A long conducting cylinder of radius a carrying a total charge +q is surrounded by a
concentric thin conducting cylindrical shell of radius b carrying a total charge -2q.
Draw the cross section of this arrangement. Use Gauss’s law to find the electric field
strength at a point r away from the axis,

Answer:

E = K Q / R^2     by Gauss' Law where Q is the  charge enclosed by the surface of revolution and R is the distance from the enclosed charge

Since Q = d 4 pi r^2   where d is the charge density and r the radius of the inner sphere

E = K / R^2 * (4 d pi r^2) = 4 K pi d (r / R)^2 = 4 K pi d * 1/4

E =  9 * 10E9 * 3.14 * 40 * 10E-12 = 1.13 N / C

Answer:

The correct answer is "3899.92 N".

Explanation:

The given values are:

Force,

[tex]F_{app}=58 N[/tex]

Ratio,

[tex]\frac{R_2}{R_1}=8.2[/tex]

As we know,

Area, [tex]A=\pi r^2[/tex]

or,

⇒  [tex]\frac{F_2}{F_1} =\frac{A_2}{A_1}[/tex]

On substituting the value of "A", we get

⇒  [tex]\frac{F_2}{F_1} =\frac{\pi r_2^2}{\pi r_1^2}[/tex]

⇒  [tex]\frac{F_2}{F_1} =\frac{r_2^2}{r_1^2}[/tex]

On applying cross-multiplication, we get

⇒  [tex]F_2=F_1\times (\frac{r_2}{r_1} )^2[/tex]

On substituting the given values, we get

⇒       [tex]=58\times (8.2)^2[/tex]

⇒       [tex]=58\times 67.2[/tex]

⇒       [tex]=3899.92 \ N[/tex]

Answer:

0.174 plasma

[tex]$5.85 \times 10^{-4}$[/tex] flame

Explanation:

Given :

Energy :

[tex]$\Delta E=3.371 \times 10^{-19} $[/tex] J per atom

[tex]$g^*=2$[/tex] (degenraci of excited state)

g = 1  (degenraci of excited state)

Boltzmann Distribution

[tex]$\frac{N^*}{N}=\frac{g^*}{g}e^{-\frac{\Delta E}{kT}}$[/tex]

where,

[tex]$N^*$[/tex] = atoms in excited state

N = atoms in lower energy level

k = [tex]$1.38 \times 10^{-23}$[/tex]  J/K

Therefore,

Relative population in plasma

T = 10,000 K

[tex]$\frac{N^*}{N}=\frac{g^*}{g}e^{-\frac{\Delta E}{kT}}$[/tex]

[tex]$\frac{N^*}{N}=\frac{2}{1}e^{-\frac{-3.37\times 10^{-19}}{1.38 \times 10^{-23} \times 10000}}$[/tex]

[tex]$\frac{N^*}{N}=2 \times e^{-2.44275}$[/tex]

[tex]$\frac{N^*}{N}=2 \times 0.8692$[/tex]

[tex]$\frac{N^*}{N}=0.1738$[/tex]

Relative population in flame    

T = 3000

[tex]$\frac{N^*}{N}=\frac{g^*}{g}e^{-\frac{\Delta E}{kT}}$[/tex]

[tex]$\frac{N^*}{N}=\frac{2}{1}e^{-\frac{-3.371\times 10^{-19}}{1.38 \times 10^{-23} \times 3000}}$[/tex]

[tex]$\frac{N^*}{N}=2 \times e^{-8.1425}$[/tex]

[tex]$\frac{N^*}{N}=2 \times 2.9090 \times 10^{-4}$[/tex]

[tex]$\frac{N^*}{N}=5.85 \times 10^{-4}$[/tex]

Answer:C

Explanation:

I think its c im pretty sure

Explanation:

a hot toothbrush. ......lol

Answer:

         v₃ = - (3 i ^ + 4 j ^) m / s

v₃ = 5 m / s,  θ = 233º

Explanation:

This is a momentum problem. Let us form a system formed by the three objects so that the forces during the collisions have been internal and the moment is conserved.

