A long straight solenoid has 800 turns. When the current in the solenoid is 2.90 amperes the average flux through each turn is 3.25×10−3Wb.
A. What is the inductance of the coil?
B. What must be the magnitude fo the rate of change of the current (di/dt) in order for the self-induced emf to equal 7.50 mV?

Answer:

F = 2.7×10¯⁶ N.

Explanation:

From the question given:

F = (9×10⁹ Nm/C²) (3.2×10¯⁹ C × 9.6×10¯⁹ C) /(0.32)²

Thus we can obtain the value value of F by carrying the operation as follow:

F = (9×10⁹) (3.2×10¯⁹ × 9.6×10¯⁹) /(0.32)²

F = 2.7648×10¯⁷ / 0.1024

F = 2.7×10¯⁶ N.

Therefore, the value of F is 2.7×10¯⁶ N.

Answer:

b

Explanation:

Answer:

Explanation:

In the given case for destructive interference , the condition is,

path difference = (2n+1)λ /2  where n is an integer and λ is wavelength

2 μ d = (2n+1)λ /2

Putting λ = 653 nm

for minimum thickness n = 0

2 μ d = 653 / 2 nm

= 326.5 nm

For constructive interference the condition is

2 μ d = n λ₁

326.5 nm = n λ₁

λ₁ = 326.5 / n  

For n = 1

λ₁ = 326.5 nm ,

or , 326.5nm .

Longest wavelength possible is 326.5

Answer:

1. Energy: ultraviolet>> visible> infrared

2. Frequency: ultraviolet>> visible > infrared

3. Wavelength: infrared >> visible > ultraviolet

Explanation:

Electromagnetic waves are a class of waves that do not require material medium for their propagation, and travel at the same speed. They are arranged with respect to either their decreasing wavelength or increasing frequency to form a spectrum called an electromagnetic spectrum.

Comparing the energy, frequency and wavelength of ultraviolet, infrared and visible regions, it can be deduced that:

1. Energy: ultraviolet has the highest energy, then followed by visible, and infrared has the lowest energy.

i.e energy: ultraviolet>> visible> infrared

2. Frequency: ultraviolet radiation has the highest frequency, visible region has a greater frequency than that of infrared.

i.e frequency: ultraviolet>> visible > infrared

3. Wavelength: infrared radiation has the highest wavelength, followed by visible region, and ultraviolet radiation has the lowest wavelength.

i.e  wavelength: infrared >> visible > ultraviolet

In terms of lowest to the highest energy,frequency, and wavelength is;

Energy: infrared > visible light > ultraviolet

Frequency: infrared > visible light > ultraviolet

Wavelength: ultraviolet > visible light > infrared

The electromagnetic spectrum is made up of all the electromagnetic waves (ultraviolet, infrared, and visible) arranged according to their energy,frequency, and wavelength.

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Answer:

b. Third Law entropy  

Explanation:

Third law entropy: In physics, the term "third law entropy" or "the third law of thermodynamics" states that the specific entropy of a particular system at "absolute zero" is considered as a "well-defined constant". It occurs because any system at "zero temperature"  tends to exists or persists in its "ground state" in order for the entropy to be determined or described only by the "degeneracy" of the given ground state.

In the question above, the correct answer is option b.

Answer:

λmax = 372 nm

Explanation:

First we find the energy of photon:

E = hc/λ

where,

E = Energy of Photon = ?

λ = Wavelength of Light = 233 nm = 2.33 x 10⁻⁷ m

c = speed of light = 3 x 10⁸ m/s

h = Planks Constant = 6.626 x 10⁻³⁴ J.s

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.33 x 10⁻⁷ m)

E = 8.5 x 10⁻¹⁹ J

Now, from Einstein's Photoelectric Equation:

E = Work Function + Kinetic Energy

8.5 x 10⁻¹⁹ J = Work Function + (1.98 eV)(1.6 x 10⁻¹⁹ J/1 eV)

Work Function = 8.5 x 10⁻¹⁹ J - 3.168 x 10⁻¹⁹ J

Work Function = 5.332 x 10⁻¹⁹ J

Since, work function is the minimum amount of energy required to emit electron. Therefore:

Work Function = hc/λmax

λmax = hc/Work Function

where,

λmax = maximum wavelength of light that will produce photoelectrons = ?

