What is the maximum wavelength of incident light for which photoelectrons will be released from gallium

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:

c. because the rod polarizes the metal.

Explanation:

Bringing the negatively charged rod close to the neutral metal ball causes the neutral metal ball to be polarized with induced positive charge on it. The polarizing of the formally neutral metal ball is due to the negative charge on the metal rod (bodies induce a charge opposite of their own charge on a nearby neutral body). The ball and rod then attract themselves because bodies with opposite charges attract each other, unlike bodies with same charges that repel each other.

Explanation:

It took 7.5 ms or [tex]7.5×10^{-3}\:\text{s}[/tex] for the laser beam to reach the planet's surface. Since the speed of light is [tex]3×10^8\:\text{m/s}[/tex], the distance is

[tex]d = vt = (3×10^8\:\text{m/s})(7.5×10^{-3}\:\text{s})[/tex]

[tex]\:\:\:\:=2.25×10^{6}\:\text{m}[/tex]

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:

A. Ef/ Ei = 1/2

B. EF/ Ei = 1

C Ef / Ei = 4

Explanation:

To solve this we apply Coulomb's law which States that

E = Kq / r^2

Where

q = charge r = straight line distance from q to the point in question and

K = Coulomb's constant

Then

Ei = K Q / r^2

So

A) If Q is halved then

Ef = K Q / (2 r^2)

Ef/Ei = 1/2

B) If R is halved, the value of the E-f

at a distance r remains unchanged. So

Ef/Ei = 1

C) if r is now r/2 then

Ef = K Q / (r/2)^2 = K Q / r^2/4 = 4 K Q / r^2

Ef / Ei = 4

Answer:

The distance is [tex]d = 193.6 \ m[/tex]

Explanation:

From the question we are told that

   The time interval between the sounds is  k[tex]t_1 = k + t_2[/tex] =  0.50 s

    The  speed of sound in air is  [tex]v_s = 343 \ m/s[/tex]

    The  speed of sound in the concrete is [tex]v_c = 3000 \ m/s[/tex]

Generally the distance where the collision occurred is  mathematically represented as

          [tex]d = v * t[/tex]

Now from the question we see that d is the same for both sound waves

 So

        [tex]v_c t = v_s * t_1[/tex]

Now  

So [tex]t_1 = k + t[/tex]

      [tex]v_c t = v_s * (t+ k)[/tex]

=>     [tex]3000 t = 343* (t+ 0.50)[/tex]

=>    [tex]3000 t = 343* (t+ 0.50)[/tex]

=>    [tex]t = 0.0645 \ s[/tex]

So

     [tex]d = 3000 * 0.0645[/tex]

     [tex]d = 193.6 \ m[/tex]

Answer:

The magnetic field at the center of the solenoid is 2.1  × 10⁻³ T

Explanation:

The magnetic field B at the center of the solenoid is given by

B = μ₀ni where μ₀ = permeability of free space = 4π × 10⁻⁷H/m, n = number of turns per unit length of the solenoid = 1100 turns per meter and i = current in the solenoid = 1.5 A.

So B = μ₀ni

= 4π × 10⁻⁷H/m × 1100 × 1.5 A

= 4π × 10⁻⁷H/m × 1650 A-turns/m

= 20734.5 × 10⁻⁷T  

= 2.07345 × 10⁻³ T

≅ 2.1  × 10⁻³ T

So the magnetic field at the center of the solenoid is 2.1  × 10⁻³ T

Answer:

221 lines per millimetre

Explanation:

We know that for a diffraction grating, dsinθ =mλ where d = spacing between grating, θ = angle to maximum, m = order of maximum and λ = wavelength of light.

Since the grating is 42.0 cm from the screen and its first order maximum (m = 1) is at 6.09 cm from the center of the pattern,

tanθ = 6.09 cm/42.0 cm = 0.145

From trig ratios, cot²θ + 1 = cosec²θ

cosecθ = √((1/tanθ)² + 1) = √((1/0.145)² + 1) = √48.562 = 6.969

sinθ = 1/cosecθ = 1/6.969 = 0.1435

Also, sinθ = mλ/d at the first-order maximum, m = 1. So

sinθ = (1)λ/d = λ/d

Equating both expressions we have  

0.1435 = λ/d

d = λ/0.1435

Now, λ = 650 nm = 650 × 10⁻⁹ m

d = 650 × 10⁻⁹ m/0.1435

d = 4529.62 × 10⁻⁹ m per line

d = 4.52962 × 10⁻⁶ m per line

d = 0.00452962 × 10⁻³ m per line

d = 0.00452962 mm per line

Since d = width of grating/number of lines of grating

Then number of lines per millimetre = 1/grating spacing

= 1/0.00452962

= 220.77 lines per millimetre

≅ 221 lines per millimetre since we can only have a whole number of lines.

