A rocket takes off from Earth's surface, accelerating straight up at 69.2 m/s2. Calculate the normal force (in N) acting on an astronaut of mass 87.4 kg, including his space suit. (Assume the rocket's initial motion parallel to the +y-direction. Indicate the direction with the sign of your answer.)

Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Explanation: To find the answer, we need to know about the different equations of planetary motion.

                    [tex]m_g=7.88*10^{18}kg\\r_g=6.32*10^4 m.\\h_{max}=1.44*10^3m.\\[/tex]

                    [tex]v=\sqrt{2gh}[/tex]

                  [tex]g_g=\frac{GM_g}{r_g^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2} =0.132m/s^2.[/tex]

                  [tex]v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s.[/tex]

                [tex]v=\sqrt{ \frac{GM}{r}}=\sqrt{\frac{7.88*10^{18}*6.67*10^{-11}}{1.45*10^5 }} =3.624km/s.[/tex]

Thus, we can conclude that,

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

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1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.

2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.

Understanding the planetary motion equations is necessary in order to determine the solution.

                             [tex]v=\sqrt{2gh}[/tex]  

                    [tex]a=\frac{GM}{R^2} =0.132m/s^2[/tex]

                  [tex]v=\sqrt{2*9.8*0.132} =19.46m/s[/tex]

                        [tex]v=\sqrt{\frac{GM}{R} } =3.624km/s\\where, M=7.88*10^{18}kg[/tex]

Consequently, we can say that

1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.

2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.

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Solutions for the three problems are is mathematically given as

(a) First cosmic speed (arbitral velocily)

[tex]v=\sqrt{\frac{G M}{r}}\\v=\sqrt{\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24}}{5870 \times 10^{3}}}[/tex]

v=7338.9349

(b) Second cosmic speed (escape velo.)

$$

\begin{gathered}

[tex]V=\sqrt{\frac{2 G M}{r}}\\\\V=\sqrt{2} \sqrt{\frac{G M}{r}}\\\\=V\sqrt{2} \times 7338.9349 \\[/tex]

[tex]V=14677.86986 \mathrm{~m} / \mathrm{s}[/tex]

(c) In conclusion, in a circular orbit, the gravitational force is gets balanced by centripetal force

[tex]&m_{q \omega \omega^{2}}=\frac{G M M}{r^{2}} \\&r^{3}=\frac{G M}{\omega^{2}}=\frac{G M}{4 \pi^{2}} T^{2} \\&r^{3}=\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24} \times(16.6 \times 3600)^{7}}{4 \pi^{2}} \\[/tex]

[tex]&r=30581248.06 \times 10^{4} \mathrm{~m} / \mathrm{s}[/tex]

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The stretching force acting on the second wire, given the data is 588 N

F₁ / r₁ = F₂ / r₂

450 / 3.9×10⁻³ = F₂ / 5.1×10⁻³

cross multiply

3.9×10⁻³ × F₂ = 450 × 5.1×10⁻³

Divide both side by 3.9×10⁻³

F₂ = (450 × 5.1×10⁻³) / 3.9×10⁻³

F₂ = 588 N

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The law that relates the absolute pressure to atmospheric pressure and gauge pressure is Pascal law.

Pascal's law states that when an object is immersed in a fluid, it experiences equal pressure on all surfaces.

P = Patm + pgh

where;

Thus, the law that relates the absolute pressure to atmospheric pressure and gauge pressure is Pascal law.

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The speed of the ball at the given distance and time of motion is 4 m/s.

The speed of an object is the rate of change of distance traveled by the object with time.

Speed is a scalar quantity because it has only magnitude and no direction. It is measured in meters per second.

For a ball that travelled a distance of 20 m and displacement of 10 m in 5 seconds​, the speed of the ball at the given distance and time of motion is calculated as follows;

Speed of the ball = distance traveled by the ball / time of motion

speed of the ball = 20 m / 5 s

speed of the ball = 4 m/s

Thus, the speed of the ball at the given distance and time of motion is 4 m/s.

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The two units for measuring the diameter of nucleus atom are femtometre and metre.

How do you measure the size of the nucleus ?

Nucleus size is expressed in fermi, often known as femtometers. between a lighter and a heavier nucleus. Despite its modest size, the nucleus contains the majority of an atom's mass. The weight or mass of the atom's nucleus and neutrons are determined by neutrons.

femtometre (fm), which equals [tex]10^{-15}[/tex] metre.