Let's start working with the first two objects.  As each object moves in a different direction let's work with the components in an xy coordinate system

X axis

initial instant. Before the shock

       p₀ₓ = m₁ v₁₀ + 0

final instant. After the crash

      p_{fx} = (m1 + m2) vₓ

the moment is preserved

       p₀ₓ = p_{fx}

       m₁ v₀₁ = (m₁ + m₂) vₓ

       vₓ = [tex]\frac{m_1}{m_1+m_2} \ v_{o1}[/tex]

Y axis  

initial instant

       p_{oy} = 0 + m₂ v₀₂

final moment

       p_{fy} = (m₁ + m₂) v_y

the moment is preserved

      p_{oy} = p_{fy}

      m₂ v₀₂ = (m₁ + m₂) v_y

      v_y = [tex]\frac{m_2}{m_1 +m_2 } \ v_{o2}[/tex]

We already have the speed of the set of these two cars, now let's work on this set and vehicle 3

X axis    

initial instant

      p₀ₓ = (m₁ + m₂) vₓ + m₃ v₃ₓ

final instant

      p_{fx} = 0

      p₀ₓ = p_{fx}

      (m₁ + m₂) vₓ + m₃ v₃ₓ = 0

       v₃ₓ = [tex]- \frac{m_1+m_2 }{m_3} \ v_x[/tex]

Y Axis

initial instant

        p_{oy} = (m₁ + m₂) v_y + m₃ v_{3y}

final moment

        p_{fy} = 0

        p_{oy} = p_{fy}

        (m₁ + m₂) v_y + m₃ v_{3y} = 0

        v_{3y} = [tex]- \frac{m_1+m_2}{m_3} \ v_y[/tex]

now we substitute the values ​​of the speeds

       v₃ₓ = [tex]- \frac{m_1+m_2}{m_3} \ \frac{m_1}{m_1+m_2} \ v_{o1}[/tex]  

       v₃ₓ = [tex]- \frac{m_1}{m_3} \ v_{o1}[/tex]

       v_{3y} = [tex]- \frac{m_1+m_2}{m_3} \ \frac{m_2}{m_1+m_2} \ v_{o2}[/tex]

       v_{3y} = [tex]- \frac{m_2}{m_3} \ v_{o2}[/tex]

let's calculate

         v₃ₓ = - ⅓ 9

         v₃ₓ = - 3 m / s

         v_{3y} = - ⅔ 6

         v_{3y} = - 4 m / s

therefore the speed of vehicle 3 is

         v₃ = - (3 i ^ + 4 j ^) m / s

It can also be given in the form of modulus and angles using the Pythagorean theorem

         v₃ = [tex]\sqrt{v_{3x}^2 + v_{3y}^2}[/tex]

          v₃ = [tex]\sqrt{3^2+4^2}[/tex]

          v₃ = 5 m / s

let's use trigonometry for the angle

          tan θ' = [tex]\frac{v_{3y}}{v_{3x}}[/tex]

          θ' = tan⁻¹ (\frac{v_{3y}}{v_{3x}})

          θ' = tan⁻¹ (4/3)

          θ'  = 53º

That the two speeds are negative so this angle is in the third quadrant, measured from the positive side of the x axis

          θ = 180 + θ'

          θ = 180 +53

          θ = 233º

Answer:

Explanation:

Money is used to cover the explicit cost. They're what we're used to seeing, and they're easy to identify. There are several classifications for explicit costs; we can distinguish between fixed and variable costs, as well as direct and indirect costs. Raw materials, manpower, indirect production costs, and so on are all factors that contribute to its expense.

The opportunity cost of using a resource is the amount of money paying for it that might have been used on something else.

Alternative benefit options or money that we miss earning by doing such business acts are referred to as opportunity costs.

When a corporation foregoes an alternative action but does not make a bill, it incurs implicit costs. These are a company's hidden costs:

1. The use of the firm's own funds (money or assets).

2. The owner's capital, savings, and financial services are used.

From the given information:

The total explicit cost = Rent + office supplies + office staff + telephone expense

=$( 14000 + 1000 + 24000 + 4500)

= $43500

The total implicit cost = forgone salary + forgone interest

= $40000 + 7% of $5000

= $40000 + 350

=$40350

Answer:

The angular speed of the blade is of 166.67 rad/min

If a radial saw has a blade with a 9-in. radius. Suppose that the blade spins at 1000 rpm then the angular speed of the blade would be 104.72 rad/seconds.

The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.

The mathematical expression for speed is given by

speed = total distance /Total time

As given in the problem If a radial saw has a blade with a 9-in. radius. Suppose that the blade spins at 1000 rpm then we have to find the angular speed of the blade,

the angular speed of the blade (ω) = 2πN/60

                                                          =2×π×1000/60

                                                          = 104.72 rad/seconds

Thus, the angular speed of the radial saw would be 104.72 rad/seconds

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Answer:

Current

Explanation:

hope it helps and your day will be full of happiness

Answer: long pipe is used to increase torque and reduce force needed.

Explanation: A torque is needed to open bolt. torque = F·r, if

R increases, Force F needed to open bold is smaller.

Problem is a worn bolt may break down. It sometimes I send useful to heat bolt instead using too much power.

Answer:

Potential energy only occurs when an object is in motion.

Explanation:

Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;

a. Kinetic energy (K.E): it is an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

[tex] K.E = \frac{1}{2}MV^{2}[/tex]

Where;

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

b. Potential energy (P.E): it is an energy possessed by an object or body due to its position above the earth.

Mathematically, potential energy is given by the formula;

[tex] P.E = mgh[/tex]

Where,

P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Additionally, the mechanical energy of a physical object or body is the sum of the potential energy and kinetic energy possessed by the object or body.

Mathematically, it is given by the formula;

Mechanical energy = P.E + K.E

Answer:

It's important

Explanation:

Mental wellness is critical to overall physical health. Participating in recreational activities helps manage stress. Taking time to nurture oneself provides a sense of balance and self-esteem, which can directly reduce anxiety and depression.

Answer:

Mental wellness is critical to overall physical health. Participating in recreational activities helps manage stress. Taking time to nurture oneself provides a sense of balance and self-esteem, which can directly reduce anxiety and depression.

Explanation:

Answer:

0.525 m

Explanation:

From the question,

In an open pipe,

v = λf................... Equation 1

Where v = speed of sound in the pipe, f = frequency, λ = wave length.

But,

λ = 2l................ Equation 2

Where l = length of the pipe

Substitute equation 2 into equation 1

v = 2lf.............. Equation 3

make l the subject of the equation

l = v/2f............. Equation 4

Given:

v ≈ (331+0.60T) m/s at T = 26°C

v ≈ (331+0.60×26) ≈ 346.6 m/s

f = 330 Hz.

Substitute these values into equation 4

l = 346.6/(2×330)

l = 346.6/660

l = 0.525 m

Hence the length of the pipe is 0.525 m

Answer:

I1 = ε/R1

I2 = ε/R2

I3 = ε/R3

Explanation:

From the image, we see that the resistors are connected in parallel. This means that the voltage passing through them is the same.

Now, formula for current is; I = V/R

In this case, V which is voltage is denoted by ε.

Thus;

I1 = ε/R1

I2 = ε/R2

I3 = ε/R3

Answer:

a. I = 30 A

b. E = 1080000 J = 1080 KJ

c. ΔT = 12.86°C

d. Cost = $ 4.32  

Explanation:

a.

The current in the coil is given by Ohm's Law:

[tex]V = IR\\I = \frac{V}{R}[/tex]

where,

I = current = ?

V = Voltage = 120 V

R = Resistance = 4 Ω

Therefore,

[tex]I = \frac{120\ V}{4 \Omega}\\[/tex]

I = 30 A

b.

The energy can be calculated as:

[tex]E = VIt\\E = (120\ V)(30\ A)(5\ min)(\frac{60\ s}{1\ min})\\[/tex]

E = 1080000 J = 1080 KJ

c.

For the increase in the temperature of water:

[tex]E = mC\Delta T\\[/tex]

where,

m = mass of water = 20 kg

C = specific heat of water = 4.2 KJ/kg.°C

Therefore,

[tex]1080\ KJ = (20\ kg)(4.2\ KJ/kg.^oC)\Delta T[/tex]

ΔT = 12.86°C

d.

First, we will calculate the total energy consumed:

[tex]E=(Power)(Time)\\E=VI(Time)\\E = (120\ V)(30\ A)(0.5\ h/d)(30\ d)\\E = 54000\ Wh\\E = 54 KWh[/tex]

Now, for the cost:

[tex]Cost = (Unit\ Cost)(Energy)\\Cost = (\$ 0.08\KWh)(54\ KWh)[/tex]

Cost = $ 4.32

Answer:

The new velocity of the string is 100 centimeters per second (1 meter per second).

Explanation:

The speed of a wave through a string ([tex]v[/tex]), in meters per second, is defined by the following formula:

[tex]v = \sqrt{\frac{T\cdot L}{m} }[/tex] (1)

Where:

[tex]T[/tex] - Tension, in newtons.

[tex]L[/tex] - Length of the string, in meters.

[tex]m[/tex] - Mass of the string, in kilograms.

The expression for initial and final speeds of the wave are:

Initial speed

[tex]v_{o} = \sqrt{\frac{T_{o}\cdot L_{o}}{m_{o}} }[/tex] (2)

Final speed

[tex]v = \sqrt{\frac{(4\cdot T_{o})\cdot (0.5\cdot L_{o})}{2\cdot m_{o}} }[/tex]

[tex]v = \sqrt{\frac{T_{o}\cdot L_{o}}{m_{o}} }[/tex] (3)

By (2), we conclude that:

[tex]v =v_{o}[/tex]

If we know that [tex]v_{o} = 1\,\frac{m}{s}[/tex], then the new speed of the wave in the string is [tex]v = 1\,\frac{m}{s}[/tex].

Answer:

120 kg•m/s.

Explanation:

From the question given above, the following data were obtained:

Case 1

Mass of object = M

Velocity of object = V

Momentum = 15 kg•m/s

Case 2

Mass of object = 2M

Velocity of object = 4V

Momentum = ?

