A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The potential for rb1 < r < rb2 is:________

Answer:

In sof the friction with the nylon is very small and the current with the line e is largeummary

Explanation:

When we rub the shoes against the carpet, static electricity is produced, when you touch the metal knob you close the circuit and the current can circulate to three of the body, the value of this current is of the order of micro volts, for which a small discharge, the power that circulates through the body is very small of the order of 0.005 A

When you touch the power line, the voltage may be small, but the amount of current that can generate them is of the order of tens of amps, the electric shock is much greater per location.

In general there is a rule that if the body resumes more than P = 4000W the discharge could be fatal.

In sof the friction with the nylon is very small and the current with the line e is largeummary, the difference is that the current at the stop , so the paper that passes through the body is large and can be dangerous.

In contact with metal doorknob, get a mild shock  while with power line of same voltage, fatal the body as the amount of current is more.

When the two materials are rubbed each other, then the electric charged generated between them.

This charging of materials, due to the rubbing of two materials against each other, is called the charging by friction.

It is easy to produce a potential difference of several thousand volts between the body and the floor by scuffing your shoes across a nylon carpet.

In this case, the potential difference may be higher, but the value of current is very low. Thus, when the body touches a metal doorknob, it will get a mild shock.

Now, in another case, the contact with a power line of comparable voltage would probably be fatal. This is because in the power line the amount of current is much higher.

Hence, in contact with metal doorknob, get a mild shock  while with power line of same voltage, fatal the body as the amount of current is more.

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Answer:

The constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²

Explanation:

From the question, the car is traveling at 118 km/h, that is the initial velocity, u = 118km/h

The distance between the car and the accident at the moment when the driver sees the accident is 85 m, that is s = 85 ,

Since the driver slams on the brakes and the car will come to a stop, then the final velocity, v = 0 km/h = 0 m/s

First, convert 118 km/h to m/s

118 km/h = (118 × 1000) /3600 = 32.7778 m/s

∴ u = 32.7778 m/s

Now, to determine the deceleration, a, required to stop,

From one of the equations of motion for linear motion,

v² = u² + 2as

Then

0² = (32.7778)² + 2×a×85

0 = 1074.3841 + 170a

∴ 170a = - 1074.3841

a = - 1074.3841 / 170

a = - 6.3199

a ≅ - 6.32 m/s²

Hence, the constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²

Answer:

rock go down

Explanation:

what comes up must come down.

A concave mirror is an example of curved mirrors. So that the appropriate answer to the given question is option D. The rays will cross at the focal point.

A concave mirror is a type of mirror in which its inner part is the reflecting surface, while its outer part is the back of the mirror.  This mirror reflects all parallel rays close to the principal axis to a point of convergence. It can also be referred to as the converging mirror.

In this type of mirror, all rays of light parallel to the principal axis of the mirror after reflection will cross at the focal point.

Therefore, the required answer to the given question is option D. i.e The rays will cross at the focal point.

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The upward force on each supporting plank is 200 N

The given parameters include;

To show that the upward force of each supporting plank is 200 N, make the following sketch.

                 W₂                                                           W₃

                  ↑                                                              ↑                                                              

           -----------------------------------------------------------------------

              0.2m                         ↓                               0.2m

                                              400 N

W₂  +   W₃ = 400 N

But the distance of each supporting plank from the end is equal, (0.2m).

Then, W₂  =  W₃

2W₂ = 400 N

W₂ = 400N/2

W₂ = 200 N

W₃ = 200 N

Therefore, the upward force on each supporting plank that keeps the plank in equilibrium position is 200 N.

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Answer:

prove that Sin^6 ϴ-cos^6ϴ=(2Sin^2ϴ-1)(cos^2ϴ+sin^4ϴ)

please sove step by step with language it is opt maths question

Answer:

test her prototype and collect data about its flight

The magnetic field at 11 cm from the center of the plates is 2.364 x 10⁻⁷ T.

Given;

radius of the circular plate, d = 25 cm = 0.25 m

current in the plate, I = 1.3 A

distance from the center of the circular plate, r = 11 cm = 0.11 m

To find:

The magnetic field from the given distance is calculated as from Biot Savart equation:

[tex]B = \frac{\mu_o I}{2\pi r} \\\\where;\\\\\mu_o \ is \ permeability \ of \ free \ space \ 4\pi \times 10^{-7} \ T.m/A\\\\B = \frac{(4\pi \times 10^{-7} ) \times (1.3)}{2\pi \times 0.11} \\\\B = 2.364 \ \times 10^{-6} \ T[/tex]

Therefore, the magnetic field 11 cm from the center of the plates is 2.364 x 10⁻⁷ T.