Therefore,

λmax = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5.332 x 10⁻¹⁹ J)

λmax = 3.72 x 10⁻⁷ m

λmax = 372 nm

Assuming constant speeds, the P-wave covers a distance d in time t such that

9000 m/s = d/(60 t)

while the S-wave covers the same distance after 1 more minute so that

5000 m/s = d/(60(t + 1))

Now,

d = 540,000 t

d = 300,000(t + 1) = 300,000 t + 300,000

Solve for t in the first equation and substitute it into the second equation, then solve for d :

t = d/540,000

d = 300,000/540,000 d + 300,000

4/9 d = 300,000

d = 675,000

So the earthquake center is 675,000 m away from the seismic station.

Answer:

T_object = 380.35 K

Explanation:

From Stefan–Boltzmann law, the power output is given by the formula:

P = σAT⁴

where;

σ is Stefan-Boltzmann constant

A is area of the radiating surface.

T is temperature of the body

Now, we are told that the power the object emitted is 3 times the power absorbed from the room.

Thus, we have;

P_e = 3P_a

Where P_e is power emitted and P_a is power absorbed.

So, we have;

σA(T_object)⁴ = 3σA (T_room)⁴

σA will cancel out to give;

(T_object)⁴ = 3(T_room)⁴

We are given T_room = 289 K

Thus;

(T_object)⁴ = 3 × 289⁴

(T_object) = ∜(3 × 289⁴)

T_object = 380.35 K

Answer:

B. 1.6 x 10⁻³ m

Explanation:

The magnetic field at the center of the loop is given by;

[tex]B = \frac{\mu_o I }{2R}[/tex]

Where;

μ₀ is the permeability of free space

I is the current in the loop

R is the radius of the circular loop

B is the magnetic field

Given;

diameter of the loop = 1cm

radius of the loop, r = 0.5 cm = 0.005 m

magnetic field, B = 2.5mT = 2.5 x 10⁻³ T

The current in the loop is calculated as;

[tex]I = \frac{2BR}{\mu_o} \\\\I = \frac{2*2.5*10^{-3}*0.005}{4\pi*10^{-7}} \\\\I = 19.89 \ A[/tex]

The magnetic at a distance from the long straight wire is calculated as;

[tex]B = \frac{\mu_o I}{2\pi d}[/tex]

where;

d is the distance from the wire;

[tex]d = \frac{\mu_o I}{2\pi B} \\\\d = \frac{4\pi *10^{-7} * 19.89}{2\pi *2.5*10^{-3}} \\\\d = 1.6 *10^{-3} \ m[/tex]

Therefore, the distance from the wire where the magnetic field is 2.5 mT is 1.6 x 10⁻³ m.

B. 1.6 x 10⁻³ m

This question involves the concepts of the magnetic field due to a loop and a  current-carrying wire and current.

A long straight wire carrying the same current as the loop of wire has a magnetic field of 2.5 mT at a distance of b "B. 1.5 x 10⁻³ m".

The magnetic field at the center of a loop of wire is given by the following formula:

[tex]B=\frac{\mu_o I}{2r}[/tex]

where,

B = Magnetic Field = 2.5 mT = 2.5 x 10⁻³ T

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

I = current = ?

r = radius = diameter/2 = 1 cm/2 = 0.5 cm = 0.005 m

Therefore,

[tex]I = \frac{(2.5\ x\ 10^{-3}\ T)(2)(0.005\ m)}{4\pi\ x\ 10^{-7}\ N/A^2}[/tex]

I = 19.9 A

Now, the magnetic field at a distance from the straight wire is given by the following formula:

[tex]B=\frac{\mu_o I}{2\pi R}[/tex]

where,

R = distance from wire = ?

Therefore,

[tex]R = \frac{(4\pi \ x \ 10^{-7}\ N/A^2)(19.9\ A)}{2\pi(2.5\ x\ 10^{-3}\ T)}[/tex]

R = 1.6 x 10⁻³ m

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Explanation:

According to Ohms Law :

V = I * R

(A) R (Resistance) = 0.022 / 0.75 = 0.03 Ohms

Also,

[tex]r = \alpha \frac{length}{area} = \alpha \frac{5.8}{3.14 \times 0.001 \times 0.001} [/tex]

(B)

[tex] \alpha(resistivity) = 1.62 \times {10}^{ - 8} [/tex]

Drift speed is missing. It is given as;

1.7 × 10^(-5) m/s

A) R = 0.0293 ohms

B) ρ = 1.589 × 10^(-8)

C) n = 8.8 × 10^(28) electrons

This is about finding, resistance and resistivity.

Length; L = 5.8 m

Diameter; d = 2mm = 0.002 m

Radius; r = d/2 = 0.001 m

Voltage; V = 22 mv = 0.022 V

Current; I = 750 mA = 0.75 A

Area; A = πr² = 0.001²π

Drift speed; v_d = 1.7 × 10^(-5) m/s

R = V/I

R = 0.022/0.75

R = 0.0293 ohms

ρ = RA/L

ρ = (0.0293 × 0.001²π)/5.8

ρ = 1.589 × 10^(-8)

J = n•e•v_d

Where;

J = I/A = 0.75/0.001²π A/m² = 238732.44 A/m²

e is charge on an electron = 1.6 × 10^(-19) C

v_d = 1.7 × 10^(-5) m/s

n is number of free electrons per unit volume

Thus;

238732.44 = n(1.6 × 10^(-19) × 1.7 × 10^(-5))

238732.44 = (2.72 × 10^(-24))n

n = 238732.44/(2.72 × 10^(-24))

n = 8.8 × 10^(28)

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The correct answer is B. Calcite

Explanation:

Mohs hardness scale indicates the hardness of minerals using a scale from 1 to 10 as well as defining the objects or tools that can be used to scratch the minerals. These two features of minerals are shown in the table of the image. About this, it is shown gypsum and talc can be scratched by just a fingernail, considering minerals with a hardness of 2.5 or below can be scratched by a fingernail. In the case of calcite that has a hardness of 3, this cannot be scratched by a fingernail, but it can be scratched by a penny, which works for minerals with a hardness of 3.5 or below. Thus, the correct answer is Calcite.

Answer:

The power of the mirror in diopters is 58.8 D

Explanation:

Given;

radius of curvature of the spoon, R = 3.4 cm = 0.034 m

The focal length of a mirror is given by;

[tex]f = \frac{R}{2} \\\\f = \frac{0.034}{2} \\\\f = 0.017 \ m[/tex]

The focal length of the mirror is 0.017 m

(b) The power of the mirror is given by;

[tex]P = \frac{1}{f}[/tex]

where;

P is the power of the mirror

f is the focal length

[tex]P = \frac{1}{f}\\\\P= \frac{1}{0.017}\\\\P = 58.8 \ D[/tex]

Thus, the power of the mirror in diopters is 58.8 D

Answer:

0.5639m

Explanation:

For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ

=sin⁻¹(mλ/d)

mλ /d =y/L

for the first order,

y= mλL/d

For ratio separation y₀/yD=1 and d= 1

y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]

1=λ ₀L₀/λD.LD.

λD.LD= λ ₀L₀

L₀= λD.LD/ λ ₀..............(1)

Then substitute the given values into (1) we have

L₀=471 *0.497/611

= 0.3831m

Distance by which the screen has to be moved towards the slit is

LD- Lo

0.947-0.3831= 0.5639m

Answer:

p = -q  

he distance is equal to the current distance, so the distance does not change

Explanation:

For this exercise we can solve it using the equation of the constructor

            1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

For a flat surface the radius is at infinity, therefore 1 / f = 0, which implies

          1 / p = - 1 / q

           p = -q

Therefore the distance is equal to the current distance, so the distance does not change

Answer:

The charge is  [tex]Q = 2.177 *10^{-9} \ C[/tex]

Explanation:

From the question we are told that

     The electric potential is  [tex]V = 25.0 \ kV = 25.0 *10^{3}\ V[/tex]

     The  mass of the drop is  [tex]m = 0.168 \ m g = 0.168 *10^{-3} \ g = 0.168 *10^{-6}\ kg[/tex]

      The  speed is  [tex]v = 18.8 \ m/s[/tex]

Generally the charge on the paint drop due to the electric potential which will give it the speed stated in the question  is mathematically represented as

       [tex]Q = \frac{m v^2 }{ 2 * V }[/tex]

Substituting values

      [tex]Q = \frac{0.168 *10^{-6} (18)^2 }{ 2 * 25*10^3 }[/tex]

       [tex]Q = 2.177 *10^{-9} \ C[/tex]

Explanation:

gravitional force attracts objects with mass toward each other.

Answer:

lateral splaying of echoes in the far field can be improved by Increasing the maximum number of transmit focal zones and optimize their location.

Answer:

a

   [tex]\lambda = 1.667 nm[/tex]

b

     [tex]\theta = 0.8681^o[/tex]

Explanation:

From the question we are told that

   The distance of separation is [tex]d = 0.220 \ mm = 0.00022 \ m[/tex]

    The  is distance of the screen from the slit is  [tex]D = 2.60 \ m[/tex]

    The distance between the central bright fringe and either of the adjacent bright   [tex]y = 1.97 cm = 1.97 *10^{-2}\ m[/tex]

Generally  the condition for constructive interference is  

      [tex]d sin \tha(\theta ) = n \lambda[/tex]

From the question we are told that small-angle approximation is valid here.

So    [tex]sin (\theta ) = \theta[/tex]

=>        [tex]d \theta = n \lambda[/tex]

=>        [tex]\theta = \frac{n * \lambda }{d }[/tex]

Here n is the order of maxima and the value is  n =  1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright  is mathematically represented as

         [tex]y = D * sin (\theta )[/tex]

From the question we are told that small-angle approximation is valid here.

So

       [tex]y = D * \theta[/tex]

=>   [tex]\theta = \frac{ y}{D}[/tex]

So

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

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

substituting values

       [tex]\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }[/tex]

        [tex]\lambda = 1.667 *10^{-6}[/tex]

        [tex]\lambda = 1.667 nm[/tex]

In the b part of the question we are considering the next set of bright fringe so  n=  2

    Hence

     [tex]dsin (\theta ) = n \lambda[/tex]

    [tex]\theta = sin^{-1}[\frac{ n * \lambda }{d} ][/tex]

    [tex]\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ][/tex]

    [tex]\theta = 0.8681^o[/tex]

Answer:

Unbalenced

Explanation:

when balenced forces are applied to an object there is no motion. When you apply unbalenced force the object you are applying the force to will move in the opposite direction of the force.

Answer:

im pretty sure it unbalenced

Explanation:

i just am

Answer:

8 mm

Explanation:

From the information given:

The Ampere circuital law can be used to estimate the magnetic field strength at two points when the distance is less than the radius and when the distance is greater than the radius.

when the distance is less than the radius ; we have:

[tex]B_1 = \dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2}[/tex]

when the distance is greater than the radius; we have:

[tex]B_2 = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]

Equating both equations together ; we have :

[tex]\dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2} = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]

[tex]\dfrac{1}{R}= \dfrac{r}{d^2}[/tex]

[tex]R= \dfrac{d^2}{r}[/tex]

where; d = radius of the wire and r = distance;

[tex]R =\dfrac{4^2}{2}[/tex]

[tex]R =\dfrac{16}{2}[/tex]

R = 8 mm

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:

The question is not complete, here is the other part.

Assume that n = 1.36.

Express your answer to three significant figures and include the appropriate units.

λ = 707.2nm

Explanation:

n = 1.36

t = 130

2 n t= (m+1/2) λ

To solve for λ.

λ = 2 n t ÷ m+1/2

λ = 2 × 1.36 × 130 ÷ m +1/2

λ =

If m= 0

λ= 353.6 ÷ 0+ 1/2

λ = 353.6 × 2

= 707.2nm

Answer

The effect is that it Decreases the field current IF and increases slope K1

Answer:

a) the distance between the interference fringes is reduced by half

b) the distance between stripes is doubled

Explanation:

Interference experiments constructive interference is described by the expression

          d sin θ = m λ

let's use trigonometry to find the distance between the interference fringes

              tan θ=  y / L

dodne y is the distance from the central maximum, L the distance from the slit to the observation screen. In general these experiments are carried out at very small angles

            tan θ = sin θ / cos θ = sin θ

we substitute

             sin θ = y / L

            d y / L = m  λ

           y = m λ / d L

a) it asks us when the screen doubles its distance

           L ’= 2 L

subtitute in the equation

           y ’= m λ / (d 2L)

           y ’=( m λ / d L) /2

           y ’= y / 2

the distance between the interference fringes is reduced by half

b) the density of the network doubles

      if the density doubles in the same distance there are twice as many slits, so the distance between them is reduced by half

            d ’= d / 2

we substitute

          y ’= m λ (L d / 2)

          y ’= m λ / (L d) 2

          y ’= y 2

the distance between stripes is doubled

Answer:

The radius is  [tex]r = 3.1905 \ m[/tex]

Explanation:

From the question we are told that

        The  distance  beneath the liquid  is  [tex]d = 2.70 \ m[/tex]

