Which observation have scientists used to support Einstein's general theory of relativity?
The orbital path of Mercury around the Sun has changed.
O GPS clocks function at the same rate on both Earth and in space.
O The Sun has gotten more massive over time.
Objects act differently in a gravity field than in an accelerating reference frame.

Answer:

he correct answers are a, b

Explanation:

In the two-slit interference phenomenon, the expression for interference is

          d sin θ= m λ                       constructive interference

          d sin θ = (m + ½) λ             destructive interference

in general this phenomenon occurs for small angles, for which we can write

           tanθ = y / L

           tan te = sin tea / cos tea = sin tea

           sin θ = y / La

un

derestimate the first two equations.

Let's do the calculation for constructive interference

         d y / L = m λ

the distance between maximum clos is and

         y = (me / d) λ

this is the position of each maximum, the distance between two consecutive maximums

         y₂-y₁ = (L   2/d) λ - (L 1 / d) λ₁          y₂ -y₁ = L / d λ

examining this equation if the wavelength decreases the value of y also decreases

the same calculation for destructive interference

         d y / L = (m + ½) κ

         y = [(m + ½) L / d] λ

again when it decreases the decrease the distance

the correct answers are a, b

Answer:

The length of the rod for the condition on the question to be met is [tex]L = 1.5077 \ m[/tex]

Explanation:

The  Diagram for this  question is  gotten from the first uploaded image  

From the question we are told that

          The mass of the rod is [tex]M = 3.41 \ kg[/tex]

           The mass of each small bodies is  [tex]m = 0.249 \ kg[/tex]

           The moment of inertia of the three-body system with respect to the described axis is   [tex]I = 0.929 \ kg \cdot m^2[/tex]

             The length of the rod is  L  

     Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

        [tex]I = I_r + 2 I_m[/tex]

Where  [tex]I_r[/tex] is the moment of inertia of the rod about the describe axis which is mathematically represented as  

        [tex]I_r = \frac{ML^2 }{12}[/tex]

And   [tex]I_m[/tex] the  moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented  as

           [tex]I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}[/tex]

Thus  [tex]2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}[/tex]

Hence

       [tex]I = M * \frac{L^2}{12} + m * \frac{L^2}{2}[/tex]

=>   [tex]I = [\frac{M}{12} + \frac{m}{2}] L^2[/tex]

substituting vales  we have  

        [tex]0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2[/tex]

       [tex]L = \sqrt{\frac{0.929}{0.40867} }[/tex]

      [tex]L = 1.5077 \ m[/tex]

Answer:

The new voltage between the plates of the capacitor is 18 V

Explanation:

The charge on parallel plate capacitor is calculated as;

q = CV

Where;

V is the battery voltage

C is the capacitance of the capacitor, calculated as;

[tex]C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}[/tex]

[tex]q = \frac{\epsilon _0A V}{d}[/tex]

where;

ε₀ is permittivity of free space

A is the area of the capacitor

d is the space between the parallel plate capacitors

If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;

[tex]q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V[/tex]

Therefore, the new voltage between the plates of the capacitor is 18 V

Answer:

Explanation:

The current through the  resistor is 0.83 A

.

Part b

The current through  resistor is 0.53 A

.

Part c

The current through  resistor is 0.30 A

Answer:

From the image, the force as shown in option A will exert the biggest torque on the cylinder about its central axes.

Explanation:

The image is shown below.

Torque is the product of a force about the center of rotation of a body, and the radius through which the force acts. For a given case such as this, in which the cylinders are identical, and the forces are of equal magnitude, the torque at the maximum radius away from the center will exert the maximum torque. Also, the direction of the force also matters. To generate the maximum torque, the force must be directed tangentially away from the circle formed by the radius through which the force acts away from the center. Option A satisfies both condition and hence will exert the most torque on the cylinder.

Answer:

C. The atom loses 1 electron to have a total of 36.

Explanation:

Cations have a positive charge. Cations lose electrons.

The number of electrons in a Rubidium atom is 37. If the atom loses 1 electron, then it has 36 left.

Given info

d = 0.000250 meters = distance between slits

L = 302 cm = 0.302 meters = distance from slits to screen

[tex]\theta_8 = 1.12^{\circ}[/tex] = angle to 8th max (note how m = 8 since we're comparing this to the form [tex]\theta_m[/tex])

[tex]x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}[/tex] (n = 5 as we're dealing with the 5th minimum )

---------------

Method 1

[tex]d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}[/tex]

Make sure your calculator is in degree mode.

-----------------

Method 2

[tex]\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\[/tex]

-----------------

Method 3

[tex]\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\[/tex]

There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.

The  lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is: in the near-ultraviolet.

The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.

Longer wavelengths of light are diffracted at a greater angle than shorter ones, with the amount of diffraction being dependent on the wavelength of the light. Hence, among the light waves of  the visible, near-infrared, and near-ultraviolet range, near-ultraviolet waves have the shortest wavelengths. So,  The best resolution of this lens from a diffraction standpoint is in the near-ultraviolet, where diffraction is minimum.

Learn more about diffraction here:

https://brainly.com/question/11176463

#SPJ5

Answer:

  F ’= F 0.25

Explanation:

This problem refers to the electric force, which is described by Coulomb's law

        F = k q₁ q₂ / r²

where k is the Coulomb constant, q the charges and r the separation between them.

The initial conditions are

       F = k q_A q_B / d²

they indicate that the loads are reduced to ¼ q and the distance is reduced to ½ d

       F ’= k (q / 4 q / 4) / (0.5 d)²

       F ’= k q / 16 / 0.25 d²

       F ’= k q² / d²    0.0625 / 0.25

       F ’= F 0.25

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. This can be explained through Coulomb's law.

Coulomb's law is a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.

[tex]F = k \frac{q_Aq_B}{d^{2} } = k \frac{q^{2} }{d^{2} } [/tex]

where,

If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, the new force will be:

[tex]F_2 = k \frac{(0.25q_A)(0.25q_B)}{(0.5d)^{2} } = 0.25k\frac{q^{2} }{d^{2} } = 0.25 F[/tex]

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.

Learn more about Coulomb's law here: https://brainly.com/question/506926

Answer:

The speed is  [tex]v =10.27 *10^{7} \ m/s[/tex]

Explanation:

From the question we are told that

      The  voltage  is  [tex]V = 30 kV = 30*10^{3} V[/tex]

      The  initial velocity of the electron is  [tex]u = 0 \ m/s[/tex]

Generally according to the law of energy conservation

    Electric potential Energy  =  Kinetic energy of the electron

So  

      [tex]PE = KE[/tex]

Where  

      [tex]KE = \frac{1}{2} * m* v^2[/tex]

Here  m is the mass of the electron with a value of  [tex]m = 9.11 *10^{-31} \ kg[/tex]

     and  

         [tex]PE = e * V[/tex]

      Here  e is the charge on the electron with a value  [tex]e = 1.60 *10^{-19} \ C[/tex]

=>    [tex]e * V = \frac{1}{2} * m * v^2[/tex]

=>      [tex]v = \sqrt{ \frac{2 * e * V}{m} }[/tex]

substituting values  

           [tex]v = \sqrt{ \frac{2 * (1.60*10^{-19}) * 30*10^{3}}{9.11 *10^{-31}} }[/tex]

          [tex]v =10.27 *10^{7} \ m/s[/tex]

Answer:

t = 3.4 s

The box will come to rest in 3.4 s

Explanation:

For the block to come to rest, the friction force must become equal to the unbalanced force. Therefore:

Unbalanced Force = Frictional Force

but,

Unbalanced Force = ma

Frictional Force = μR = μW = μmg

Therefore,

ma = μmg

a = μg

where,

a = acceleration of box = ?

μ = coefficient of sliding friction = 0.45

g = 9.8 m/s²

Therefore,

a = (0.45)(9.8 m/s²)

a = -4.41 m/s²  (negative sign due to deceleration)

Now, for the time to stop, we use first equation of motion:

Vf = Vi + at

where,

Vf = Final Speed = 0 m/s (since box stops at last)

Vi = Initial Speed = 15 m/s

t = time to stop = ?

Therefore,

0 m/s = 15 m/s + (-4.41 m/s²)t

(-15 m/s)/(-4.41 m/s²) = t

t = 3.4 s

The box will come to rest in 3.4 s

Answer:

b. only ray that reflects back in the same direction it came from

Explanation:

C-rays can be said to be a ray that comes from the center of the curvature. It is known that any ray that comes from the center of the curvature reflects back in the same direction it came from, this is because the line joining from the center of the curvature to any point in the mirror is perpendicular to the mirror.

Correct answer is option B.

C-ray is the only ray that reflects back in the same direction it came from.

Option A is incorrect because for other rays, angle of incidence = angle of reflection. This is not a property of c-ray.

Answer:

(a) The power wasted for 0.289 cm wire diameter is 15.93 W

(b) The power wasted for 0.417 cm wire diameter is 7.61 W

Explanation:

Given;

diameter of the wire, d = 0.289 cm = 0.00289 m

voltage of the wire, V = 120 V

Power drawn, P = 1850 W

The resistivity of the wire, ρ = 1.68 x 10⁻⁸ Ω⋅m

Area of the wire;

A = πd²/4

A = (π x 0.00289²) / 4

A = 6.561 x 10⁻⁶ m²

(a) At 26 m of this wire, the resistance of the is

R = ρL / A

R = (1.68 x 10⁻⁸  x 26) / 6.561 x 10⁻⁶

R = 0.067 Ω

Current in the wire is calculated as;

P = IV

I = P / V

I = 1850 / 120

I = 15.417 A

Power wasted = I²R

Power wasted = (15.417²)(0.067)

Power wasted = 15.93 W

(b) when a diameter of 0.417 cm is used instead;

d = 0.417 cm = 0.00417 m

A = πd²/4

A = (π x 0.00417²) / 4

A = 1.366 x 10⁻⁵ m²

Resistance of the wire at 26 m length of wire and  1.366 x 10⁻⁵ m² area;

R = ρL / A

R = (1.68 x 10⁻⁸  x 26) / 1.366 x 10⁻⁵

R = 0.032 Ω

Power wasted = I²R

Power wasted = (15.417²)(0.032)

Power wasted = 7.61 W

Answer:

Increasing population

Explanation:

As we can see that the death rate is decreasing while at the same time the birth rate is increasing due to which it increased the population that directly impact the planet so great

Day by day the population of the villages, cities, states, the country is increasing which would create a direct human impact on the planet  

Therefore the increasing population is the one and single most important reason

Answer:

A, C

Explanation:

Since the frequency is inversely proportional to the length of a string, then I want to increase the frequency of the lowest

A. Replacing the string with a thicker string.

Thicker strings have more density. The more density the string has, the lower the sound.

Mathematically, we can see the proportionality (direct and inverse) by looking at those formulas for Frequency and Speed, when combined:

For:

[tex]f=\frac{v}{\lambda}[/tex]

[tex]f=\frac{v}{\lambda}*\sqrt{\frac{T}{D} }[/tex]

See above, how density (D) and [tex](\lambda)[/tex] wave length are inversely proportional.

C. Doubling the length of the string.

Because the length of the string is inversely proportional to the frequency.

The longer the string, the lower the frequency.

So, if we double string, we'll hear lower sounds in any string instrument

--

In short,  for A, and C  We can justify both since length and density are inversely proportional to the Frequency, we need longer or thicker string.

Answer:

The number of bright spot is  m =4

Explanation:

From the question we are told that

    The number of lines is  [tex]s = 500 \ lines / mm = 500 \ lines / 10^{-3} m[/tex]

     The wavelength of the laser is  [tex]\lambda = 480 nm = 480 *10^{-9} \ m[/tex]

Now the the slit is mathematically evaluated as

        [tex]d = \frac{1}{s} = \frac{1}{500} * 10^{-3} \ m[/tex]

Generally the diffraction grating is mathematically represented as

        [tex]dsin\theta = m \lambda[/tex]

Here m is the order of fringes (bright fringes) and at maximum m  [tex]\theta = 90^o[/tex]

    So

          [tex]\frac{1}{500} * sin (90) = m * (480 *10^{-3})[/tex]

=>        [tex]m = 4[/tex]

This  implies that the number of bright spot is  m =4

Answer:

(a) force is 1066.56N

Explanation:

(a) MV²/R

Answer:

The main points of difference between a citizen and alien are: (a) A citizen is a permanent resident of a state, while an alien is a temporary resident, who comes for a specific duration of time as a tourist or on diplomatic assignment. ... Aliens do not possess such rights in the state where they reside temporarily

Explanation:

Answer:

The definition of that same given problem is outlined in the following section on the clarification.

Explanation:

The Q seems to be endless (hardly any R on the circuit). So energy equations to describe and forth through the inducer as well as the condenser.  

Presently take a gander at the energy stored in your condensers while charging is Q.

⇒  [tex]U =\frac{Qmax^2}{C}[/tex]

So conclude C doesn't change substantially as well as,

When,

⇒  [tex]Q=\frac{Qmax}{2}[/tex]

⇒  [tex]Q^2=\frac{Qmax^2}{4}[/tex]

And therefore only half of the population power generation remains in the condenser that tends to leave this same inductor energy at 3/4 U.

Answer:

9,375

Explanation:

Data provided

The mass of the car m = 1500 Kg.

The diameter of the circular track D = 200 m.