Answer:

Alp decay.

Explanation:

From the above equation, the parent nucleus 185 79Au produces a daughter nuclei 181 77 Ir.

A careful observation of the atomic mass of the parent nucleus (185) and the atomic mass of the daughter nuclei (181) shows that the atomic mass of the daughter nuclei decreased by a factor of 4. Also, the atomic number of the daughter nuclei also decreased by a factor of 2 when compared with the parent nucleus as shown in the equation given above.

This simply means that the parent nucleus has undergone alpha decay which is represented with a helium atom as 4 2He.

Therefore, the equation is an example of alpha decay.

Answer:

About 23 meters

Explanation:

To do this, you'll want to apply one of the kinematic equations to find the time it takes for the cabin to reach max velocity from rest. (Use the max velocity as V_f and V_i=0)

Then, you can find the distance travelled during the acceleration by equating the acceleration to the change in distance of the time squared.

My work is in the attachment, comment if you have any questions.

Answer:

The percentage change in resistance of the wire is 69%.

Explanation:

Resistance of a wire can be determined by,

R = (ρl) ÷ A

Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.

When the wire is stretched, its length and area changes but its volume and resistivity remains constant.

[tex]l_{o}[/tex] = 1.3l, and [tex]A_{o}[/tex] = [tex]\frac{A}{1.3}[/tex]

So that;

[tex]R_{o}[/tex] = (ρ[tex]l_{o}[/tex]) ÷ [tex]A_{o}[/tex] = (ρ × 1.3l) ÷ ([tex]\frac{A}{1.3}[/tex])

    = (1.3lρ) ÷ ([tex]\frac{A}{1.3}[/tex])

    = [tex](1.3)^{2}[/tex] × [(ρl) ÷ A]

   = 1.69R               (∵ R = (ρl) ÷ A)

[tex]R_{o}[/tex] = 1.69R

Where [tex]R_{o}[/tex] is the new resistance, [tex]l_{o}[/tex] is the new length, and [tex]A_{o}[/tex] is the new area after stretching the wire.

The change in resistance of the wire = [tex]R_{o}[/tex] - R

                                      = 1.69R  - 1R

                                      = 0.69R

The percentage change in resistance = [tex]\frac{0.69R}{R}[/tex] × 100

                                                               = 0.69 × 100

                                                              = 69%

The percentage change in resistance of the wire is 69%.

Answer:

λ = 428.6 nm

Explanation:

Hello,

In this case, we must remember that the Young's double slit experiment is described by the expression :

d sin θ = m λ

For constructive interference , and:

d sin θ = (m + ½) λ          

For destructive interference , whereas d accounts for the distance between the slits, λ for the wavelength and m for an integer that describes the order of interference . Thus, for the given angle 30º, the distance between the slits is 3.00 μm or 3.00 10⁻⁶ m and the order of interference is 3; we therefore use the destructive interference equation  to compute the wavelength as shown below:

λ = 3x10⁻⁶ sin (30) / (3 +1/2)

λ = 4.286 10⁻⁷ m

Or in manometers:

λ = 428.6 nm

Best regards.

Answer:

attached below is the free body diagram of the missing  illustration

Initial kinetic energy of the electron = 3 eV

Explanation:

The conclusion that can be drawn about the kinetic energy of the electron is

[tex]E_{e} = E_{3} - E_{1}[/tex]

E[tex]_{e}[/tex] = initial kinetic energy of the electron

E[tex]_{1}[/tex] = -4 eV

E[tex]_{3}[/tex] = -1 eV

insert the values into the equation above

[tex]E_{e}[/tex] = -1 -(-4)  eV

   = -1 + 4 = 3 eV

Answer:

y = 12.82 m

Explanation:

We can solve this exercise using the energy work theorem

          W = ΔEm

friction force work is

          W = fr . s = fr s cos θ

the friction force opposes the movement, therefore the angle is 180º

           W = - fr s

we write Newton's second law, where we use a reference frame with one axis parallel to the plane and the other perpendicular

           N -Wy = 0

           N = mg cos θ

the friction force remains

            fr = μ N

            fr = μ mg cos θ

work gives

           W = - μ mg s cos θ

initial energy

           Em₀ = ½ m v²

the final energy is zero, because it stops

we substitute

          - μ m g s cos θ = 0 - ½ m v²

          s = ½ v² / (μ g cos θ)

let's calculate

          s = ½ 20² / (0.55 9.8 cos 20)

          s = 39.49 m

this is the distance it travels along the plane, to find the vertical distance let's use trigonometry

            sin 20 = y / s

           y = s sin 20

           y = 37.49 sin 20

           y = 12.82 m

Answer:

red blood cell, also called erythrocyte

Explanation:

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Answer:

The point of contraflexure (PoC) occurs where bending is zero and at the point of change between positive and negative (or between compression and tension). In a beam that is flexing (or bending), the point where there is zero bending moment is called the point of contraflexure.

The height below the tower at which coin 1 pass coin 2 is 89.04 m.

The given parameters:

height of the tower, h = 100 m

initial velocity of coin 1, v = 15 m/s

time spent in air by coin 1 before coin 2 was dropped = 2s

To find:

Find the maximum height attained by coin 1 before falling to the ground:

[tex]v^2 = u^2 - 2gh\\\\where;\\\\v \ is \ the \ final \ velocity \ of \ coin \ 1 \ at \ maximum \ height, v \ = 0\\\\0 = (15^2) - 2(10)h\\\\20h = 225\\\\h = \frac{225}{20} \\\\h = 11.25 \ m[/tex]

Find the time taken for coin 1 to fall to the ground:

Total height of coin 1 above the ground, H = 11.25 m + 100 m = 111.25 m

[tex]t = \sqrt{\frac{2H}{g} } \\\\t = \sqrt{\frac{2\times 111.25}{10} } \\\\t = 4.72 \ s[/tex]

But the time taken for the coin 1 to reach 11.25 m above the tower:

[tex]t_1 = \sqrt{\frac{2h}{g} } \\\\t_1 = \sqrt{\frac{2\times 11.25}{10} } \\\\t_1 = 1.5 \ s[/tex]

Total time spent by coin 1 before reaching ground with respect to coin 2:

time = (1.5 s + 4.72 s) - 2 s

time = 4.22 s

Note: the 2 s was subtracted to keep both coins at a fair starting time below the tower.

Find the total time taken for coin 2 to fall to the ground:

Height of coin 2 above the ground = 100 m

Total time taken by coin 2 before falling to the ground is calculated as:

[tex]t_2 = \sqrt{\frac{2(100)}{10} } \\\\t_2 = 4.47s[/tex]

The time  at which coin 1 will pass coin 2 is 4.22 s.

Find the height below the tower when the time is 4.22 s.

[tex]h = \frac{1}{2} (10)(4.22)^2\\\\h = 89.04 \ m[/tex]

Thus, the height below the tower at which coin 1 pass coin 2 is 89.04 m.

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Answer:

The speed of the rod is 2.169 m/s.

Explanation:

Given that,

Mass = 0.100 kg

Current = 15.0 A

Distance = 2 m

Length = 0.550 m

Kinetic friction = 0.120

(a). We need to calculate the magnetic field

Using relation of frictional force and magnetic force

[tex]F_{f}=F_{B}[/tex]

[tex]\mu mg=Bli[/tex]

[tex]B=\dfrac{\mu mg}{li}[/tex]

Where, l = length

i = current

m = mass

Put the value into the formula

[tex]B=\dfrac{0.120\times0.1\times9.8}{0.550\times15.0}[/tex]

[tex]B=0.01425\ T[/tex]

[tex]B=1.425\times10^{-2}\ T[/tex]

(b). If the friction between the rod and rail is reduced zero.

So, [tex]f_{f}=0[/tex]

We need to calculate the acceleration

Using formula of force

[tex]F_{net}=f_{f}+F_{B}[/tex]

[tex]F_{net}=0+Bil[/tex]

[tex]ma=Bil[/tex]

[tex]a=\dfrac{Bil}{m}[/tex]

Put the value into the formula

[tex]a=\dfrac{1.425\times10^{-2}\times15\times0.55}{0.1}[/tex]

[tex]a=1.176\ m/s^2[/tex]

We need to calculate the speed of the rod

Using equation of motion

[tex]v^2=u^2+2as[/tex]

Put the value into the formula

[tex]v^2=0+2\times1.176\times2[/tex]

[tex]v^2=\sqrt{4.704}\ m/s[/tex]

[tex]v=2.169\ m/s[/tex]

Hence, The speed of the rod is 2.169 m/s.

Explanation:

(a) Given that,

Area of a parallel plate capacitor, [tex]A=1.8\ cm^2=1.8\times 10^{-4}\ m^2[/tex]

The separation between the plates of a capacitor, [tex]d=0.01\ mm = 10^{-5}\ m[/tex]

The dielectric constant of, k = 2.1

When a dielectric constant is inserted between parallel plate capacitor, the capacitance is given by :

[tex]C=\dfrac{k\epsilon_o A}{d}[/tex]

Putting all the values we get :

[tex]C=\dfrac{2.1\times 8.85\times 10^{-12}\times 1.8\times 10^{-4}}{0.01\times 10^{-3}}\\\\C=3.345\times 10^{-10}\ F\\\\C=334.5\ pF[/tex]

(b) We know that the Teflon has dielectric strength of 60 MV/m, [tex]E=60\times 10^6\ V/m[/tex]

The voltage difference between the plates at this critical voltage is given by :

[tex]V=Ed\\\\V=60\times 10^6\times 0.01\times 10^{-3} \\\\V=600\ V[/tex]

or

V = 0.6 kV

We have that the Capacitance and potential difference is mathematically given as



Question Parameters:

having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm

having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.

a)

Generally the equation for the Capacitance  is mathematically given as

[tex]C=\frac{ke_0A}{d}\\\\Therefore\\\\C=\frac{2.1*1.80e-4*8.85e12}{0.01e-3}\\\\[/tex]

C=334.68pF

b)

Generally the equation for the Capacitance  is mathematically given as

[tex]Vmax=\frac{Q}{C}[/tex]

Where

Q is the charge on the plates, and hence not given

Therefore, maximum potential difference is

[tex]Vmax=\frac{Q}{334.68pF}[/tex]

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Answer:

The width is [tex]w_c = 0.00252 \ m[/tex]

Explanation:

From the question we are told that

  The  width of the single slit is  [tex]a = 1.5 \ mm = 1.5 *10^{-3} \ m[/tex]

   The  wavelength is  [tex]\lambda = 420 *10^{-9} \ m[/tex]

   The distance of the screen is  [tex]D = 4.5 \ m[/tex]

Generally the width of the central maximum is  

        [tex]w_c = 2 * y[/tex]

where y is the width of the first maxima which is mathematically represented as

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

=>   [tex]y = \frac{ 420 *10^{-9} * 4.5}{ 1.5*10^{-3}}[/tex]

=>  [tex]y = 0.00126 \ m[/tex]

So

    [tex]w_c = 2 *0.00126[/tex]

    [tex]w_c = 0.00252 \ m[/tex]

Answer:

v = 6.28 m/s

Explanation:

It is given that,

A windmill on a farm rotates at a constant speed and completes one-half of a rotation in 0.5 seconds,

Number of revolution is half. It means angular velocity is 3.14 radians.

Let v is the angular speed. So,

[tex]v=\dfrac{\omega}{t}\\\\v=\dfrac{3.14}{0.5}\\\\v=6.28\ m/s[/tex]

So, the rotation speed is 6.28 m/s.

The angular velocity is the rotation speed, which is the angle of rotation

of the windmill per second, which is 2·π radians.

Response:

The given parameter are;

The angle of rotation the windmill rotates in 0.5 seconds = One-half a

rotation.

Required:

The rotational speed (angular velocity)

Solution:

The angle of one rotation = 2·π radians

Angle of one-half ration = [tex]\frac{1}{2}[/tex] × 2·π radians = π radians

[tex]Rotational \ speed = \mathbf{\dfrac{Angle \ of \ rotation}{Time}}[/tex]

Which gives;

[tex]Rotational \ speed, \omega = \dfrac{\pi}{0.5 \ s} = \mathbf{2 \cdot \pi \ rad/s}[/tex]

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Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

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:

The  charge on the dust particle is  [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Explanation:

From the question we are told that

    The length is  [tex]l = 2.0 \ m[/tex]

    The width is  [tex]w = 4.0 \ m[/tex]

   The charge is  [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]

    The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]

Generally the electric field on the carpet is mathematically represented as

           [tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]

Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

           [tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]

           [tex]E = -70621.5 \ N/C[/tex]

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        [tex]F__{E}} = F__{G}}[/tex]

=>     [tex]q_d * E = m * g[/tex]

=>      [tex]q_d = \frac{m * g}{E}[/tex]

=>      [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]

=>     [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Answer:

statement 1 with answer C

statement 2 with answer F

statement 3 with answer B

Statement 1 with E

Statement 2 with A

Statement 3 with D

Explanation:

In this exercise you are asked to relate each with the answers

In general, in the optics diagram,

* Ray 1 is a horizontal ray that after stopping by the optical system goes to the focal point

* Ray 2 is a ray that passes through the intercept point between the optical axis and the system and does not deviate

* Ray 3 is a ray that passes through the focal length and after passing the optical system, it comes out horizontally.

With these statements, let's review the answers

statement 1 with answer C

statement 2 with answer F

statement 3 with answer B

Statement 1 with E

Statement 2 with A

Statement 3 with D

Answer:

The direction of net magnetic force on the circular section of the loop is in the positive y-axis

The net magnetic force on the circular section of the loop is 3.74 N

Explanation:

The magnetic field strength [tex]B[/tex] = 1.7 T

the current [tex]I[/tex] = 0.7 A

The diameter of the loop = 2 m

the length of the circular section of the semi-circular loop [tex]l[/tex] = πd/2

==> [tex]l[/tex] = (3.142 x 2)/2 = 3.142 m

The force on the semi-circular is given as

F = [tex]BIl[/tex] sin ∅

but the loop is perpendicular to the field, therefore

sin ∅ = sin 90° = 1

F = 1.7 x 0.7 x 3.142 x 1 = 3.74 N

The right hand rule states that "if the fingers of the right hand are held parallel to each other in the direction of the magnetic field, and the thumb is held at right angle to the other fingers in the direction of the flow of current. The palm will push in the direction of the magnetic force on the conductor".

According to the right hand rule, the direction of net magnetic force on the circular section of the loop is in the positive y-axis

Answer:

e. the point about which the torque is to be calculated

Explanation:

torque is the product of a force and a distance

the point about which the torque is calculated is required to know the distance.

None of the other terms are relevant as they refer to mass or its equivalent, and velocity. Force is not mentioned in any of them.

A particle moves along the x-axis. In order to calculate the torque on the particle, you need to know the point about which the torque is to be calculated. Therefore, option E is correct.

The rotating equivalent of force is torque. Depending on the subject of study, it is also known as the moment, moment of force, rotating force, or turning effect. It illustrates how a force can cause a change in the body's rotational motion.

Ancient Romans gave these necklaces the term "torque" by describing them as twisted and spiral screw-shaped using the Latin word "torquere," which also means "twisting" and "turning."

It's critical to realize that torque, which has to do with your motor's power in terms of rotational force, is not the same thing as speed. Find a motor with a top speed if you require more motor speed, and a motor with a motor torque that is maximized if you need more rotational force.

Thus, option E is correct.

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Answer:

t(min) = 94nm

Explanation:

The wavelength of the light incident is 500 nm.

The refractive index of the film is 1.33.

The minimum thickness of the soap film required for constructive interference.

The thickness of the film for the constructive interference is given by:

2*t= (m + 1/2) λ′

Now, λ′ = λ/μ = 500/1.33 = 376nm

The minimum thickness of the film

′t′ will be at m=0 :

2*t(min) = (0 + 1/2) 376

t(min) = 94nm

Answer:

only in or near star-forming clouds

Explanation: