*Look at attachment for photo of object**

An object, whose mass is 0.520 kg, is attached to a spring with a force constant of 106 N/m. The object rests upon a frictionless, horizontal surface (shown in the figure below).

The object is pulled to the right a distance A = 0.150 m from its equilibrium position (the vertical dashed line) and held motionless. The object is then released from rest.

(a) At the instant of release, what is the magnitude of the spring force (in N) acting upon the object?
__________ N

(b) At that very instant, what is the magnitude of the object's acceleration (in m/s2)?
________ m/s2


(c)
In what direction does the acceleration vector point at the instant of release?
- The direction is not defined (i.e., the acceleration is zero).
- Toward the equilibrium position (i.e., to the left in the figure).
- Away from the equilibrium position (i.e., to the right in the figure).
- You cannot tell without more information.

Answer:

A and C

Explanation:

drag (the area of lower air pressure behind the car when moving) and mostly air resistance (the work to push the air in front of us away to move through - the faster we go, the stronger the air resists to move aside).

(a) The magnitude of the magnetic field at point P1 , midway between the wires is  1.005 x 10⁻⁴ T and the direction will be out of the page.

(b) The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.

(c) The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1  is 3.88 x 10⁻⁶ T and the direction is downwards.

B = μ/2π[I₁/0.5r + I₂/0.5r]

B = (μ/2π) x (I/0.5r + I/0.5r)

B = (μ/2π) x (2I/0.5r)

B = μI/0.5r

B = 2μI/r

where;

B = (2 x 4π x 10⁻⁷ x 4)/(0.1)

B = 1.005 x 10⁻⁴ T

The direction of the magnetic field is out of the page.

B = μI/2πd

d = 5 cm + 25 cm = 30 cm

B =  (4π x 10⁻⁷ x 4)/(2π x 0.3)

B = 2.67 x 10⁻⁶ T

The direction of the magnetic field is into the page towards P1.

B = μI/2πd

d = √(20² + 5²)

d = 20.62 cm

B =  (4π x 10⁻⁷ x 4)/(2π x 0.2062)

B = 3.88 x 10⁻⁶ T

The direction of the magnetic field is downwards towards P1.

Thus, the magnitude of the magnetic field at point P1 , midway between the wires is  1.005 x 10⁻⁴ T and the direction will be out of the page.

The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.

The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1  is 3.88 x 10⁻⁶ T and the direction is downwards.

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The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.

The work done is given by the use of the formula;

W = F * x

Where;

F = force applied

x = distance covered

W = 150 N *  4.95 m = 742.5 J

Now;

The coefficient of kinetic friction is given by;

μ = F/mg

μ = 150/ 33 * 9.8

μ = 0.46

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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

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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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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Based on the calculations, the sound intensity level is equal to 1.0 × 10⁻⁵ W/m².

Mathematically, sound intensity level can be calculated by using this formula:

[tex]\beta = 10 log(\frac{I}{I_{ref}} )[/tex]

Where:

I is the intensity of the sound.

Note: The reference value of sound intensity is equal to 1.0 × 10⁻¹² W/m².

Rewriting the formula, we have:

β/10 = logI - logIo

Substituting the parameters into the formula, we have;

60/10 = logI - log(1.0 × 10⁻¹²)

6 = logI + 12

logI = 6 - 12

logI = -6

I = 1.0 × 10⁻⁵ W/m².

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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 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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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 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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Answer:

Wc = 7.84    weight of crown

Ww = 7.84 - 6.86 = .98       weight of water displaced

Density = 7.84 / .98 = 8     crown is 8 X that of water

Since gold has a density of 19.3 that of water the crown is certainly not 100 percent (if any) gold  

(a) The  acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

(b) The velocity of the ball when it reaches its maximum height is zero.

(c) The initial velocity of the ball is 17.36 m/s.

(d) The maximum height it reaches is 15.36 m.

The acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

The velocity of the ball decreases as the ball moves upwards and eventually becomes zero at maximum height.

v = u - gt

at maximum height, final velocity, v = 0

0 = u - gt

u = gt

u = 9.81 x 1.77

u = 17.36 m/s

h = ut - ¹/₂gt

h = 17.36(1.77) - ¹/₂(9.81)(1.77²)

h = 15.36 m

Thus, the  acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

The velocity of the ball when it reaches its maximum height is zero.

The initial velocity of the ball is 17.36 m/s.

The maximum height it reaches is 15.36 m.

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Answer:

The type of lens involved is a convex lens and the object is positioned at the center of the curvature of the convex lens.

According to the question, a real, inverted, and same-sized image of the object is formed.

A concave lens is a diverging lens and always forms virtual images. But convex lenses are converging lenses and are the only type of lens that produce real and inverted images of the corresponding objects.

When an object is placed at the center of the curvature of a convex lens, its corresponding image is formed on the opposite side of the convex lens. The image formed is real and inverted.

The distance of the image from the lens is equal to the distance of the object from the lens.

For a convex lens, the distance of the center of curvature from the lens is double the focal length of the lens.

That's why the convex lens and center of curvature are the correct answer to this question.

Explanation:

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 initial momentum of the yellow and the orange train is 1000kgm/s.

Momentum is the product of the mass and velocity of any object.

Momentum is denoted by P.

Momentum P = mv , where m = mass and v = velocity.

Mass of the orange train = 200kg

Velocity of the orange train = 1m/s

So, the momentum of the orange train will be,

                            ∴    P = mv

                                  P = 200 x 1

                                  P = 200 kgm/s

∴   The initial momentum of the orange train is 200kgm/s.

Mass of the yellow train = 100kg

Velocity of the yellow train = 8m/s

So, the momentum of the yellow train will be,

                            ∴    P = mv

                                  P = 100 x 8

                                  P = 800 kgm/s

∴ The initial momentum of the yellow train is 800kgm/s.

Therefore, the initial momentum of the yellow and the orange train is 1000kgm/s.

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The magnitude of the cannons velocity after the ball is fired will be 2.4m/s.

To find the answer, we have to know more about the law of conservation of linear momentum.

                      [tex]M_B=8kg\\M_C=1*10^3 kg\\P_f=2.4*10^3 kgm/s.\\[/tex]

                   [tex]P_f=M_CV_C\\V_C=\frac{P_f}{M_C}=\frac{2.4*10^3}{1*10^3} =2.4m/s \\west[/tex]

Thus, we can conclude that, the magnitude of the cannons velocity after the ball is fired is 2.4m/s.

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After the ball is shot, the cannon will move at a speed of 2.4 m/s.

We must learn more about the rule of conservation of linear momentum in order to locate the solution.

How can I determine the cannon's post-ball velocity magnitude?

                          [tex]m=8kg\\M=1000kg\\P_f=2.4*10^3kgm/s[/tex]

                               [tex]P_f=MV\\V=\frac{P_f}{M} =2.4m/s[/tex]

Thus, we can infer that the cannon's maximum speed once the ball is fired is 2.4 m/s.

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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 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 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 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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The output current for the given step down transformer is 30A.

What is the step down transformer?

A step down transformer is a passive device that converts high voltage power to low voltage power, while the output current is higher than the  input current. They are used in power adaptors and rectifiers to decrease the voltage to the desired level. It works according to Faraday's law of Electromagnetic induction.

The current in the windings of a step down transformer is inversely proportional to the voltage in windings as:

Input voltage / Output voltage = Output current / Input current

220 / 110 = Outout current / 15

Output current = 30A

Hence, the output current (30A) obtained is higher than the given input current (15A).

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The vertical component of the force being applied to the beam by the high-tension cable is 36.37 N.

Calculation of the cable's tension-

Utilizing the moment principle, determine the cable's tension:

Torque clockwise = TL sin∅

Counterclockwise torque = 1/2WL

TL sin∅ = 1/2WL

⇒T sin∅ = 1/2W

⇒T = W/2sin∅

⇒T = (29* 9.8)/ (2*sin57)

⇒T = 169.43 N

Calculation of the vertical component of the force-

Forces that operate perpendicular to the surface vertical plane are called the vertical force. Gravity always pulls objects straight down to the earth's core.

T+F = W

⇒F = W-T

⇒F = (21*9.8)-169.43

⇒F = 36.37 N

So, the force on the beam has a vertical component of 36.37 N.

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according to definition of acceleration

a=v-u/t

t=v-u/a(equation 1)

according to the formula of average velocity

v+u/2*s/t

s=v+u/2*t(equation 2)

now putting the value of t in equation 2

s=v+u/2*v-u/a

s=v^2-u^2/2a

v^2=u^2+2as

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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:

(a) The induced current in the loop will be counterclockwise.

(b) The rate at which electrical energy is being dissipated by the resistance of the loop is 0.012 W.

The induced current in the loop will be counterclockwise to the direction of magnetic field.

emf = -NdФ/dt

emf = -NBA/dt

where;

A is area of the loop

A = πr² = π(0.041)² = 5.28 x 10⁻³ m²

emf = -(-0.605 - 7.78) x 5.28 x 10⁻³

emf = 8.385 x 5.28 x 10⁻³

emf = 0.0442 V

P = emf²/R

P = (0.0442)²/0.169

P = 0.012 W

Thus, the rate at which electrical energy is being dissipated by the resistance of the loop is 0.012 W.

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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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According to mathematics, the planet's angle is stated as

dY=704 degrees.

We may demonstrate this by using Kepler's third law, which asserts that a planet's orbit squared is a function of cubed radius.

The equation for the period is often expressed numerically as

[tex](periodX / periodY)^2 = (radius X / radius Y)^3[/tex]

Therefore

(pX / pY)^2 = 4^3

(pX / pY)^2 = 64

[tex]\sqrt{(pX / pY )^2}= \sqrt{64}[/tex]

pX / pY=8

In conclusion, planet Y travels 8 times further than planet X does in the same amount of time since one orbit on planet X takes 8 times longer to complete.

planet Y travels ;

dY=8 * 88.0

dY= 704 degrees

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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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