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Answer:

The initial velocity is 1.27 m/s.

Explanation:

distance, s = 1.8 m

acceleration, a = - 0.45 m/s^2

final velocity, v = 0

let the initial velocity is u.

Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times 0.45\times 1.8\\\\u = 1.27 m/s[/tex]

We have that the Initial velocity  is mathematically given as

u=1.27m/s

Question Parameters:

a slight push toward an end stop 1.80 meters away

he fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2

Generally the equation for the  third equation of motion    is mathematically given as

Vf^2 = Vi^2 + 2ad

Therefore

0=u^2+0.45*1.8

u=1.27m/s

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Answer:

a)  fem = - 2.1514 10⁻⁴ V,  b) I = - 64.0 10⁻³ A, c)    P = 1.38  10⁻⁶ W

Explanation:

This exercise is about Faraday's law

         fem = [tex]- \frac{ d \Phi_B}{dt}[/tex]

where the magnetic flux is

        Ф = B x A

the bold are vectors

        A = π r²

we assume that the angle between the magnetic field and the normal to the area is zero

         fem = - B π 2r dr/dt = - 2π B r v

linear and angular velocity are related

        v = w r

        w = 2π f

        v = 2π f r

we substitute

        fem = - 2π B r (2π f r)

        fem = -4π² B f r²

For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T

we reduce the magnitudes to the SI system

       f = 2 rev / s (2π rad / 1 rev) = 4π Hz

we calculate

       fem = - 4π² 428 10⁻⁶ 4π 0.10²

       fem = - 16π³ 428 10⁻⁶ 0.010

       fem = - 2.1514 10⁻⁴ V

for the current let's use Ohm's law

        V = I R

        I = V / R

         I = -2.1514 10⁻⁴ / 0.00336

         I = - 64.0 10⁻³ A

Electric power is

        P = V I

        P = 2.1514 10⁻⁴ 64.0 10⁻³

        P = 1.38  10⁻⁶ W

Answer:

Explanation:

The best way to do this is to remember the rule about the halfway mark in a parabolic path. At a trajectory's half way point in its travels, it will be at its max height. To get the total time in the air, we take that time at half way and double it. Here's what we know that we are told:

initial velocity is 20 m/s

Here's what we know that we are NOT told:

a = -9.8 m/s/s and

final velocity is 0 at an object's max height in parabolic motion.

We will use the equation:

[tex]v=v_0+at[/tex] where v is final velocity and v0 is initial velocity. Filling in:

0 = 20 + (-9.8)t and

-20 = -9.8t so

t = 2 seconds. The stone reaches its max height 2 seconds after it is thrown; that means that after another 2 seconds it will be on the ground. Total air time is 4 seconds.

Answer:

cobina

Explanation:

me 2

Answer:

[tex]P_d=6203.223062W/mm^2[/tex]

Explanation:

From the question we are told that:

Voltage [tex]V=45kV[/tex]

Current [tex]I=50mAmp[/tex]

Diameter  [tex]d=0.50mm[/tex]

Heat transfer factor [tex]\mu= 0.87.[/tex]

Generally the equation for  Power developed is mathematically given by

[tex]P=VI\\\\P=45*10^3*50*10^{-3}[/tex]

[tex]P=2.250[/tex]

Therefore

Power in area

[tex]P_a=1400*0.87[/tex]

[tex]P_a=1218watt[/tex]

Power Density

[tex]P_d=\frac{P_a}{Area}[/tex]

[tex]P_d=\frac{1218}{\pi(0.5^2/4)}[/tex]

[tex]P_d=6203.223062W/mm^2[/tex]

Answer:

  E = 1.25 MV / m

Explanation:

For this exercise let's use Newton's second law

          F = m a

where the force is electric

          F = q E

we substitute

          q E = m a

          E = m a / q

indicate there are 500,000 excess electrons

          q = 500000 e

          q = 500000 1.6 10⁻¹⁹

          q = 8 10⁻¹⁴ C

the mass is m = 10⁻¹¹ kg and the acceleration a = 10⁴ m / s²

let's calculate

          E = 10⁻¹¹ 10⁴ / 8 10⁻¹⁴

          E = 0.125 10⁷ V / m = 1.25 10⁶ V / m

          E = 1.25 MV / m

Answer:

thanks for da 5points hoi

Explanation: thanks dawg

There can be uncertainty in calculating the wavelength of a laser light  due to experimental errors

All measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.

