A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 5.0 μg dust particle is suspended in midair just above the center of the carpet.

Required:
What is the charge on the dust particle?

Answer:

K = 227.04 W/m.°C

Explanation:

First we need to find the heat required to melt the ice:

q = m H

where,

q = heat required = ?

m = mass of the ice = 8.5 g = 8.5 x 10⁻³ kg

H = Latent heat of fusion of ice = 3.34 x 10⁵ J/kg

Therefore,

q = (8.5 x 10⁻³ kg)(3.34 x 10⁵ J/kg)

q = 2839 J

Now, we find the heat transfer rate through rod:

Q = q/t

where,

t = time = (10 min)(60 s/1 min) = 600 s

Q = Heat Transfer Rate = ?

Therefore,

Q = 2839 J/600 s

Q = 4.73 W

From Fourier's Law of Heat Conduction:

Q = KA ΔT/L

where,

K = Thermal Conductivity = ?

A = cross sectional area = 1.25 cm² = 1.25 x 10⁻⁴ m²

L = Length of rod = 60 cm = 0.6 m

ΔT = Difference in temperature = 100°C - 0°C = 100°C

Therefore,

4.73 W = K(1.25 X 10⁻⁴ m²)(100°C)/0.6 m

K = (4.73 W)/(0.0208 m.°C)

K = 227.04 W/m.°C

Answer:

a) 3.7 x 10^-3 s

b) 230.41 rad/s

Explanation:

The angular speed N = 2200 rpm (revolution per minute)

==> 2200/60 revolutions per sec = 36.67 rps

The total angle turned in one second = 36.67 x 360° = 13201.2°

if it takes 1 sec to revolve 13201.2°

then it will take t sec to rotate 49.0°

time t = 49/13201.2 = 3.7 x 10^-3 s

conversion to rad/s = 2πN/60 = (2 x 3.142 x 2200)/60 = 230.41 rad/s

Answer:

3x10⁴v

Explanation:

Using

Wavelength= h/ √(2m.Ke)

880nm = 6.6E-34/√ 2.9.1E-31 x me

Ke= 6.6E-34/880nm x 18.2E -31.

5.6E-27/18.2E-31

= 3 x 10⁴ Volts

Answer:

11.2 Ω

Explanation:

The impedance of a circuit is given by;

Z= √R^2 +(XL-XC)^2

Since

Resistance R= 10 Ω

Inductive reactance XL= 12 Ω

Capacitive reactance XC= 7 Ω

Z= √10^2 + (12-7)^2

Z= √100 + 25

Z= √125

Z= 11.2 Ω

Answer:

1.The temperature of each sample will increase by the same amount

Explanation:

This is because, since their specific heat capacities are the same and we have the same mass of each substance, and the same amount of energy due to heat flow is supplied to both the glass and brick at room temperature, their temperatures would thereby increase by the same amount.

This is shown by the calculation below

Q = mcΔT

ΔT = Q/mc where ΔT = temperature change, Q = amount of heat, m = mass of substance and c = specific heat capacity of substance.

Since Q, m and c are the same for both substances, thus ΔT will be the same.

So, the temperature of each sample will increase by the same amount

Answer:

The wavelength is  [tex]\lambda = 1805 nm[/tex]

Explanation:

From the question we are told that

    The wavelength of the light is  [tex]\lambda = 601 \ nm = 601 *10^{-9} \ m[/tex]

     The  distance of the screen is  D  =  3.0  m

     The  fringe width is  [tex]y = 4.84 \ mm = 4.84 *10^{-3} \ m[/tex]

Generally the fringe width for a bright fringe  is mathematically represented as

          [tex]y = \frac{ \lambda * D }{d }[/tex]  

=>     [tex]d = \frac{ \lambda * D }{ y }[/tex]

=>     [tex]d = \frac{ 601 *10^{-9} * 3}{ 4.84 *10^{-3 }}[/tex]

=>     [tex]d = 0.000373 \ m[/tex]

Generally the fringe width for a dark fringe  is mathematically represented as

      [tex]y_d = [m + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]

Here  m = 0  for  first order dark fringe

   So  

         [tex]y_d = [0 + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]

looking at which we see that   [tex]y_d = y[/tex]

         [tex]4.84 *10^{-3} = [0 + \frac{1}{2} ] * \frac{\lambda * 3 }{ 0.000373 }[/tex]

=>    [tex]\lambda = 1805 *10^{-9} \ m[/tex]

=>    [tex]\lambda = 1805 nm[/tex]

Answer and Explanation:

The computation of the object distance related to two quantities is shown below:

It could find out by using the lens formula which is shown below:

[tex]\frac{1}{v} - \frac{1}{u} = \frac{1}{f}[/tex]

where,

v = image distance

u = object distance

f = focal length

It could be found by applying the above formula i.e considering the image distance, object distance and the focal length

Answer:

The magnetic field produced by an electric current is always oriented perpendicular to the direction of the current flow. And.Direction of magnetic field is governed by the 'right hand thumb rule, The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of Force . Similar to the situation with electric field lines, the greater the number of lines (or the closer they are together) in an area the stronger the magnetic field.

Answer:

[tex]v_1 = 301 Hz[/tex]

[tex]v_2 = 601 \ \ Hz[/tex]

[tex]v_3 = 901 \ Hz[/tex]

Explanation:

From the question we are told that

     The  length of the string is  [tex]l = 90 \ cm = 0.9 \ m[/tex]

     The mass of the string is  [tex]m_s = 3.5 \ g =0.0035 \ kg[/tex]

     The  distance  from the bridge to the support post [tex]L = 62 \ c m = 0.62 \ m[/tex]

    The tension is [tex]T = 540 \ N[/tex]

Generally the frequency is mathematically represented as

        [tex]v = \frac{n}{2 * L } [\sqrt{ \frac{T}{\mu} } ][/tex]

Where n is and integer that defines that overtones

i.e  n =   1 is for fundamental frequency

      n =  2   first overtone

       n =3   second overtone

Also  [tex]\mu[/tex] is the linear density of the string which is mathematically represented as

           [tex]\mu = \frac{m_s}{l}[/tex]

=>        [tex]\mu = \frac{0.0035 }{ 0.9 }[/tex]

=>       [tex]\mu = 0.003889 \ kg/m[/tex]

So for   n = 1

     [tex]v_1 = \frac{1}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]

     [tex]v_1 = 301 \ Hz[/tex]

So for  n =  2

     [tex]v_2 = \frac{2}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]

     [tex]v_2 = 601 \ Hz[/tex]

So for  n =  3

     [tex]v_3 = \frac{3}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]

     [tex]v =901 \ Hz[/tex]

Answer:

The distance is  [tex]D = 0.000712 \ m[/tex]

Explanation:

From the question we are told that

    The wavelength of  the  light source is  [tex]\lambda = 700 \ nm = 700 *10^{-9} \ m[/tex]

     The distance from a pin hole is  [tex]x = 9\ m[/tex]

       The  diameter of the pin  hole is  [tex]d = 1.2 \ mm = 0.0012 \ m[/tex]

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

              [tex]D = \frac{1.22 \lambda }{d }[/tex]

substituting values

             [tex]D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }[/tex]

             [tex]D = 0.000712 \ m[/tex]

Answer:

 M₁₂ = 1.01 10⁻⁴ H ,   Fem = 3.54 10⁻³ V

Explanation:

The mutual inductance between two systems is

        M₁₂ = N₂ Ф₁₂ / I₁

where N₂ is the number of turns of the inner solenoid N₂ = 21.0, i₁ the current that flows through the outer solenoid I₁ = 35.0 A / s and fi is the flux of the field of coil1 that passes through coil 2

the magnetic field of the coil1 is

   B = μ₀ n I₁ = μ₀ N₁/l   I₁

the flow is

             Φ = B A₂

the area of ​​the second coil is

             A₂ = π d₂ / 4

             Φ = μ₀ N₁ I₁ / L  π d² / 4

we substitute in the first expression

            M₁₂ = N₂ μ₀ N₁ / L    π d² / 4

            M₁₂ = μ₀ N₁ N₂ π d² / 4L

           d = 0.170 cm = 0.00170 m

            L = 4.00 cm = 0.00400 m

let's calculate

            M₁₂ = 4π 10⁻⁷ 6750  21 π 0.0017²/ (4 0.004)

             M₁₂ = π² 0.40966 10⁻⁷ / 0.004

             M₁₂ = 1.01 10⁻⁴ H

The electromotive force is

              Fem = - M dI₁ / dt

              Fem = - 1.01 10⁻⁴ 35.0

              Fem = 3.54 10⁻³ V

Answer:

10m

Explanation:

Since Given frequency f= 1/t

and velocity ν=10 m/s

We know ν=λf

λ= ν/f

​ = 10/1/0.5

=5m

Since both the waves are similar but moves in opposite direction its total wavelength of the wave will be 10 m

Answer:

 cost = $ 243.00

Explanation:

This exercise must assume that it uses a complete table for each piece, we can use a direct ratio of proportions, if 1 table is 0.20 m wide, how many tables will be 3.00 m

                 #_tables = 3 m (1 / 0.20 m)

                #_tables = 15 tables

Let's use another direct ratio, or rule of three, for cost. If a board costs $ 16.20, how much do 15 boards cost?

