60\\\\\theta = 376.99 \;\rm rad[/tex]

And, angular speed is,

[tex]\omega = 2\pi n\\\omega = 2\pi \times 7\\\omega = 43.98 \;\rm rad/s[/tex]

Solving as,

[tex]376.99 = 43.98 \times 10-\dfrac{1}{2} \alpha \times 10^{2}\\\\\alpha = 1.25 \;\rm rad/s^{2}[/tex]

Apply the first rotational equation of motion to obtain the value of time to reach zero final velocity.

[tex]\omega' = \omega - \alpha t\\\\0 = 43.98 - 1.25 \times t\\\\t = 35.18 \;\rm s[/tex]

Then total time is,

T = t - t'

T = 35.18 - 10

T = 25.18 s

Now, use the standard formula to obtain the value of braking torque as,

[tex]T = m r^{2} \alpha\\\\T = 0.625 \times (0.315)^{2} \times 1.25\\\\T = 0.0775 \;\rm Nm[/tex]

Thus, we can conclude that the total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.

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Answer:

12.2 m

Explanation:

Given:

v₀ = 15.6 m/s

v = 0 m/s

a = -10 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (15.6 m/s)² + 2 (-10 m/s²) Δy

Δy = 12.2 m

[tex] \LARGE{ \boxed{ \rm{ \green{Answer:}}}}[/tex]

Given,

Finding the initial kinetic energy,

[tex]\large{ \boxed{ \rm{K.E. = \frac{1}{2}m {v}^{2}}}}[/tex]

⇛ KE = (1/2)mv²

⇛ KE = (1/2)(0.042)(15.6)²

⇛ KE = 5.11 J

|| ⚡By conservation of energy, the potential energy at the highest point will also be 5.11 J, since there is no kinetic energy at the highest point because the ball is not moving (we neglect energy lost due to air resistance, heat, sound, etc.) ⚡||

So, we have:

[tex] \large{ \boxed{ \rm{P.E. = mgh}}}[/tex]

⇛ h = PE/(mg)

⇛ h = 5.11 J /(0.042 × 9.8)

⇛ h = 12.41 m

✏The ball will rise upto a height of 12.41 m

━━━━━━━━━━━━━━━━━━━━

Answer:

Apparent depth (Da) = 60.15 cm (Approx)

Explanation:

Given:

Distance from fish (D) = 80 cm

Find:

Apparent depth (Da)

Computation:

We know that,

Refractive index of water (n2) = 1.33

So,

Apparent depth (Da) = D(n1/n2)

Apparent depth (Da) = 80 (1/1.33)

Apparent depth (Da) = 60.15 cm (Approx)

The apparent depth of the fish is 60 cm.

To calculate the apparent depth of the fish, we use the formula below.

Formula:

Where:

Make D' The subject of the equation.

From the question,

Given:

Substitute these values into equation 2

Hence, the apparent depth of the fish is 60 cm

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Answer:

386.13 N

Explanation:

The kinetic energy of the baseball is converted into workdone in moving the glove backward( work energy theorem).

Therefore, KE of the ball

[tex]\frac{1}{2} mv^2 =\frac{1}{2}(0.145)35^2\\ = 88.81 \text{J}[/tex]

Now, workdone in moving the glove

W= Fd

where F = Force applied, d = displacement of the glove= 0.23 cm.

88.81 = F×0.23

F= 88.81/0.23 = 386.13 N

Explanation:

unit is necessary to communicate values of the physical quantity for example can main to someone a particular length without using some sort of unit is impossible because a length cannot be described without a reference used to make sense of the value given

Answer:

t(min) = 94nm

Explanation:

The wavelength of the light incident is 500 nm.

The refractive index of the film is 1.33.

The minimum thickness of the soap film required for constructive interference.