Uncertainty  means the range of possible values within which the true value of the measurement lies.

The deviation  in the value of the measured quantity from the actual quantity or true value is called an error

(a) For the calculation of wavelength of laser light , the sources which can lead to uncertainty are

1. least count of measuring instruments like spectrometer or interferometer

2. Parallax error in the measurement

3. Error in identifying the order of fringes

4.. unable to identify the accurate  reading of Vernier or circular scales present in the measuring instruments.

5. Propagating errors

The least count of a measuring instrument is the smallest and accurate value in the measured quantity that can be measured by instrument.

When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is that  all the errors add up. which increases the uncertainty

b. The uncertainty in measurement due to  least count depends on the instrument used for measurement f wavelength. A  Michelson's

interferometer has the least count of .0001mm. whereas spectrometer has a least count of 0.5⁰. Hence uncertainty in the measurement by Michelson's interferometer is very less as compared to any other instrument.

C. The maximum uncertainty arises due to the least count , as all other errors can be minimized by taking an average value of many observations but the least count of an instrument do not change so uncertainty within the least count arises.

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Answer:

2400kgm²

Explanation:

Rotational inertia=mass x radius²

Answer:

p = 60.6N*s

Explanation:

v_f = v_0+a*t

a = (v_f-v_0)/t

a = (1.8m/s)/2.35s

a = 0.77m/s²

F = m*a

F = (25kg+8.5kg)*0.77m/s²

F = 25.8N

^p = F*t

p = 25.8N*2.35s

p = 60.6N*s

Answer:

it will drop simultaneously

Answer:

The angle is 153.9 degree.

Explanation:

Let the magnetic field is B and the charge is q. Angle = 26.1 degree

The force is F.

Let the angle is A'.

Now equate the magnetic forces

[tex]q v B sin 26.1 = q v B sin A'\\\\A' = 180 - 26.1 = 153.9[/tex]

Answer:

The correct answer is:

(a) 27.5 Joules

(b) 141.5 Joules

Explanation:

Given:

Energy,

[tex]Q_c = 110 \ J[/tex]

Coefficient of performance refrigerator,

[tex]Cop(refrig)=4[/tex]

(a)

As we know,

⇒ [tex]Cop(refrig) = \frac{Q_c}{Work}[/tex]

or,

⇒ [tex]Work=\frac{Q_c}{Cop(refrig)}[/tex]

              [tex]=\frac{110}{4}[/tex]

              [tex]=27.5 \ Joules[/tex]

(b)

⇒ [tex]Heat \ expelled = Heat \ removed +Work \ done[/tex]

or,

⇒ [tex]Q_h = Q_c+Work[/tex]

         [tex]=114+27.5[/tex]

         [tex]=141.5 \ Joules[/tex]

Answer:

1.2 seconds

Explanation:

Answer to the following question is 1.2 seconds

Because light from the moon takes 1.2 seconds to reach Earth, the light released from the crescent immediately before it vanishes will also take 1.2 seconds to reach Earth. As a result, the earth-shine portion of the moon will vanish 1.2 seconds after the crescent has vanished.

Answer:

The amplitude of vibration of the spring is "24 cm"

The periodic vibrating body's motion follows a sinusoidal path. This sinusoidal path is illustrated in the attached picture.

From the picture, it can be clearly seen that the amplitude of the periodic vibration motion is the distance from its mean position to the highest point.

Since the distance of both the highest and the lowest points from the mean position is the same. Therefore, the distance between the lowest and the highest point must be equal to two times the amplitude of the wave.

Amplitude = 24 cm

Explanation:

mặt đất với vận tốc ban đầu 20m/s. Bỏ qua mọi ma sát, lấy g = 10 m/s2. Độ cao cực đại mà vật đạt được là.

Answer:

record is information created, received and maintained as evidence and information by an organization or person.in simpler terms it's a collection of of fields probably of different data types.

sound is however something loud or soft.which can be defined as vibrations that travel through the air or another medium.