              Cost = 15 (16.20 / 1)

              cost = $ 243.00

Answer:

a)  A = 1.99 μm , b) λ = 0.4134 m , c)  v = 57.2 m / s , d)   s = - 1,946 nm ,

e)      v_max = 1,739 mm / s

Explanation:

A sound wave has the general expression

           s = s₀ sin (kx - wt)

where s is the displacement, s₀ the amplitude of the wave, k the wave vector and w the angular velocity, in this exercise the expression given is

           s = 1.99 sin (15.2 x - 869 t)

a) the amplitude of the wave is

        A = s₀

        A = 1.99 μm

b) wave spectrum is

      k = 2π /λ

in the equation k = 15.2 m⁻¹

      λ = 2π / k

      λ = 2π / 15.2

     λ = 0.4134 m

c) the speed of the wave is given by the relation

       v = λ f

angular velocity and frequency are related

       w = 2π f

        f = w / 2π

        f = 869 / 2π

        f = 138.3 Hz

        v = 0.4134 138.3

         v = 57.2 m / s

d) To find the instantaneous velocity, we substitute the given distance and time into the equation

       s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)

       s = 1.99 sin (0.77368 - 2.55486)

remember that trigonometry functions must be in radians

       s = 1.99 (-0.98895)

       s = - 1,946 nm

The negative sign indicates that it shifts to the left

e) the speed of the oscillating part is

           v = ds / dt)

           v = - s₀(-w) cos (kx -wt)

the maximum speed occurs when the cosines is 1

           v_maximo = s₀w

           v_maximum = 1.99 869

           v_maximo = 1739.31 μm / s

let's reduce to mm / s

          v_maxio = 1739.31 miuy / s (1 mm / 103 mu)

          v_max = 1,739 mm / s

a) A is = 1.99 μm , b) λ is = 0.4134 m , c) v is = 57.2 m / s , d) s is = - 1,946 nm, e) v_max is = 1,739 mm / s

When A sound wave has the general expression is:

Then, s = s₀ sin (kx - wt)

Now, where s is the displacement, Then, s₀ is the amplitude of the wave, k the wave vector, and w the angular velocity, Now, in this exercise the expression given is

s is = 1.99 sin (15.2 x - 869 t)

a) When the amplitude of the wave is

A is = s₀

Thus, A = 1.99 μm

b) When the wave spectrum is

k is = 2π /λ

Now, in the equation k = 15.2 m⁻¹

Then, λ = 2π / k

After that, λ = 2π / 15.2

Thus, λ = 0.4134 m

c) When the speed of the wave is given by the relation is:

Then, v = λ f

Now, the angular velocity and frequency are related is:

w is = 2π f

Then, f = w / 2π

After that, f = 869 / 2π

Now, f = 138.3 Hz

Then, v = 0.4134 138.3

Thus, v = 57.2 m / s

d) Now, To find the instantaneous velocity, When we substitute the given distance and time into the equation

Then, s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)

After that, s = 1.99 sin (0.77368 - 2.55486)

Then remember that trigonometry functions must be in radians

After that, s = 1.99 (-0.98895)

Thus, s = - 1,946 nm

When The negative sign indicates that it shifts to the left

e) When the speed of the oscillating part is

Then, v = ds / dt)

Now, v = - s₀(-w) cos (kx -wt)

When the maximum speed occurs when the cosines is 1

Then, v_maximo = s₀w

After that, v_maximum = 1.99 869

v_maximo = 1739.31 μm / s

Now, let's reduce to mm / s

Then, v_maxio = 1739.31 miuy / s (1 mm / 103 mu)

Therefore, v_max = 1,739 mm / s

Finf more informmation about Wavelength here:

https://brainly.com/question/6352445

Answer:

Impedance increases for frequencies below resonance and decreases for the frequencies above resonance

Explanation:

See attached file

Explanation:

Answer:

18.5 cm

Explanation:

From;

1/u + 1/v = 1/f

Where;

u= object distance = 86cm

image height = 23 cm

Radius of curvature = 37 cm

The radius of curvature (r) is the radius of the sphere of which the mirror forms a part.

Focal length (f) = radius of curvature (r)/2 = 37cm/2 = 18.5 cm

Therefore, the focal length of the mirror is 18.5 cm

Answer:

42000N

Explanation:

First you calculate how much it would contract, and secondly you then calculate the force to stretch it by that amount.

1) linear thermal expansion coef brass 19e-6 /K

∆L = αL∆T = (19e-6)(1.85)(110) = 0.00387 meter or 3.87 mm

Second part involves linear elasticity.

for brass, young's modulus is 15e6 psi or 100 GPa

cross-sectional area of rod is π(0.008)² = 0.0002 m²

F = EA∆L/L

F = (100e9)(0.0002)(0.00387) / (1.85)

F = 42000 or 42 kN

Answer:

79.7 days

Explanation:

Half-life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is the amount of time,

and T is the half life.

If 90% decays, then 10% is left.

A = A₀ (½)^(t / T)

0.1 A₀ = A₀ (½)^(t / 24)

0.1 = ½^(t / 24)

ln(0.1) = (t / 24) ln(0.5)

t ≈ 79.7 days

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

[tex]F_{pull}=m_{T}.a[/tex]

[tex]m_{T}=\frac{F_{pull}}{a}[/tex]

[tex]m_{T}=\frac{3615}{1.516}[/tex]

[tex]m_{T}=2384.5[/tex]

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For [tex]m_{1}[/tex]:

The only force acting On the [tex]m_{1}[/tex] box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

[tex]m_{1} = \frac{F_{12}}{a}[/tex]

[tex]m_{1} = \frac{1387}{1.516}[/tex]

[tex]m_{1}[/tex] = 915kg

For [tex]m_{2}[/tex]:

There are two forces acting on [tex]m_{2}[/tex]: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

[tex]m_{2} = \frac{F_{23}-F_{12}}{a}[/tex]

[tex]m_{2} = \frac{2304-1387}{1.516}[/tex]

[tex]m_{2}[/tex] = 605kg

For [tex]m_{3}[/tex]:

[tex]m_{3} = m_{T} - (m_{1}+m_{2})[/tex]

[tex]m_{3} = 2384.5-1520.0[/tex]

[tex]m_{3}[/tex] = 864.5kg

Answer:

The position on the y-axis where the magnetic field is zero is at y = 10 cm

Explanation:

The magnetic field B due to a long straight wire carrying a current, i at a distance R from the wire is given by

B = μ₀i/2πR

Now, let y be the point where the magnetic fields of both wires are equal.

So, the magnetic field due to wire 1 carrying current i₁ = 73 A is

B₁ = μ₀i₁/2π(y - 3) and

the magnetic field due to wire 2 carrying current i₂ = 31 A is

B₂ = μ₀i₂/2π(13 - y)

At the point where the magnetic field is zero, B₁ = B₂. So,

μ₀i₁/2π(y - 3) = μ₀i₂/2π(13 - y)

cancelling out μ₀ and 2π, we have

i₁/(x - y) = i₂/(13 - y)

cross-multiplying, we have

(13 - y)i₁ = (y - 3)i₂

Substituting the values of i₁ and i₂, we have

(13 - y)73 = (y - 3)31

949 - 73y = 31y - 93

Collecting like terms, we have

949 + 93 = 73y + 31y

1042 = 104y

dividing through by 104, we have

y = 1042/104

y = 10.02 cm

y ≅ 10 cm

So, the position on the y-axis where the magnetic field is zero is at y = 10 cm

Answer:

81 N

7.1 m/s²

Explanation:

Draw a free body diagram of the sphere.  There are two forces:

Weight force mg pulling straight down,

and tension force T pulling up along the rope.

Sum of forces in the centripetal direction:

∑F = ma

T − mg sin 45° = m v² / r

T = m (g sin 45° + v² / r)

T = (7 kg) (10 m/s² sin 45° + (3 m/s)² / 2 m)

T = 81 N

Sum of forces in the tangential direction:

mg cos 45° = ma

a = g cos 45°

a = (10 m/s²) cos 45°

a = 7.1 m/s²

Explanation:

F = ma

85 N = m (15 m/s²)

m ≈ 5.7 kg

Explanation:

u = 95 m/sec ( Initial speed)

t = 7 sec ( Time of ascent)

According to Equations of Motion :

[tex]s = ut - \frac{1}{2} g {t}^{2} [/tex]

Max. Height = 95 * 7 - 4.9 * 49 = 424. 9 = 425 m

Answer:

332.5 m

Explanation:

At the maximum height, the velocity is 0.