        The refractive index of the liquid is  [tex]n_i = 1.31[/tex]

Now the critical value is mathematically represented as

         [tex]\theta = sin ^{-1} [\frac{1}{n_i} ][/tex]

substituting values

         [tex]\theta = sin ^{-1} [\frac{1}{131} ][/tex]

         [tex]\theta = 49.76^o[/tex]

Using SOHCAHTOA rule we have that

         [tex]tan \theta = \frac{ r}{d}[/tex]

=>     [tex]r = d * tan \theta[/tex]

substituting values  

        [tex]r = 2.7 * tan (49.76)[/tex]

        [tex]r = 3.1905 \ m[/tex]

Answer:static friction and kinetic friction in a system always act in the same direction as each other and n the opposite direction of the applie force . Is the correct answer

Explanation:

Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the applied force. The correct option is B.

Friction is the force that prevents one hard material from scooting or rolling over the other.

Frictional forces, such as the locomotion required to walk without dropping, are advantageous, but they also create a significant amount of resistance to motion.

We can control cars because of friction between the tires and the road: more precisely, because there are three types of friction: rolling friction, starting friction, and sliding friction.

Friction reduces the speed of moving objects and can even stop them from moving. The friction between the objects generates heat. As a result, energy is wasted in the machines. Friction will cause wear and tear on the machine parts.

In a system, static and kinetic friction always act in the same direction and in the opposite direction of the applied force.

Thus, the correct option is B.

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Answer:

to overcome the out of friction we must increase the angle of the plane

Explanation:

To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.

X axis

       fr - Wₓ = m    a                      (1)

Y axis  

       N- [tex]W_{y}[/tex] = 0

       N = W_{y}

let's use trigonometry to find the components of the weight

        sin θ = Wₓ / W

        cos θ = W_{y} / W

        Wₓ = W sin θ

        W_{y} = W cos θ

the friction force has the formula

         fr = μ N

         fr = μ Wy

         fr = μ mg cos θ

from equation 1

at the point where the force equals the maximum friction force

in this case the block is still still so a = 0

           F = fr

           F = (μ  mg) cos θ

We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.

This is the force that balances the friction force, any force slightly greater than F initiates the movement.

Consequently, to overcome the out of friction we must increase the angle of the plane

the correct answer is to increase the angle of the plane

Answer:

Time, [tex]t=1.07\times 10^{-23}\ s[/tex]

Explanation:

Given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel [tex]3.2\times 10^{-15}\ m[/tex] before interacting.

Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,

[tex]t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s[/tex]

So, the time interval is [tex]1.07\times 10^{-23}\ s[/tex]

Answer:

5.465 × 10^10 revolutions

Explanation:

Formula for Magnetic Field = m. v/ q . r

M = mass of electron = mass of positron = 9.11 x 10^-31 kg,

radius of the positron = 5.03 mm

We convert to meters.

1000mm = 1m

5.03mm = xm

Cross multiply

x = 5.03/1000mm

x = 0.00503m

q = Electric charge = -1.6 x 10^-19 C

Magnetic field (B) = 0.85 T

Speed of the positron is unknown

0.85 = 9.11 x 10^-31 kg × v/ -1.6 x 10^-19 C × 0.00503

0.85 × 1.6 x 10^-19 C × 0.00503 = 9.11 x 10^-31 kg × v

v = 0.85 × -1.6 x 10^-19 C × 0.00503/9.11 x 10^-31 kg

v = 6.8408 ×10-22/ 9.11 x 10^-31 kg

v = 750911086.72m/s

Formula for complete revolutions =

Speed × time / Circumference

Time = 2.30s

Circumference of the circular path = 2πr

r =0.00503

Circumference = 2 × π × 0.00503

= 0.0316044221

Revolution = 750911086.72 × 2.30/0.0316044221

= 1727095499.5/0.0316044221

= 546541562294 revolutions

Approximately = 5.465 × 10^10 revolutions

Answer:

The actual height is 3.308 m.

Explanation:

The person is swimming below the water surface at distance = 0.80 m  

The height of the diving board appears at a distance or height = 5.20 m

Now we have to find the actual distance of the diving board from the water surface.

We know the refractive index of water is 1.33.

Therefore, the actual height = (Distance that appears – distance below the water surface) / Refractive index.

The actual height = ( 5.20 - 0.80 ) / 1.33 = 3.308 m