I hope this helps

Answer:

15.88

is the correct answer

a) y(max)  = 337.76 m

b) t₁ = 5.30 s  the time for y maximum

c)t₂ =  13.60 s  time for y = 0 time when the fly finish

d) vₓ = 30 m/s        vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα          v₀ₓ = 60*(1/2)     v₀ₓ = 30 m/s   ( constant )

v₀y = v₀ * sinα           v₀y = 60*(√3/2)     v₀y = 30*√3  m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y  =  y₀ + v₀y*t - (1/2)*g*t²      (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt  = v₀y  - g*t

dy/dt  = 0           v₀y  - g*t₁  = 0    t₁ = v₀y/g

v₀y = 60*sin60°  = 60*√3/2  = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s  the time for y maximum

And y maximum is obtained from the substitution of t₁  in equation (1)

y (max) = 200 + 30*√3 * (5.30)  - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max)  = 337.76 m

Total time of flying (t₂)  is when coordinate y = 0

y = 0 = y₀  + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂  - 4.9*t₂²            4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for  t₂

t =  [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t =  [51.96 ±√2700 + 3920]/9.8

t =  [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8         we dismiss this solution ( negative time)

t₂ =  13.60 s  time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60°       vₓ = 30 m/s  as we said before v₀ₓ is constant

vy = v₀y - g *t        vy = 30*√3  - 9.8 * (13.60)

vy = 51.96 - 133.28         vy = - 81.32 m/s

The sign minus means that vy  change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60  m

x = 408 m

Answer:

a)   θ = 14.23º, b)   θ₂ = 75.77,  c) t = 0.6019 s

Explanation:

This is a missile throwing exercise.

a) the reach of the ball is the distance traveled for the same departure height

          R = [tex]\frac{v_o^2 \ sin 2 \theta }{g}[/tex]

          sin 2θ = [tex]\frac{Rg}{v_o^2}[/tex]

          sin 2θ = 7.00 9.8 / 12.0²

          2θ = sin⁻¹ (0.476389) = 28.45º

           θ = 14.23º

the complementary angle that gives the same range is the angle after 45 that the same value is missing to reach 90º

          θ ’= 90  -14.23

          θ’= 75.77º

b) the two angles that give the same range are

         θ₁ = 14.23

         θ₂ = 75.77

the greater angle has a much greater height so the time of the movement is greater and has a greater chance of being intercepted by the other team.

C) the time of the pass can be calculated with the expression

           x = v₀ₓ t

           t = x / v₀ₓ

           t = 7 / 11.63

           t = 0.6019 s

Answer: Below is the complete question

A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s)

answer:

mass transfer coefficient = 9.56 * 10^-5 m/s

Explanation:

Candy density = 1950 kg/m^3

Candy diameter = 1 cm

Velocity of water = 1 m/s

water density = 1000 kg/m^3

Viscosity of water = 1 * 10^-3 kg/m/s

diffusion coefficient of candy in water = 2 * 10^-9 m^2/s

solubility of candy = 2 kg/m^3

Determine the mass transfer coefficient ( m/s )

( Sh) mass transfer coefficient ( flow across sphere ) = 2 + 0.6Re^1/2 * SC^1/3

where : Re = vdp / μ ,   Sh = KLd / Deff

attached below is the remaining solution .

mass transfer coefficient =  9.56 * 10^-5 m/s

Explanation:

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The angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is  27.29°.

Waves spreading outward around obstructions are known as diffraction. Sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as very small moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities all exhibit diffraction.

Given:

The number of lines = 6000 per cm,

The Wavelength, λ = 500 nm = 500 × 10 ⁻⁹ m

Calculate the diffraction grating,

[tex]d = 1 / no\ of\ lines[/tex]

d = 10⁻² / 6000 m,

Calculate the second-order maxima angle and third-order maxima angle by the formula given below,

[tex]dsin\theta_1 = n_1 \lambda[/tex]

[tex]sin\theta_1 = n_1\lambda / d[/tex]

[tex]\theta _1 = sin^{-1}[2\times 500\times 10 ^{-9}/10^{-2}\times 6000][/tex]

θ₁ = sin⁻¹(0.6)

θ₁ = 36.87°

Similarly, for θ₂,

θ₂ = sin⁻¹(3 × 500 × 10 ⁻⁹ / 10⁻² × 6000)

θ₂ = sin⁻¹(0.9)

θ₂ = 64.16°

Calculate the separation as follows,

θ₂ - θ₁ = 64.16° - 36.87°

θ₂ - θ₁ =  27.29°

Therefore, the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is  27.29°.

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Answer:

This causes the amplitude of the oscillation to decay over time. The damped oscillation frequency does not equal the natural frequency. Damping causes the frequency of the damped oscillation to be slightly less than the natural frequency