Momentum is defined as follow:

Momentum = mass × velocity

The momentum of object in case 2 can be obtained as follow:

From case 1

Momentum = mass × velocity

15 = M × V

15 = MV ....... (1)

From case 2:

Momentum = mass × velocity

Momentum = 2M × 4V

Momentum = 8MV ....... (2)

Finally , substitute the value of MV in equation 1 into equation 2.

Momentum = 8MV

MV = 15

Momentum = 8 × 15

Momentum = 120 kg•m/s

Therefore, an object with a mass of 2M and 4V would have a momentum of 120 kg•m/s

Answer:

Wavelength = 1.36 * 10^{-34} meters

Explanation:

Given the following data;

Mass = 0.113 kg

Velocity = 43 m/s

To find the wavelength, we would use the De Broglie's wave equation.

Mathematically, it is given by the formula;

[tex] Wavelength = \frac {h}{mv} [/tex]

Where;

h represents Planck’s constant.

m represents the mass of the particle.

v represents the velocity of the particle.

We know that Planck’s constant = 6.6262 * 10^{-34} Js

Substituting into the formula, we have;

[tex] Wavelength = \frac {6.6262 * 10^{-34}}{0.113*43} [/tex]

[tex] Wavelength = \frac {6.6262 * 10^{-34}}{4.859} [/tex]

Wavelength = 1.36 * 10^{-34} meters

Answer:

"2580 Hz" is the correct solution.

Explanation:

According to the question,

The path difference,

= [tex]4.4 - 3.6[/tex]

= [tex]0.80 \ m[/tex]

Speed,

= 344 m/s

For constructive interference,

⇒  [tex]Path \ difference =n\times \frac{Speed}{frequency}[/tex]

On substituting the values, we get

⇒  [tex]0.80=n\times \frac{344}{frequency}[/tex]

⇒  [tex]frequency=n\times 430[/tex]

⇒                [tex]n=\frac{frequency}{430}[/tex]

If the frequency range is,

f = 2193,

⇒  [tex]n=\frac{2193}{430}[/tex]

        [tex]=5.1[/tex]

If the frequency range is,

f = 2967,

⇒  [tex]n=\frac{2967}{430}[/tex]

        [tex]=6.9[/tex]

hence,

For n = 6, the frequency will be:

⇒  [tex]Frequency=n\times 430[/tex]

                       [tex]=6\times 430[/tex]

                       [tex]=2580 \ Hz[/tex]

Answer:

tbh i think hes right

Explanation:

The femur would be shortened by 0.0434 cm by a load of 600N.

The total applied force per unit of area is known as the pressure.

The pressure depends both on externally applied force as well the area on which it is applied.

The mathematical expression for the pressure

Pressure = Force /Area

As given in the problem assume the femur has a diameter of 3.0cm and a hollow center of 0.8cm diameter and a length of 50cm.if it is supporting a load of 600N,

The stress in the femur = force /area

                                       =600 / π(.004²-.0015²)

                                        = 13.89 ×10⁶ Pa

The length reduced by the load = PL/AE

                                                      = 13.89 ×10⁶ ×50/ 16 × 10⁹ cm

                                                       =0.0434 cm

The femur would be shortened by 0.0434 cm by a load of 600N.

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Answer:

dorsiflexion

Explanation:

To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as: dorsiflexion

Answer:

dorsiflexion

Explanation:

Hope this helps

Answer:

The rate of deceleration is 15 km/h^2.

Explanation:

Since the car comes in 90km/h, it will need deceleration for 15 km/h for 6 seconds to finally stop the car. 15*6=90

Answer:

X = 50 g

Explanation:

Please see attached photo for explanation.

From the attached photo,

Anticlock–wise moment = X × 20

Clockwise moment = 100 × 10

Anticlock–wise moment = clockwise moment

X × 20 = 100 × 10

X × 20 = 1000

Divide both side by 20

X = 1000 / 20

X = 50 g

Therefore, the value of X is 50 g

Answer:

Resistance in circuit = 0.53 ohm (Approx.)

Explanation:

Given:

Flow of current in circuit = 15 amp

Potential difference = 8 Volts

Find:

Resistance in circuit

Computation:

In an electrical system, resistance is a stopper of a material to electric current.

Resistance in circuit = Potential difference / Flow of current in circuit

Resistance in circuit = 8 / 15

Resistance in circuit = 0.53 ohm (Approx.)

Answer:

In a third class lever, the effort is located between the load and the fulcrum. ... If the fulcrum is closer to the effort, then the load will move a greater distance. A pair of tweezers, swinging a baseball bat or using your arm to lift something are examples of third class levers.

Explanation: