In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal:_____.

Answer:

a) 3.6 ft

b) 12.4 ft

Explanation:

Distance between mirrors = 6.2 ft

difference from from the mirror you face = 1.8 ft

a) you stand 1.8 ft in front of the mirror you face.

According to plane mirror rules, the image formed is the same distance inside the mirror surface as the distance of the object (you) from the mirror surface. From this,

your distance from your first "front" image = 1.8 ft + 1.8 ft = 3.6 ft

b) The mirror behind you is 6.2 - 1.8 = 4.4 ft behind you.

the back mirror will be reflected 3.6 + 4.4 = 8 ft into the front mirror,

the first image of your back will be 4.4 ft into the back mirror,

therefore your distance from your first "back" image = 8 + 4.4 = 12.4 ft

Answer:

a. 7.62cm

b. Real and inverted

c. 2.76 cm

d. 3450

Explanation:

We proceed as follows;

a. the lens equation that relates the object distance to the image distance with the focal length is given as follows;

1/f = 1/p + 1/q

making q the subject of the formula;

q = pf/p-f

From the question;

p = 4.70m

f = 7.5cm = 0.075m

Substituting these, we have ;

q = (4.7)(0.075)/(4.7-0.075) = 0.3525/4.625 = 0.0762 = 7.62 cm

b. The image is real and inverted since the image distance is positive

c. We want to calculate how tall the image is

Mathematically;

h1 = (q/p)h0

h1 = (7.62/4.70)* 1.7

h1 = 2.76 cm

d. We want to calculate the number of pixels that fit into this image

Mathematically:

n = h1/8 micro meter

n = 2.76cm/8 micro meter = 2.76 * 10^-2/8 * 10^-6 = 3450

Answer:

d. unchanged.

Explanation:

The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.

In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from

v = fλ

that the frequency is tied to the wave, and does not change throughout the waveform.

where v is the speed of the sound wave

f is the frequency

λ is the wavelength of the sound wave.

Answer:

   θ  = 45º

Explanation:

The light that falls on the second polarized is polarized, therefore it is governed by the law of Maluz

              I = I₀ cos² θ

in the problem they ask us

            I = ½ I₀

let's look for the angles

             ½ I₀ = I₀ cos² θ

             cos θ  = √ ½ = 0.707

            θ  = cos 0.707

           θ  = 45º

Answer:

A) 0.4 mA

B) 0.03 mA

Explanation:

Given that

voltage source, V = 120 V

to solve this question, we would be using the very basic Ohms Law, that voltage is proportional to the current and the resistance passing through the circuit, if temperature is constant.

mathematically, Ohms Law, V = IR

V = Voltage

I = Current

R = Resistance

from question a, we were given 300kΩ, substituting this value of resistance in the equation, we have

120 = I * 300*10^3 Ω

making I the subject of the formula,

I = 120 / 300000

I = 0.0004 A

I = 0.4 mA

Question said, currents above 10 mA causes involuntary muscle contraction, this current is way below 10 mA, so nothing happens.

B, we have Resistance, R = 4000kΩ

Substituting like in part A, we have

120 = I * 4000*10^3 Ω

I = 120 / 4000000

I = 0.00003 A

I = 0.03 mA

This also means nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA

The current through a person will be:

a) 0.4 mA

b) 0.03 mA

Given:

Voltage, V = 120 V

It states that the voltage or potential difference between two points is directly proportional to the current or electricity passing through the resistance, and directly proportional to the resistance of the circuit.

Ohms Law, V = I*R

where,

V = Voltage

I = Current

R = Resistance

a)

Given: Resistance=  300kΩ

[tex]120 = I * 300*10^3 ohm\\\\I = 120 / 300000\\\\I = 0.0004 A[/tex]

Thus, current will be, I = 0.4 mA

b)

Given: R = 4000kΩ

[tex]120 = I * 4000*10^3 ohm\\\\I = 120 / 4000000\\\\I = 0.00003 A[/tex]

Thus, current will be, I = 0.03 mA

From calculations, we observe that nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA.

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Answer:

30.93 cm

Explanation:

Given that:

A person with a near point of 85 cm, but excellent distance vision normally wears corrective glasses

The power of the old pair of lens p = 2.25 diopters

The focal point length = 1/p

The focal point length =  1/2.25

The focal point length = 0.444 m

The focal point length = 44.4 cm

The near point of the person from the glass = (85 -2)cm , This is because the glasses are usually 2 cm from the lens

The near point of the person from the glass = 83 cm

Let consider s' to be the image on the same sides of the lens,

∴ s' = -83 cm

We known that:

the focal length of a mirror image 1/f =1/u +1/v

Assume the near point is at an excellent distance s from the glass where the person wears the corrective glasses.

Then:

1/f = 1/s + 1/s'

1/s = 1/f - 1/s'

1/s = (s' -f)/fs'

s = fs'/(s'-f)

s =( 44.4× -83)/(-83 - 44.4)

s = - 3685.2 / - 127.4

s = 28.93 cm

Thus , the near distance point measured from the eye wearing the old glasses, if they rest 2.0 cm in front of the eye = (28.93 +2.0)cm

= 30.93 cm

Answer:

Significant

Explanation:

Valid digits in measurements are called significant digits, or also called significant figures.

These significant digits allow data and measurements to be more accurate and exact.

Answer:

Significant digits

Explanation

Took the test got it right

Answer:

The number is  [tex]Z = 216 \ fringes[/tex]

Explanation:

From the question we are told that

      The wavelength is  [tex]\lambda = 520 \ nm = 520 *10^{-9} \ m[/tex]

       The length of the glass plates is [tex]y = 21.1cm = 0.211 \ m[/tex]

      The distance between the plates (radius of wire ) =  [tex]d = 0.028 mm = 2.8 *10^{-5} \ m[/tex]

   Generally the condition for constructive  interference in a film is mathematically represented as

            [tex]2 * t = [m + \frac{1}{2} ]\lambda[/tex]

Where  t is the thickness of the separation between the glass i.e  

    t  = 0 at the edge where the glasses are touching each other and  

     t =  2d at the edge where the glasses are separated by the wire  

   m is the order of the fringe it starts from  0, 1 , 2 ...

So  

       [tex]2 * 2 * d = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]

=>   [tex]2 * 2 * (2.8 *10^{-5}) = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]

=>    

       [tex]m = 215[/tex]

given that we start counting m from zero

   it means that the number of  bright fringes that would appear is

         [tex]Z = m + 1[/tex]

=>    [tex]Z = 215 +1[/tex]

=>     [tex]Z = 216 \ fringes[/tex]

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

[tex]V(t) = \epsilon. e^{-t.\frac{L}{R} }[/tex]

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

[tex]0.8*19 = 19. e^{-4.\frac{22}{R} }[/tex]

[tex]0.8 = e^{-\frac{88}{R} }[/tex]

[tex]ln(0.8) = ln(e^{-\frac{88}{R} })[/tex]

[tex]ln(0.8) = -\frac{88}{R}[/tex]

[tex]R = -\frac{88}{ln(0.8)}[/tex]

R = 394.36

In this LR circuit, the resistance of the resistor is 394.36ohms.

Answer:

Option (D) : 0.5 kg

Explanation:

[tex]mass = density \times volume[/tex]

[tex]mass = {2500} \times 0.1 \times 0.05 \times 0.04[/tex]

Mass of block = 0.5 kg

the mass of a rectangular block of density 2.5 ×10³ k gm³ that measures 10cm by 5 cm by 4 cm is 0.5 kg.

Density is the ratio of mass to volume. it tells how much mass a body is having for its unit volume. for example egg yolk has 1027kg/m³ of density, means if we collect numbers of egg yolk and keep it in a container having volume 1 m³ then total amount of mass it is having will be 1027kg. Density is a scalar quantity. when we add egg yolk into the water, egg yolk has greater density than water( 997 kg/m³), because of higher density of egg yolk it contains higher mass in same volume as water. hence due to higher mass higher gravitational force is acting on the egg yolk therefore it goes down on the inside the water. water will float upon the egg yolk. same situation we have seen when we spread oil in the water. ( in that case water has higher density than oil. thats why oil floats on the water)

The Volume of the block is,

V = LBD, where L = length, B = breadth , D = depth of the block.

V = 10 × 5 × 4 = 200 cm³

Density of Block = 2.5 ×10³ kg/m³

Density = Mass / Volume

2.5 ×10³ kg/m³ =  Mass /  200 cm³

2.5 ×10³ kg/m³ × 200 cm³ =  Mass

2.5 ×10³ kg/m³ × 0.2 × 10⁻³ m³ =  Mass

Mass = 0.5 kg

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Answer:

The output force of the piston is 105 N.

Explanation:

Given;

the area of the input piston, A₁ = 0.5 m²

the input force of the piston, F₁ = 15 N

the area of the output piston, A₀ = 3.5 m²

the output force of the piston, F₀ = ?

The pressure of the  hydraulic lift is given by;

[tex]P = \frac{F}{A}[/tex]

where;

P is the hydraulic pressure

F is the piston force

A is the area of the piston

[tex]P = \frac{F}{A} \\\\\frac{F_o}{A_o} = \frac{F_i}{A_i} \\\\F_o = \frac{F_iA_o}{A_i} \\\\F_o = \frac{15*3.5}{0.5} \\\\F_o = 105 \ N[/tex]

Therefore,  the output force of the piston is 105 N.

Thw question is not complete. The complete question is;

Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC

Answer:

Option C: E = 75 N/C

Explanation:

We are given;

Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m

Distance on the y-axis; d = 1.6 m

Now, the formula for electric field with uniform linear density is given as;

E = λ/(2•π•r•ε_o)

Where;

E is electric field

λ is uniform linear density = 6.7 × 10^(-9) C/m

r is distance = 1.6m

ε_o is a constant = 8.85 × 10^(-12) C²/N.m²

Thus;

E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))

E = 75.31 N/C ≈ 75 N/C

Answer:

Torque = 0.012 N.m

Explanation:

We are given;

Mass of wheel;m = 750 g = 0.75 kg

Radius of wheel;r = 25 cm = 0.25 m

Final angular velocity; ω_f = 0

Initial angular velocity; ω_i = 220 rpm

Time taken;t = 45 seconds

Converting 220 rpm to rad/s we have;

220 × 2π/60 = 22π/3 rad/s

Equation of rotational motion is;

ω_f = ω_i + αt

Where α is angular acceleration

Making α the subject, we have;

α = (ω_f - ω_i)/t

α = (0 - 22π/3)/45

α = -0.512 rad/s²

The formula for the Moment of inertia is given as;

I = ½mr²

I = (1/2) × 0.75 × 0.25²

I = 0.0234375 kg.m²

Formula for torque is;

Torque = Iα

For α, we will take the absolute value as the negative sign denotes decrease in acceleration.

Thus;

Torque = 0.0234375 × 0.512

Torque = 0.012 N.m

Answer:

A) P_rad.abs = 2.4 × 10^(-8) Pa and P_rad.ref = 4.8 × 10^(-8) Pa

B) P_rad.abs = 2.369 × 10^(-13) atm and P_rad.ref = 4.738 × 10^(-13) atm

Explanation:

A) The formula for radiation pressure for absorbed light is given as;

P_rad = I/c

Where I is the intensity = 7.20 W/m² and c is the speed of light = 3 × 10^(8) m/s

Thus;

P_rad = 7.2/(3 × 10^(8))

P_rad.abs = 2.4 × 10^(-8) Pa

Now formula for radiation pressure for reflected light is given as;

P_rad = 2I/c

Thus;

P_rad = (2 × 7.2)/(3 × 10^(8))

P_rad.ref = 4.8 × 10^(-8) Pa

B) Now, 1.013 × 10^(5) Pa = 1 atm

Thus, for the absorbed surface, we have;

P_rad.abs = (2.4 × 10^(-8))/(1.013 × 10^(5))

P_rad.abs = 2.369 × 10^(-13) atm

For the reflecting surface, we have;

P_rad_ref = (4.8 × 10^(-8))/(1.013 × 10^(5))

P_rad.ref = 4.738 × 10^(-13) atm

Answer:

Δx = 4.68 x 10⁻³ m = 4.68 mm

Explanation:

The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:

Δx = λD/d

So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:

Δx = 4λD/d

where,

Δx = distance between eighth order maximum and fourth order maximum=?

λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m

d = slit separation = 0.2 mm = 2 x 10⁻⁴ m

D = Distance between slits and screen = 48 cm = 0.48 m

Therefore,

Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)

Δx = 4.68 x 10⁻³ m = 4.68 mm

Answer:

  t = 1.81 min ,     the correct answer is c

Explanation:

This is a missile throwing exercise

The object is thrown horizontally, so its vertical speed is zero (voy = 0), let's use the equation

             y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

the final height is y = 0 and the initial height is y₀ = 22000 m

            0 = y₀ + 0 - ½ g t²

            t = √y 2y₀ / g

let's calculate

           t = √(2  22000 / 3.72)

           t = 108.76 s

let's reduce to minutes

           t = 108.76 s (1 min / 60 s)

           t = 1.81 min

The correct answer is c

Answer:

Therefore maximum stretch is y2 = 32.36 m

Explanation:

In this problem let's use the initial data to find the string constant, let's apply Newton's second law when in equilibrium

        [tex]F_{e}[/tex] - W = 0

         k Δx = mg

         k = mg / Δx

         k = 80 9.8 / (30-20)

         k = 78.4 N / m

now let's use conservation of energy to find the velocity of the body just as the string starts to stretch y = 20 m

starting point. When will you jump

         Em₀ = U = mg y

final point. Just when the rope starts to stretch

         [tex]Em_{f}[/tex] = K = ½ m v²

         Em₀ = Em_{f}

          mg y = ½ m v²

          v = √ 2g y

          v = √ (2 9.8 20)

          v = 19.8 m / s

now all kinetic energy is transformed into elastic energy

starting point

            Em₀ = K = ½ m v²

final point

            Em_{f} = [tex]K_{e}[/tex] + U = ½ k y² + m g y

            Emo = Em_{f}

           ½ m v² = ½ k y² + mgy

            k y² + 2 m g y - m v² = 0

we substitute the values ​​and solve the quadratic equation

            78.4 y² + 2 80 9.8 y - 80 19.8² = 0

            78.4 y² + 1568 y - 31363.2 = 0

              y² + 20 y - 400 = 0

              y = [- 20 ±√ (20 2 +4 400)] / 2

              y = [-20 ± 44.72] / 2

the solutions are

              y₁ = 12.36 m

              y₂ = 32.36 m

These solutions correspond to the maximum stretch and its rebound.

Therefore maximum stretch is y2 = 32.36 m

Answer:

constructive interferencia  0, 1 , 2 m

destructive inteferencia   1/4, 3/4. 5/4, 7/4 m

Explanation:

This exercise is equivalent to the double slit experiment, the two sources are in phase and separated by a distance, therefore the waves observed in the line between them have an optical path difference and a phase difference, given by the expression

            Δr / λ = Φ / 2π

            Δr = Φ/2π   λ

let's apply this expression to our case

λ = 1 m

            Δr = Φ 1 / 2π

We have constructive interference for angle of  Φ = 0, 2π, ...

let's find the values ​​where they occur

  Φ         Δr

   0          0

  2π         1

  4π        2

Destructive interference occurs by    Φ = π /2, 3π / 2, ...

 Φ          Δr

 π/2       ¼ m

 3π /2    ¾ m

5π /2     5/4 m

7π /2      7/4 m

Answer: R=24.2Ω

Explanation: Power is rate of work being done in an electric circuit. It relates to voltage, current and resistance through the following formulas:

P=V.i

P=R.i²

[tex]P=\frac{V^{2}}{R}[/tex]

The resistance of the system is:

[tex]P=\frac{V^{2}}{R}[/tex]

[tex]R=\frac{V^{2}}{P}[/tex]

[tex]R=\frac{110^{2}}{500}[/tex]

R = 24.2Ω

For the device, resistance is 24.2Ω.

Answer:

A = 6.8 km²

Explanation:

A) The sail should be reflective. This is so that, it can produce the maximum radiation pressure.

B) let's begin with the formula used to calculate the average solar sail in orbit around the sun. Thus;

F_rad = 2IA/c

I is given by the formula;

I = P/(4πr²)

Thus;

F_rad = (2A/c) × (P/(4πr²)) = PA/2cπr²

Where;

A is the area of the sail

r is the distance of the sail from the sun

c is the speed of light = 3 × 10^(8) m/s

P is total power output of the sun = 3.90 × 10^(26) W

Now,F_rad = F_g

Where F_g is gravitational force.

Thus;

PA/2cπr² = G•m•M_sun/r²

r² will cancel out to givw;

PA/2cπ = G•m•M_sun

Making A the subject, we have;

A = (2•c•π•G•m•M_sun)/P

Now, m = 1.06 × 10⁴ kg and M_sun has a standard value of 1.99 × 10^(30) kg

G is gravitational constant and has a value of 6.67 × 10^(-11) Nm²/kg²

Thus;

A = (2 × 3 × 10^(8) × π × 6.67 × 10^(-11) × 1.06 × 10^(4) × 1.99 × 10^(30))/(3.90 × 10^(26))

A = 6.8 × 10^(6) m² = 6.8 km²

Answer:

0.09N, attractive

Explanation:

It can be deducted from the question that the currents are arranged in parallel settings, then it is obvious that the force on each of the wire will be attractive toward the other wire.

the magnitude of force can be determined by using below formula;

F2 = (μ₀/2π)(I₁I₂/d)I₂

μ₀ = constant = 4π × 10^-7 H/m,

I₁, I₂ = currents= 30A

L = the length o the wire=30m

d = distance between these two wires= 0.06m

Since the current are arranged in the same direction, they exhibit attractive force on each other.

Then plugging the values Into the formula above we have

F₂ = (4π × 10^-7 T.m/A)/2π) × ((30A)²/ 0.06m)× 30 m

= 0.09 N, attractive

Therefore, the magnitude and direction of the force is 0.09 N, attractive

Answer:

v = 0.89 c = 2.67 x 10⁸ m/s

Explanation:

The time dilation consequence of the special theory of relativity shall be used here, From time dilation formula we have:

t = t₀/√[1 - v²/c²]

where,

t = time measured by the person on earth = 2.50 s

t₀ = rest time of the intergalactic rock star = 1.30 s

v = relative speed of the rock star = ?

Therefore,

2.5 s = (1.3 s)/√[1 - v²/c²]

√[1 - v²/c²] = 1.3/2.5

√[1 - v²/c²] = 0.52

[1 - v²/c²] = 0.52²

[1 - v²/c²] = 0.2074

v²/c² = 1 - 0.2074

v²/c² = 0.7926

v/c = √0.7926

v = 0.89 c

where,

c = speed of light = 3 x 10⁸ m/s

v = (0.89)(3 x 10⁸ m/s)

v = 0.89 c = 2.67 x 10⁸ m/s

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F   ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Answer:

The length of the longer pipe is L = 2.30 m

Explanation:

Given that:

Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long.

How long is the other pipe?

From above;

The formula for the frequency of open ended pipes can be expressed as:

[tex]f = \dfrac{nv}{2L}[/tex]

where n = 1 ( since half wavelength exist between those two pipes)

v = 343 m/s  and L = 2.08 m

Thus, the shorter pipe produces a frequency of :

[tex]f = \dfrac{1*343}{2*2.08}[/tex]

[tex]f = \dfrac{343}{4.16}[/tex]

[tex]f =82.45 \ Hz[/tex]

Also; we know that the beat frequency was given as 8.0 Hz

Then,

The lower frequency of the longer pipe = ( 82.45 - 8.0 )Hz

The lower frequency of the longer pipe = 74.45 Hz

Finally;

From the above equation; make Length L the subject of the formula. Then,

The length of the longer pipe is L = [tex]\dfrac{nv}{2f}[/tex]

The length of the longer pipe is L = [tex]\dfrac{1*343}{2*74.45}[/tex]

The length of the longer pipe is L = [tex]\dfrac{343}{148.9}[/tex]

The length of the longer pipe is L = 2.30 m

Answer:

Capacitor, is the right answer.