For the computation of magnitude and direction of the net force on the car first we need to find out the radius of the circular path which is shown below:-

The radius of the circular path is

[tex]R = \frac{D}{2}[/tex]

[tex]= \frac{200}{2}[/tex]

= 100 m

after the radius of the circular path we can find the magnitude of the centripetal force with the help of below formula

[tex]Force F = \frac{mv^2}{R}[/tex]

[tex]= \frac{1500\times (25)^2}{100}[/tex]

= 9,375

Therefore for computing the magnitude of the centripetal force we simply applied the above formula.

Answer:

Momentum of the car = [tex]1.575\times 10^{-2}[/tex] kg meter per second

Explanation:

Julie is playing with a car which has mass = 6.3 g = [tex]6.3\times 10^{-3}[/tex] kg

Velocity of the car is 2.5 meter per second

Since formula to calculate the momentum of an object is,

p = mv

Where, p = momentum of the object

m = mass of the object

v = velocity of the object

By substituting these values in the formula,

p = [tex](6.3\times 10^{-3})\times 2.5[/tex]

  = [tex]1.575\times 10^{-2}[/tex] Kg meter per second

Therefore, momentum of the car will be [tex]1.575\times 10^{-2}[/tex] Kg meter per second.

Answer:

16200 km/s

270 km/min

4.5 km/h

Explanation:

Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)

Simply plug in our known variables and solve:

a = (45.0 - 0)/10

a = 45.0/10

a = 4.5 km/h

Answer:

[tex]\boxed{\mathrm{4.5 \: kmph/s \: or \: 1.25 \: m/s^2 }}[/tex]

Explanation:

[tex]\displaystyle \mathrm{acceleration = \frac{change \: in \: velocity}{time \: taken}}[/tex]

[tex]\displaystyle \mathrm{a = \frac{v - u}{t}}[/tex]

[tex]\displaystyle \mathrm{v=final \: velocity}\\\displaystyle \mathrm{u=initial \: velocity}[/tex]

[tex]\displaystyle \mathrm{a = \frac{45- 0}{10}}[/tex]

[tex]\displaystyle \mathrm{a = \frac{45}{10}}[/tex]

[tex]\displaystyle \mathrm{a = 4.5}[/tex]

[tex]\mathrm{4.5 \: kmph/s = 1.25 \: m/s^2 }[/tex]

Answer:

The answer is "[tex]\bold{7.30 \times 10^{-6}}[/tex]"

Explanation:

length of the copper wire:

L= 62.9 cm

r is the radius of the loop then:

[tex]r=\frac{L}{2 \pi}\\[/tex]

  [tex]=\frac{62.9}{2\times 3.14}\\\\=\frac{62.9}{6.28}\\\\=10.01\\[/tex]

area of the loop Is:

[tex]A_L= \pi r^2[/tex]

     [tex]=100.2001\times 3.14\\\\=314.628[/tex]

change in magnetic field is:

[tex]=\frac{dB}{dt} \\\\ = 0.01\ \frac{T}{s}[/tex]

then the induced emf is:  [tex]e = A_L \times \frac{dB}{dt}[/tex]

                                              [tex]=314.628 \times 0.01\\\\=3.14\times 10^{-5}V[/tex]

resistivity of the copper wire is: [tex]\rho =[/tex]  1.69 × 10-8Ω·m

diameter d = 1.15mm

radius (r) = 0.5mm

               [tex]= 0.5 \times 10^{-3} \ m[/tex]

hence the resistance of the wire is:

[tex]R=\frac{\rho L}{\pi r^2}\\[/tex]

   [tex]=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times (0.5 \times 10^{-3})^2}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.5 \times 0.5 \times 10^{-6}}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.25 \times 10^{-6}}\\\\=135.41 \times 10^{-2}\\=1.35\times 10^{-4}\\[/tex]

Power:

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

[tex]=\frac{3.14\times 10^{-5}\times 3.14\times 10^{-5}}{1.35 \times 10^{-4}}\\\\=7.30 \times 10^{-6}[/tex]

The final answer is: [tex]\boxed{7.30 \times 10^{-6} \ W}[/tex]

Answer:

21,920 Pascals

Explanation:

P = ρgh

P = (1096 kg/m³) (10 m/s²) (2 m)

P = 21,920 Pa

Answer:

  a = 1.15 10³ g

Explanation:

For this exercise we will use the relations of the centripetal acceleration

     a = v² / r

where is the linear speed of the rotor and r is the radius of the rotor

let's use the relationships between the angular and linear variables

          v = w r

let's replace

          a = w² r

let's reduce the angular velocity to the SI system

        w = 550 rev / min (2pi rad / 1 rev) (1 min / 60 s)

         w = 57.6 rad / s

let's calculate

       a = 57.6²  3.4

       a = 1.13 10⁴ m / s²

To calculate this value in relation to g, let's find the related

       a / g = 1.13 10⁴ / 9.8

       a = 1.15 10³ g

Answer:

Explanation:

Heat absorbed by oil

= mass x specific heat x rise in temperature

= 30 x 10⁻³ x 950 x 2.2 x 10³ x ( 50-10 )

= 25.08 x 10⁵ J  

Heat absorbed by air

= 50 x 1.2 x 1.0054 x 10³ x ( 20-10 )

= 6.03 x 10⁵ J

Total heat absorbed = 31.11 x 10⁵ J

If time required = t

heat lost from room

= .35 x 10³ t

Total heat generated in time t

= 1.8 x 10³ t

Heat generated = heat used

1.8 x 10³ t =  .35 x 10³ t  + 31.11 x 10⁵

1.45 x 10³ t = 31.11 x 10⁵

t = 31.11 x 10⁵ / 1.45 x 10³

t = 2145.5 s

Answer:

An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about  2 years.

Explanation:

Given;

orbital period of 3 years, P = 3 years

To calculate the years of an orbital with a semi-major axis, we apply Kepler's third law.

Kepler's third law;

P² = a³

where;

P is the orbital period

a is the orbital semi-major axis

(3)² = a³

9 = a³

a = [tex]a = \sqrt[3]{9} \\\\a = 2.08 \ years[/tex]

Therefore, An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about  2 years.

Answer:

Explanation:

a. Is the sphere charged negatively or positively?

The sphere us negatively charged. In a Millikan type experiment, there will be two forces that will be acting on the sphere which are the electric force which acts upward and also the gravity which acts downward.

Because the upper plate is positively charged, there'll what an attractive curve with an upward direction which will be felt by the negatively charged sphere.

b. What is the magnitude of the electric field intensity between the plates?

The magnitude of the electric field intensity between the plates is 18400v/m.

C. Calculate the magnitude of the charge on the latex sphere.

The magnitude of the charge on the latex sphere hae been solved and attached

d. How many excess or deficit electrons does the sphere have?

There are 5 excess electrons that the sphere has.

Check the attachment for further explanation.

Answer:

The astronaut and the satellite are 53.718 m apart.

Explanation:

Given;

mass of spacewalking astronaut, = 88 kg

mass of satellite, = 645 kg

force exerts by the satellite, F = 110N

time for this action, t = 0.45 s

Determine the acceleration of the satellite after the push

F = ma

a = F / m

a = 110 / 645

a = 0.171 m/s²

Determine the final velocity of the satellite;

v = u + at

where;

u is the initial velocity of the satellite = 0

v = 0 + 0.171 x 0.45

v = 0.077 m/s

Determine the displacement of the satellite after 1.4 m

d₁ = vt

d₁ = 0.077 x (1.4 x 60)

d₁ = 6.468 m

According to Newton's third law of motion, action and reaction are equal and opposite;

Determine the backward acceleration of the astronaut after the push;

F = ma

a = F / m

a = 110 / 88

a = 1.25 m/s²

Determine the final velocity of the astronaut

v = u + at

The initial velocity of the astronaut = 0

v = 1.25 x 0.45

v = 0.5625 m/s

Determine the displacement of the astronaut after 1.4 min

d₂ = vt

d₂ = 0.5625 x (1.4 x 60)

d₂ = 47.25 m

Finally, determine the total separation between the astronaut and the satellite;

total separation = d₁ + d₂

total separation = 6.468 m + 47.25 m

total separation = 53.718 m

Therefore, the astronaut and the satellite are 53.718 m apart.

Answer:

clockwise

Explanation:

when current flows through a ring in a clockwise direction, it produces the equivalent magnetic effect of a southern pole of a magnet on the coil.

Since the current is decreasing, there is a flux change on the lower ring; generating an induced current on the lower ring. According to Lenz law of electromagnetic induction, "the induced current will act in such a way as to oppose the motion or the action producing it". In this case, the induced current will have to be the same  polarity to the polarity of the current change producing it so as to repel the two rings far enough to stop the electromagnetic induction. The induced current will then be in the clockwise direction on the lower ring.

The direction of the induced current in the bottom ring is in the clockwise direction.

The given problem is based on the concept and fundamentals of the induced current and the direction of flow of the induced current.

Thus, we can conclude that the direction of the induced current in the bottom ring is in the clockwise direction.

Learn more about the concept of induced current here:

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