Given:

v₀ = 95 m/s

v = 0 m/s

t = 7 s

Find: Δy

Δy = ½ (v + v₀) t

Δy = ½ (0 m/s + 95 m/s) (7 s)

Δy = 332.5 m

Answer:

29°

Explanation:

because the refracted ray angle is small than angle of incidence

Answer:

56°

Explanation:

Brewsters angle can be simply derived from

n1sin theta1= n2sintheta2= n2costheta1

because the reflected light will be 100% polarized if it is reflected at an angle 90o to the refracted light. Hence, Brewsters angle is

Tan theta= n2/n1

1.66/1.11= 1.495

Theta = 56°

Explanation:

Answer:

The fish would appear 42.7 cm on the left side from the front of the bowl.

Explanation:

The fish (object) distance = 11.9 cm, radius of curvature of the bowl = 33 cm. The distance of image of the fish (image distance) can be determined by applying the mirror formula;

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]

where f is the focal length of the reflecting surface, u is the object distance and v is the image distance.

But, f = [tex]\frac{radius of curvature}{2}[/tex]

         = [tex]\frac{33}{2}[/tex]

       f = 16.5 cm

Substitute f = 16.5 = [tex]\frac{165}{10}[/tex], and u = 11.9 = [tex]\frac{119}{10}[/tex] in equation 1;

[tex]\frac{10}{165}[/tex] = [tex]\frac{10}{119}[/tex] + [tex]\frac{1}{v}[/tex]

[tex]\frac{1}{v}[/tex] = [tex]\frac{10}{165}[/tex] - [tex]\frac{10}{119}[/tex]

  = [tex]\frac{1190 - 1650}{19635}[/tex]

[tex]\frac{1}{v}[/tex] = [tex]\frac{-460}{19635}[/tex]

⇒ v  = [tex]\frac{19635}{-460}[/tex]

       = -42.6848

    v = 42.7 cm

The fish would appear 42.7 cm on the left side from the front of the bowl.

Answer:

the final angular velocity of the particle is approximately 38.18  Rad/s

Explanation:

To start with, let's make sure that units of angle measure are the same, converting everything into radians:

[tex]4500^o\, \frac{\pi}{180^o}= 25\,\pi[/tex]

And now we can use the kinematic formulas for rotational motion:

[tex]\theta-\theta_0=\omega_0\,t+\frac{1}{2} \alpha\,t^2[/tex]

Therefore we can find the initial angular velocity [tex]\omega_0[/tex]  of the particle:

[tex]\theta-\theta_0=\omega_0\,t+\frac{1}{2} \alpha\,t^2\\25\,\pi=\omega_0\,(3)+\frac{1}{2} (8)\,(3)^2\\25\,\pi-36=\omega_0\,(3)\\\omega_0=\frac{25\,\pi-36}{3} \\\omega_0\approx 14.18\,\,\,rad/s[/tex]

and now we can estimate the final angular velocity using the kinematic equation for angular velocity;

[tex]\omega=\omega_0\,+\alpha\,t\\\omega=14.18+8\,(3)\\\omega=38.18\,\,\,rad/s[/tex]

Answer:

The direction of the force is towards the East.

Explanation:

Using the right hand rule, the force on the current carrying conductor is east.

In the right hand rule, if the hand is held with the fingers pointed parallel to the palm representing the magnetic field, and the thumb held at right angle to the rest of the fingers representing the direction of the current, then the palm will push in the direction of the force.

In this case, the thumb is pointing downwards, with the fingers pointing north away from the body in the direction of the earth's magnetic field, the palm will push east.

Answer:

The  value is  [tex]B = 0.0043 \ T[/tex]

Explanation:

From the question we are told that

   The  length of the solenoid is  [tex]l = 63.5 = 0.635 \ m[/tex]

    The number of turns is  [tex]N = 960 \ turns[/tex]

    The  current is  [tex]I = 2.28 \ A[/tex]

Generally the magnetic field is mathematically represented as

      [tex]B = \mu _o * n * I[/tex]

Where  n is the number of turn per unit length which is mathematically evaluated as

      [tex]n = \frac{N}{l}[/tex]

     [tex]n = \frac{960}{0.635}[/tex]

     [tex]n = 1512 \ turns /m[/tex]

and  [tex]\mu_o[/tex] is the permeability of free space with value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

So  

    [tex]B = 4\pi * 10^{-7} * 1512 * 2.28[/tex]

    [tex]B = 0.0043 \ T[/tex]