Explanation:

The unknown element is a Capacitor.

Below is the calculation that proves that it is a capacitor.

We know that for the Capacitor

i = Imax × sin(wt+(pi/2)).

i = Imax × sin ((2 × pi/T) × (T/4) + (pi/2))

i = Imax × sin(3.142) = 0 A

at, t = T/2

wt = (2 × pi/T) × (T/2) = pi

wt + (pi/2) = pi + (pi/2) = ( 3 × pi/2) =

i = Imax × sin(3 × pi/2) = -Imax

Which is in a correct agreement with capacitor  therefore, the answer is a Capacitor.

Answer

3.340J

Explanation;

Using the relation. Energy stored in capacitor = U = 7.72 J

U =(1/2)CV^2

C =(eo)A/d

C*d=(eo)A=constant

C2d2=C1d1

C2=C1d1/d2

The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.

Initial separation between the plates =d1= 3.30mm .

Final separation = d2 = 1.45 mm

(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same

Energy=U =(1/2)q^2/C

U2C2 = U1C1

U2 =U1C1 /C2

U2 =U1d2/d1

Final energy = Uf = initial energy *d2/d1

Final energy = Uf =7.72*1.45/3.30

(A) Final energy = Uf = 3.340J

Answer:

450 x10^-6 T

Explanation:

We know that the magnetic of each wire is derived from

ByB= uoi/2pir

Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)

= 0.2 x 10^ -3T

B2=

4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)

= 0.4 x 10^ -3T

So

(Bnet)² = (Bx)² + ( By)²

= (0.2² + 0.4²)mT

= 450 x10^-6T

The magnitude of magnetic field at the origin is required.

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

x = Location at x axis = 5 cm

y = Location at y axis = 4 cm

[tex]I_x[/tex] = Current at the x axis point = 50 A

[tex]I_y[/tex] = Current at the y axis point = 80 A

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]

Magnitude of the magnetic field is given by

[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]

Finding the net magnetic field using the Pythagoras theorem

[tex]B^2=B_x^2+B_y^2\\\Rightarrow B^2=\left(\dfrac{\mu_0I_x}{2\pi x}\right)^2+\left(\dfrac{\mu_0I_y}{2\pi y}\right)^2\\\Rightarrow B=\dfrac{\mu_0}{2\pi}\sqrt{\left(\dfrac{I_x}{x}\right)^2+\left(\dfrac{I_y}{y}\right)^2}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}}{2\pi}\sqrt{\left(\dfrac{50}{0.05}\right)^2+\left(\dfrac{80}{0.04}\right)^2}\\\Rightarrow B=0.0004472=447.2\ \mu\text{T}[/tex]

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

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Answer:

The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

Explanation:

The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.

When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.

So the centripetal acceleration, a' = 16v²/R.

To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16

a'/a = 16

a' = 16a.

So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

Answer:

269 lb

Explanation:

We first find the tangential acceleration, a on the drums

a = Δv/Δt since the speed of the belt is 11 ft/s after 0.16 s, Δv = 11 ft/s and Δt = 0.16 s

a = Δv/Δt = 11 ft/s ÷ 0.16 s = 68.75 ft/s²

Since torque τ = Tk = Iα where I = moment of inertia of larger drum = Mk² where m = mass of larger drum = 4 lbs, k = radius of gyration of larger drum = 1.25 inches, T = tension due to larger drum and α = angular acceleration of larger drum.

So, T = Iα/k = Mk²α/k = Mαk = Ma (since a = αk )

T = 4 lbs × 68.75 ft/s² = 275 lb

The tension due to the smaller drum is T' = 6 lb  .

So the net tension in the belt is T'' = T - T' = 275 lb - 6 lb = 269 lb

Answer:

The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

[tex]Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}[/tex]

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

[tex]Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa[/tex]

Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.