The thickness of the film for the constructive interference is given by:

2*t= (m + 1/2) λ′

Now, λ′ = λ/μ = 500/1.33 = 376nm

The minimum thickness of the film

′t′ will be at m=0 :

2*t(min) = (0 + 1/2) 376

t(min) = 94nm

Answer:

Explanation:

The phase angle of an RLC circuit  ϕ is expressed as shoen below;

ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]

Xc is the capacitive reactance = 1/2πfC

Xl is the inductive reactance = 2πfL

R is the resistance = 25.0Ω

Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz

Xl = 2π * 70*0.0940

Xl = 41.32Ω

For the capacitive reactance;

Xc = 1/2π * 70*35.5*10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]

ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]

[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]

Since tan is negative in the 2nd quadrant;

[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]

Hence the phase angle ϕ of the circuit in degrees is 137.69°

The phase angle ϕ of the series RLC circuit that is driven at a frequency of 70.0 Hz is ϕ = 137.69°

Given that:

capacitance C = 35.5 μF,

Inductance L = 0.0940 H,

The resistance R = 25.0Ω

and frequency f = 70.0Hz

The capacitive reactance is given by:

Xc = 1/2πfC

Xc = 1/2π × 70 × 35.5× 10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

The inductive reactance is given by:

Xl = 2πfL

Xl = 2π × 70 × 0.0940

Xl = 41.32Ω

The phase angle of an RLC circuit ϕ  is given by:

[tex]\phi=tan^{-1}\frac{X_l-X_c}{R}\\\\\phi=tan^{-1}\frac{41.32-64.08}{25}[/tex]

Ф = -42.31°

Since tan is negative in the 2nd quadrant, thus:

ϕ = 180° - 42.31°

ϕ = 137.69°

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The color that is reflected when a red card is illuminated by red light is white.

The color an object is perceived to have, depends on the frequency of light it reflects.

If white light incidents on a red filter, red is transmitted while blue and green are absorbed.

Consequently, when a red card is illuminated by red light, the red card will  reflect back almost all the incident light on it, causing it to appear brighter which creates an  illusion of white color to the eyes.

Thus, we can conclude the color that is reflected when a red card is illuminated by red light is white.

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Answer:

Explanation:

To solve for the power consumed by the trains motor we have to employ the formula for power which is

Power= current * voltage

Given that

voltage V= 800 V

current I= 2130 A

Substituting in the formula for power we have

Power= 2130*800=  1704000 watt

Power = 1704 kW

This is the amount of energy consumed, transferred or converted per unit of time

Hence the power consumed  by the trains motor is 1704 kW

Answer:

To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to a value 2D that is twotimes D

Answer:

Explanation:

Formula used for calculating power P = current * voltage

P = IV

From ohms law, V = IR where R is the resistance. Substituting V = IR into the formula for calculating power, we will have;

P = IV

P =(V/R)V

P = V²/R

Given parameters

Power rating of the bulb P = 100 Watts

Source voltage V = 110V

Required

Resistance of the bulb R

Substituting the given parameters into the formula for calculating power to get Resistance R;

P = V²/R

100 = 110²/R

R = 110²/100

R = 110 * 110/100

R = 12100/100

R = 121 ohms

Hence, the resistance of this bulb is 121 ohms

Answer:

m = 9.1 x 10⁶ kg

Explanation:

First, we need to find the rate of heat transfer through the box to the ice. For this purpose, we use Fourier's Law of Heat Conduction:

Q = KA ΔT/L

where,

Q = Rate Of Heat Transfer = ?

K = Thermal Conductivity = 0.16 KW/m.°C = 160 W/m.°C

A = Area = 1.5 m²

ΔT = Difference in Temperature = 29°C - 0°C = 29°C

L = Thickness of wall = 2 cm = 0.002 m

Therefore,

Q = (160 W/m °C)(1.5 m²)(29°C)/(0.002 m)

Q = 3.48 x 10⁶ W

Now, we find the amount of heat transferred in one day to the ice:

q = Qt

where,

q = amount of heat = ?

t = time = (1 day)(24 h/1 day)(3600 s/1 h) = 86400 s

Therefore,

q = (3.48 x 10⁶ W)(8.64 x 10⁴ s)

q = 3 x 10¹¹ J

Now, for mass of ice melted in a day:

q = m H

m = q/H

where,

m = mass of ice melted in a day = ?

H = latent heat of fusion of ice = 3.3 x 10⁵ J/kg

Therefore,

m = (3 x 10¹¹ J)/(3.3 x 10⁵ J/kg)

m = 9.1 x 10⁶ kg

Answer:

The average emf induced in the coil is 175 mV

Explanation:

Given;

number of turns of the coil, N = 1060 turns

diameter of the coil, d = 20.0 cm = 0.2 m

magnitude of the magnetic field,  B = 5.25 x 10⁻⁵ T

duration of change in field, t = 10 ms = 10 x 10⁻³ s

The average emf induced in the coil is given by;

[tex]E = N\frac{\delta \phi}{dt} \\\\E = N\frac{\delta B}{\delta t}A[/tex]

where;

A is the area of the coil

A = πr²

r is the radius of the coil = 0.2 /2 = 0.1 m

A = π(0.1)² = 0.03142 m²

[tex]E = \frac{NBA}{t} \\\\E = \frac{1060*5.25*10^{-5}*0.03142}{10*10^{-3}} \\\\E = 0.175 \ V\\\\E = 175 \ mV[/tex]

Therefore, the average emf induced in the coil is 175 mV

Answer:

v = 6.28 m/s

Explanation:

It is given that,

A windmill on a farm rotates at a constant speed and completes one-half of a rotation in 0.5 seconds,

Number of revolution is half. It means angular velocity is 3.14 radians.

Let v is the angular speed. So,

[tex]v=\dfrac{\omega}{t}\\\\v=\dfrac{3.14}{0.5}\\\\v=6.28\ m/s[/tex]

So, the rotation speed is 6.28 m/s.

The angular velocity is the rotation speed, which is the angle of rotation

of the windmill per second, which is 2·π radians.

Response:

The given parameter are;

The angle of rotation the windmill rotates in 0.5 seconds = One-half a

rotation.

Required:

The rotational speed (angular velocity)

Solution:

The angle of one rotation = 2·π radians

Angle of one-half ration = [tex]\frac{1}{2}[/tex] × 2·π radians = π radians

[tex]Rotational \ speed = \mathbf{\dfrac{Angle \ of \ rotation}{Time}}[/tex]

Which gives;

[tex]Rotational \ speed, \omega = \dfrac{\pi}{0.5 \ s} = \mathbf{2 \cdot \pi \ rad/s}[/tex]

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Answer:

34°

Using the relation

θᶜ = sin^-1(n₂/n₁),

where n1= the refractive index of light is propagating from a medium

And n2 = refractive index of medium into which light is entering

So we know that

refractive index of diamond at 589nm = 2.41= n₁

refractive index of ethanol at 589nm and 20°C = 1.36= n₂

Thus. θᶜ = sin^-1(1.361/2.417) = 0.58radians = 34°

Explanation:

Answer:

The  radius of the earth is [tex]r = 6365.4 \ km[/tex]

Explanation:

From the question we are told that

     The distance at  Alexandria is  [tex]d_a = 800 \ km = 800 *10^{3} \ m[/tex]

      The angle of the sun is  [tex]\theta = 7.2 ^o[/tex]

So we want to first obtain the circumference of the earth

   So let assume that the earth is  circular ([tex]360 ^o[/tex])

  Now from question we know that the sun made an angle of [tex]7.2 ^o[/tex] so with this we will obtain how many  [tex](7.2 ^o)[/tex]  are in [tex]360^o[/tex]

 i.e    [tex]N = \frac{360}{7.2}[/tex]

=>      [tex]N = 50[/tex]

     With this  value we can evaluate the circumference as

             [tex]c = 50 * 800[/tex]

              [tex]c = 40000 \ km[/tex]

Generally circumference is mathematically represented as

        [tex]c = 2\pi r[/tex]

         [tex]40000 = 2 * 3.142 * r[/tex]

=>        [tex]r = 6365.4 \ km[/tex]

Answer:

Explanation:

1 )

An object is placed 46.9 cm away from a converging lens. The lens has a focal length of 10.0 cm.

Since the object is placed at a distance more than twice the focal length , its image will be inverted , real  and will be of the size less than the size of object . So option B is applied .

B)  real, inverted and smaller than the object.

2 )

An object is placed 46.9 cm away from a spherical convex mirror. The radius of curvature of the mirror is 20.0 cm.

The object is placed at a point beyond its radius of curvature, its image will be formed at a point between f and C   or between focal point and centre of curvature . Its size will be smaller than size of object and it will be real and inverted .

B)  real, inverted and smaller than the object.

Answer:

At the moment contact is made with the battery, the voltage across the resistor is equal to the batteries terminal voltage

Explanation;

Because at series connection the battery and resistor have equal voltage