A nucleus' diameter largely depends as to how many particles it contains, from about 4 fm for a light nucleus like carbon to 15 fm for a heavy nucleus as lead.

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The magnitude of the force on the left-hand pole is mathematically given as

f'=0.167N

Q=45 degrees

Generally, the equation for Force is  mathematically given as

F=mg/sinФ

Therefore

F(17.1*10^{-3})*9.8/sin 45

F=0.237N

Considering horizontal axis or plane

f'-fcos=0

Therefore

f'=0.237*cos45

f'=0.167N

In conclusion, the slope

tanФ=30/30

tanФ=1

Ф=45

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Answer:

Explanation:

The angular speed of the rotor is 200 rad/s.

The torque needed to be transmitted by the engine is 180 Nm.

The power of the rotor required to transmit energy to apply a torque τ to rotate a motor with angular speed ω,

P=τω

=180×200W

=36kW

These given conditions satisfy the second law of thermodynamics.

As the process is isobaric

So there will be a straight line of P= 200kPa in P-v and P-T planes

P1 = P2 = 100kPa

For perfect ideal gas, v-T plane:

[tex]v = (\frac{R}{P}) T[/tex]

[tex]v_{1} = (\frac{R}{P_{1} }) T_{1}[/tex] = 287 × 500/200000 = 0.717 m³/kg

[tex]v_{2} = (\frac{R}{P_{2} }) T_{2}[/tex] = 287 × 600/200000 = 0.861m³/kg

As it is the calorically perfect gas

de = [tex]c_{v}[/tex]dT

Integration on both sides

e2 - e1 = [tex]c_{v}[/tex](T2 - T1)

           = ( 716.5J/kg/K) (600-500)

           = 71650 J/kg

also,

Tds = de + Pdv

Tds = [tex]c_{v}[/tex]dT +Pdv

For ideal gas

V = RT/P        

dv = Rdt/P - RTdp/P²

Tds = [tex]c_{v}[/tex]dT + Rdt - RTdp/P

ds = ([tex]c_{v}[/tex] + R)dT/T - RdP/P

ds = ([tex]c_{v} + c_{p} -c_{v}[/tex])dT/T - RdP/P

ds = [tex]c_{p}[/tex]dT/T - RdP/P

Integration on both sides

s2 - s1 = [tex]c_{p}[/tex]ln (T2/T1) - R ln (P2/P1)

Since P is constant

s₂ - s₁ = [tex]c_{p}[/tex] ln (T2/T1)

           = 1003.5 ln (600/500)

           = 1003.5 × 0.182

           = 182.95 J/kg/K

w = Pdv

[tex]w_{12}[/tex] = P(v₂ - v₁)

     = 2,00,000 ( 0.861 - 0.717)

     = 28,800 J/kg

de = δq -δw

δq = de + δw

q₁₂ = (e₂ - e₁) +  w₁₂

    =  71,650 + 28,800 = 1,00,450 J/kg

Now in this process, the gas is heated from 500 K to 600 K. We would expect at a minimum that the surroundings were at 600 K.

Let’s check for second law satisfaction.

s₂ - s₁ ≥ q₁₂ / Tₓ

182.95 ≥ 1,00,450 / 600 K

182.95 J/kg/K ≥ 167.41 J/kg/K

Hence this condition satisfies the second law of thermodynamics

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A psychologist will need to have good linguistic intelligence in other to be successful.

This is referred to as a professional who specializes in the handling of mental health challenges in individuals.

It is best for such professional to have a good linguistic intelligence as the right words being said to the patient will solve the problem thereby bringing in more success.

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(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.

(b) The net work done on the crate while it is on the rough surface is 23.7 J.

(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.

F(net) = F - μFf

F(net) = 280 - 0.351(92 x 9.8)

F(net) = -36.46 N

W = F(net) x L

W = -36.46 x 0.65

W = - 23.7 J

a = F(net)/m

a = -36.46/92

a = - 0.396 m/s²

v² = u² + 2as

v² = 0.845² + 2(-0.396)(0.65)

v² = 0.199

v = √0.199

v = 0.45 m/s

Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.

The net work done on the crate while it is on the rough surface is 23.7 J.

The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.

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Answer:

The mass of Sun doesn't change with respective to the conditions.

Michael has 4 Apples, which may increase his own mass or weight but not the Sun's .

His train is 7 minutes, but this doesn't mean the Sun has been made to change. The train coming late affects the time management and delays work.

So, As the per the question, It is evident that Sun's Mass is still the same irrespective of conditions .

Hence, The required answer Sun's Mass is 2*10^30      kg

Explanation:

There is no A in the vector, so I assume you mean λ.

The magnitude of any unit vector is 1, so

[tex]\|0.4\,\vec\imath + 0.8\,\vec\jmath + \lambda \,\vec k\| = \sqrt{0.4^2 + 0.8^2 + \lambda^2} = 1[/tex]

Square both sides and solve.

[tex]0.4^2 + 0.8^2 + \lambda^2 = 1^2 \implies \lambda = \boxed{\pm \sqrt{0.2}}[/tex]

The distance from the Earth's center to the point outside the Earth is 55800 Km

g = GM / r²

Cross multiply

GM = gr²

Divide both sides by g

r² = GM / g

Take the square root of both sides

r = √(GM / g)

r = √[(6.67×10¯¹¹ × 5.97×10²⁴) / 0.163)]

r = 4.94×10⁷ m

Divide by 1000 to express in Km

r = 4.94×10⁷ / 1000

r = 4.94×10⁴ Km

Distance from the centre of the Earth = R + r

Distance from the centre of the Earth = 6400 + 4.94×10⁴

Distance from the centre of the Earth = 55800 Km

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Answer: The speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth is 4188.11 m/s.

Explanation: To find the answer, we need to know about the equation of motion of a satellite around earth.

                 [tex]F_g=\frac{GMm}{r^2}[/tex]

                 [tex]F_c=\frac{mv^2}{r} \\[/tex]

                    [tex]\frac{GMm}{r^2} =\frac{mv^2}{r} \\thus,\\v=\sqrt{\frac{GM}{r} }[/tex]

                  [tex]r=3.57 R_E=3.57*6.37*10^3=22.74*10^3 km\\M=5.98*10^24kg\\G=6.67*10^{-11}Nm^2/kg^2[/tex]

                  [tex]v=\sqrt{\frac{6.67*10^{-11}*5.98*10^{24}}{22.74*10^6m} } =4188.11 m/s[/tex]

Thus, we can conclude that the speed of satellite will be 4188.11 m/s.

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In a circular orbit with a radius that is 3.57 times the mean radius of the Earth, a satellite moves at a speed of 132.43km/s.

In order to get the solution, we must understand the satellite's planetary motion equation.

                          [tex]F_g=\frac{GMm}{r^2}[/tex]

                         [tex]F_c=\frac{MV^2}{r}[/tex]

                          [tex]\frac{GMm}{r^2}=\frac{mV^2}{r}\\V=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10{-11}*5.98*10^{24}}{22.74*10^3} } \\V=132.43*10^3m/s[/tex]

As a result, we may estimate that the satellite will move at a speed of 132.43 km/s.

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The escape velocity from the surface of the planet X is 2,249.2 m/s.

[tex]v = \sqrt{\frac{2GM}{r} } \\\\v^2 = \frac{2GM}{r}\\\\v^2r = 2GM\\\\G = \frac{v^2r}{2M}[/tex]

where;

[tex]\frac{v_x^2 \ r_x}{2M_x} = \frac{v_y^2 \ r_y}{2M_y} \\\\\frac{v_x^2 \ r_x}{M_x} = \frac{v_y^2 \ r_y}{M_y} \\\\v_x^2 = \frac{v_y^2 \ r_yM_x}{M_yr_x}\\\\v_x^2 = \frac{(1695)^2 (r_y)(1.59M_y)}{M_y(0.903r_y)} \\\\v_x^2 = 5,058,814.78\\\\v_x = \sqrt{5,058,814.78} \ \ = 2,249.2 \ m/s[/tex]

Thus, the escape velocity from the surface of the planet X is 2,249.2 m/s.

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The velocity with which the particle Q is fired is 15m/s upwards.

The vector quantity velocity (v), represented by the equation v = s/t, quantifies displacement (or change in position, s) over the change in time (t). Speed (or rate, r) is a scalar number, denoted by the equation r = d/t, that quantifies the distance traveled (d) over the change in time (t).

Assume the upward direction is positive and the downward direction is negative.

Velocity P, = 40m/s

Distance traveled by P,

Using the first equation of motion for particle P,

v = u + at

⇒ 0 = 40 + (-10)t

⇒ t = 4s

This is the time it takes for P to rise

Now, the maximum height(s) reached by the particle P is,

Using the second equation of motion,

s = ut + 1/2at²

⇒ s = 40×4 + 1/2 × (-10) × 4²

⇒ s = 80m

a) A particle P falls when Q is shot after 4 seconds from the initial time:

V² = U² + 2aH₁

⇒ H₁ = 15² - 0/ 2(-10)

⇒ H₁ = 11.25m

Particles P and Q meet at a distance from the ground (H₂).

Height, H₂ = s - H

⇒ H₂ = 80 - 11.25

         = 68.75m

Particles P and Q meet at a distance of 68.75m from the ground.

v = u + at₁

⇒ 15 = 0 + (-10) t₁

⇒ t₁ = 1.5s

It is equal to P's fall time and Q's rise time.

b) For particle Q

H₂ = u₂t₁ + 1/2at₁

⇒ 11.25 = u₂ × 1.5 + 1/2 (-10) × 1.5

⇒ u₂ = 15 m/s

Therefore,

The velocity with which the particle Q is fired is 15m/s upwards.

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If you speed through a construction zone while workers are present, your fines could be as much as one thousand dollars.

This is referred to as the amount which is instructed by a court or authority to be paid as a result of it being a penalty for various types of offences. each crime has its specific fine which helps to serve as deterrent for unlawful behavior in the community.

it is always best not to speed when within a construction zone which has workers present in the area. This is ideal as it helps to prevent incidences of accidents or death.

it is therefore the reason why a fine of 1000 dollars is to be paid so that people can think of such high amount before performing certain types of activities when driving and makes it the most appropriate choice.

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I will go to school tomorrow .....is this present tense or past tense or future tense

Answer:

D

Explanation:

The given vector can be broken down into two components (as shown in the figure).

Which results ; vector (s) = 3x + 4y.

Answer:

The speed of light is constant for all observers

Explanation:

As per general postulate of relativity

So

light speed c is independent of States of matter

The tension, T in the cable is equal to 323.5 N.

Tension is force exerted by a cable or string on another object usually a weight suspended from the cable or string

The tension in the cable is found this:

Angle of the boom with horizontal, θ = tan⁻¹(5/10) = 26.56°

The angle of cable with horizontal, B = tan⁻¹(4/10) = 21.80

Taking moments about the pivot:

175.5 * cos 26.56 + 94.7 * cos 26.56 * 0.5 = T (sin(26.56 + 21.80) * 1

T = 241.68/0.747

T = 323.5 N

In conclusion, the tension in the cable is determined by taking moments about the pivot.

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vr>vs because the rolling ball acquires rotational as well as translational kinetic energy.

The accelerating force acting on the ball as it goes along a smooth plane is mgsin. Its acceleration is therefore equal to gsin. The mgsin acts down the plane as the ball travels down the rough inclined plane, but friction develops that acts up the plane.

Since both balls' potential energy is lost at the same rate, their KEs are actually equal at the base of the planes. However, a ball sliding down a smooth plane has only translational kinetic energy, but a ball rolling down a rough plane contains both translational and rotational kinetic energy at the bottom of the plane. As a result, the ball's translational KE will be lower than its translational Kinetic energy.

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The voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.

The work-energy theorem states that the delta in the equation equals the change in kinetic energy plus the change in potential energy. Here, a charge's potential energy is expressed as qV, where V is the position's electric potential.  The greater the change in voltage per unit distance, the greater the electric field.

The kinetic energy of the electrons = 4.1 × [tex]10^-^1^5[/tex] J

Charge of the electron = 1.602 × [tex]10^-^1^9[/tex] coulomb

Using,

     ΔU = q × ΔV

4.1 × [tex]10^-^1^5[/tex] = 1.602 × [tex]10^-^1^9[/tex] × ΔV

      ΔV  = 3.9 × [tex]10^-^3[/tex] V

Therefore, the voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.

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Answer: the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.

Explanation: To find the answer, we need to know the Snell's law.

                      [tex]\frac{sin (i)}{sin(r)} =\frac{n_r}{n_i}[/tex]

where, i is the incident angle, r is the refracted angle, n is the refractive index.

                           [tex]\frac{sin 65}{sin48} =\frac{n_w}{n_X} \\n_X=\frac{1.33*sin48}{sin65} =1.09\\[/tex]

                         [tex]\frac{sin 48}{sin r}=\frac{1}{1.33} \\\\ sin(r)=\frac{1.33*sin 48}{1}=0.988\\\\r=sin^{-1}(0.988)=81.25 degrees.[/tex]

Thus, we can conclude that, the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.

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The light makes an angle of 81.25° with the normal in the air and the index of refraction of material X will be 1.09.

In order to determine the solution, we must understand Snell's law.

                            [tex]\frac{sin(i)}{sin(r)}=\frac{n_r}{n_i}[/tex]

where the refractive index is n and the incidence angle is i. The refracted angle is r.

                      [tex]n_X=\frac{n_w*sin 48}{sin65} =1.09[/tex]

                  [tex]sin(r)=n_w*sin48=0.988\\r=sin^{-1}(0.988)=81.25 degrees.[/tex]

As a result, we can infer that the material X will have an index of refraction of 1.09 and that the angle the light makes with the normal in the air is 81.25°.

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The amount of work done to stop a 975- kg car traveling at 105 km/h is 414,808.34J.

The amount of work done by a moving object can be calculated using the following formula:

W (Kinetic energy) = ½ mv²

Where;

According to this question, a car of 975 kg is traveling at 105 km/h. This speed in m/s is 29.17m/s.

K.E = ½ × 975 × 29.17²

K.E = 414,808.34J

Therefore, the amount of work done to stop a 975- kg car traveling at 105 km/h is 414,808.34J.

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The period of M2, in years. is mathematically given as

T2= 134.3968years

M1 = 4.00 kg

M2 = 1.00 kg

T1 = 34.0 years.

Generally, the equation for is  mathematically given as

T2 = T1 (a2/a1)3/2

T2=  34.0  * (5/2)^{3/2}

T2= 134.3968years

In conclusion, the period of M2, in years

T2= 134.3968years

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The wavelength of A2 with frequency (110hz) is 3.01 m.

Wavelength of a sound wave is the distance between successive similar points in the wave such as rarefactions or compressions.

A tuba is a musical instrument that produces sound waves.

Wavelength is related to frequency and velocity by the formula below:

The wavelength of A2 with frequency (110hz) is 3.01 m.

In conclusion, the wavelength of a wave is inversely proportional to frequency.

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The density of the planet is determined as 1,974.26 kg/m³.

√(⁴/₃πGρ) = 2π/(2.35 x 3600)

where;

√(⁴/₃πGρ) = 2π/(2.35 x 3600)

√(⁴/₃πGρ) = 2π/8460

(⁴/₃πGρ)  = (2π/8460)²

⁴/₃πGρ = 4π²/(8460)²

ρ = 12π/(8460² x 4G)

ρ = (12π) / (8460² x 4 x 6.67 x 10⁻¹¹)

ρ = 1,974.26 kg/m³

Thus, the density of the planet is determined as 1,974.26 kg/m³.

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The mutual inductance of the pair of coils is 6.67 x 10⁻³ H.

The the flux through each turn is  2.78 x 10⁻⁴ Tm².

The magnitude of the induced emf in the first coil is 2.435 x 10⁻³ V.

M = -NΦ/I

M = emf/I

M = -(1.6 x 10⁻³)/(-0.24)

M = 6.67 x 10⁻³ H

M = NΦ/I

MI = NΦ

Φ = MI/N

Φ = (6.67 x 10⁻³  x 1.25)/(30)

Φ = 2.78 x 10⁻⁴ Tm²

emf = MI

emf = 6.67 x 10⁻³ x 0.365

emf = 2.435 x 10⁻³ V

Thus, the mutual inductance of the pair of coils is 6.67 x 10⁻³ H.

The the flux through each turn is  2.78 x 10⁻⁴ Tm².

The magnitude of the induced emf in the first coil is 2.435 x 10⁻³ V.

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