A satellite is a large distance from a planet, and the gravitational force from the planet is the only significant force exerted on the satellite. The satellite begins falling toward the planet, eventually colliding with the surface of the planet. As the satellite falls, what claims is correct about how the force that the planet exerts on the satellite Fps changes and how the force that the satellite exerts on the planet Fsp changes, if at all? What reasoning supports this claim?

Answers

Answer 1

Answer:

The answer is "Option A".

Explanation:

Please find the complete question in the attached file.

The strength is equal due to the third law of Newton, and [tex]Fsp = Fps = \frac{Gmsmp}{r^2}\[/tex], therefore the force increases with decreased separation r.

A Satellite Is A Large Distance From A Planet, And The Gravitational Force From The Planet Is The Only

Related Questions

Explain how waves and their interactions with matter are used in the operation of technical devices. (i.e. solar cells, scanners, and medical imaging)

Explain energy transformations

Explain how electromagnetic induction is used in a simple motor

Relate any other terms like, speed, acceleration, momentum, friction, forces, gravity will affect the motion of the device.

Understand the 3 basic attributes of a PV cell. (Absorption of light, separation of charge carriers, and extraction of charge carriers to an external circuit.)

I know it's a lot but I really need to pass this class and I have zero passion for physics, they forced me to take this class. If I had biology I would have loved it. but you know anyone who loves Physics is welcome :)

Answers

Answer:

Examples of electromagnetic waves include radio waves, microwaves, infrared, visible light, ultraviolet, x-rays, and gamma rays. ... Microwaves are used to cook your food. Infrared waves are used in remote controls and are emitted from all warm objects, allowing them to be used to create heat-sensitive cameras.

Explanation:

brainliest please

14.According to the graph how far does the person travel in the first 5
seconds
2 points
Your answer
6
Displacement (m)
1
2
4 5 6 7 8 9 10 11 12 13
Time (s)

Answers

Answer:

d = 5 m

Explanation:

In this exercise we have a graph of displacement against time, the graph being a line, so the body has a uniform movement, the speed of the person is

         v = [tex]\frac{\Delta x}{\Delta t}[/tex]

         v =[tex]\frac{5-0}{5-0}[/tex] (5-0) / (5-0)

         v = 1 m / s

therefore the displacement is

         d = v t

         d = 1  5

          d = 5 m

Anna litic and Noah formula how place a 1.50 kg brick on a wooden board and incline the board at 34.4* above the horziontal. The coefficient of friction between the brick and the board is 0.350. determine the force of gravity, parallel component of gravity and the perpendicular component of gravity. Please also find net force and acceleration

Answers

The force due to gravity here is 14.7 N. The net force acting on the brick is 9.5 N and the acceleration of the brick is then 6.3 m/s².

What is frictional force?

Friction is a resistive force which opposes the motion of an object. The product of normal force by gravity and frictional coefficient gives the frictional force.

Given that, mass = 1.50 Kg

force by gravity = mg = 1.50 × 9.8 m/s² = 14.7 N

parallel component = 14.7 cos 34.4 = 12.12 N

perpendicular component  = 14.7  sin 34.4 = 8.30 N

The frictional force = mg × coefficient of friction

= 14.7 × 0.35 = 5.14 N

then net force = 14.7 - 5.14 = 9.5 N

Acceleration of the brick = net force/mass

a = 9.5 N/1.50 Kg = 6.30 m/s²

Hence, acceleration on the brick is 6.30 m/s².

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8:11
GCSE Science - Physics
22 of 25
14 67%
A cat accelerates from rest to 10 m/s
when it sees a dog. This takes 2
the cat?
seconds What is the acceleration of the cat ?

Answers

the acceleration of the cat is 5m/s square

To find the acceleration of the cat, you can use the formula a = (v-u)/t, where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time it takes for the acceleration to occur.

In this case, the initial velocity (u) is 0 m/s (since the cat starts at rest), the final velocity (v) is 10 m/s, and the time (t) is 2 seconds. Plugging these values into the formula, we get:

a = (10 m/s - 0 m/s) / 2 s
= 10 m/s / 2 s
= 5 m/s^2

So the acceleration of the cat is 5 m/s^2.

examine the following graph.
a) What is the amplitude of the oscillation?
b) What is the period of the oscillation?​

Answers

A) 0.2m since it's the height of the wave.

b)1s it's how long a wave is.

An astronomy class is so excited by the discovery of planets around other stars that they decide to do a library exhibit on the subject so that everyone in the school can learn about it. In this exhibit they want to pay tribute to both the astronomers of today who have done the work AND some of the scientists of the past whose work was essential to making the discoveries possible (and directly related to the techniques involved). Which of the following scientists of the past should definitely be included in the exhibit?

a. George Herbigâ
b. Ejnar Hertzsprungâ
c. Ptolemyâ
d. Gerard Kuiperâ
e. Christian Dopplerâ

Answers

Answer: e. Christian Dopplerâ

Explanation:

Based on the information given, the scientist of the past that should definitely be included in the exhibit is Christian Dopplera.

He described how the frequency of sound waves and light is being affected by the relative speed of both the source and also the observer. This was referred to as the Doppler effect.

In this scenario, the Doppler effect can be used to show how the universe is expanding. Therefore, the correct option is E.

What keeps planets in our solar system?
A. The elliptical motion of the planets around the sun
B. The gravitational force between the Earth and Sun
C. The magnetic field surrounding the Sun
D. The electromagnetic radiation from the sun​

Answers

Answer:C

Explanation:

3. Which of the forms of electromagnetic radiation listed below has the greatest energy?
Ogamma rays
O infrared
O ultraviolet
radio waves

Answers

Answer:

Choice A. gamma rays

Explanation:

You apply a net force on a soccer ball of 15 N. If the acceleration it has is 5 m/s2 what is the mass of the ball?​

Answers

Answer:

3 kg

Explanation:

The mass of the ball can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

f is the force

a is the acceleration

We have

[tex]m = \frac{15}{5} = 3 \\ [/tex]

We have the final answer as

3 kg

Hope this helps you

brandon buys a new seadoo. he goes 12km north from the beach he jumps wakes for 6km to the east what distance did he cover what was his displacement

Answers

Total distance covered by Brandon is 18 km and total displacement covered by him is 13.41 km.

Displacement: What is it?

The definition of displacement is the changing of an object's position. It has a magnitude and direction and is a vector quantity. It is shown as an arrowhead that travels from the initial location to the end. An object's position changes, for instance, if it moves from position A to position B.

Distance covered by Brandon is-

12 + 6 = 18 km

Displacement covered by Brandon is-

d²= 12² + 6²

= 144 + 36

d²= 180

d = √180

d = 13.41 km.

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The function of the ossicles in the middle ear is to transmit the vibrations of the tympanic membrane caused by sound waves propagated in air to the fluid-filled cochlea. In doing so, the ossicles change the nature of the vibrations. The ear tries to conserve the energy in the wave I,
where v is the wave speed, cramster-equation-2010731130406341613664 is the density of the medium, cramster-equation-2010731131286341613668 is the frequency of the wave and A is the amplitude of the wave.
Calculate the magnitude of the change in A that occurs. The density of air and water is 1.3x10–3g/cm3 and 1.0g/cm3, respectively, and the velocity of sound is 331 m/s in air and 1410 m/s in water. The frequency of the wave remains constant as the wave propagates. (Hint: Begin with to find .)

Answers

A I hope I helpedddd

The frequency of the wave which remains constant as the wave propagates and the magnitude of change in amplitude is 57.2.

What is Density?

Density can be defined as how much matter is contained within a particular volume of a substance. A dense object is the one which weighs more than a less dense object which is the same size. An object which is less dense than that of water will float on it, whereas an object with greater density will sink. The density equation is: D = M / V.

density: dair = 1.3 x 10⁻³ g/cm³, dwater = 1.0 g/cm³,

velocity: vair = 331 m/s, vwater = 1410 m/s

v×d×A² = constant

vair×dair×Aair² = vwater×dwater×Awater²

so,

Aair/Awater = sqrt[(vwater/vair)×(dwater/dair)] = 57.2

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A 4.51 kg object is placed upon an inclined plane which has an incline angle of 23.0*. The object slides down the inclined plane with a constant speed. Find the normal force, friction force and the coefficient of sliding friction

Answers

To find the normal force, we can use the equation: normal force = weight + friction force * cos(incline angle).

How to find the normal force ?The weight of the object is (4.51 kg) * (9.8 m/s^2) = 44.398 NTo find the friction force, we can use the equation: friction force = coefficient of friction * normal force.We can assume that the friction force is equal to the force of gravity acting against the object because it is moving down the inclined plane at a constant pace. As a result, the friction force is equal to the product of the object's weight and sin (incline angle)Friction force is equal to (9.927 N)*sin(23.0)*(44.398 N)We can use the following equation to determine the coefficient of sliding friction:friction coefficient is calculated as friction force divided by normal force.coefficient of sliding friction = 9.927 N /44.398 N = 0.224Therefore, the normal force is 44.398 N, the friction force is 9.927 N, and the coefficient of sliding friction is 0.224.

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A dog drags a 1-kg bone across the floor for 2 meters with an applied
force of 10N. How long did it take him if he used 40 Watts of power?
seconds

Answers

Answer:

0.5 seconds

Explanation:

Work = Fdcos(theta) = 10*2*cos(0) = 20 J

Power = W/t

40 = 20/t

t = 0.5s

Which of the following is the only group in mission control that gets to communicate with the astronauts in space?


Flight Director


CAPCOM


Senior Flight Controller


Lead Ground Astronaut

Answers

Capsule communicator or Capcom  is the only group in mission control that gets to communicate with the astronauts in space. Hence, option (B) is correct.

What is capsule communicator or Capcom?

The capsule communicator, or Capcom, was the only voice that spoke to the astronauts during their trip to avoid any mistake. To ensure that the men in the capsule always had a familiar individual who understood their situation and could provide the information they required, Capcom was always manned by astronauts.

Both the technical control team on the ground and the astronauts in space are represented by Capcom.

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8. A 2kg object explodes and divides into three pieces, one piece has a mass of 800g and has a velocity of [30] m/s, a second piece has a mass of 500g and has a velocity of [520] m/s. What is the velocity of the third mass?​

Answers

Answer:

v3 = 0 gm/s / 700g

Explanation:

To solve this problem, you need to use the principle of conservation of momentum, which states that the total momentum of a system remains constant unless acted upon by an external force. In this case, the total momentum of the system (the 2kg object before it explodes) is equal to the sum of the momenta of the three pieces after the explosion.

You can calculate the momentum of each piece by multiplying its mass by its velocity:

P1 = 800g * 30 m/s = 24,000 gm/s

P2 = 500g * 520 m/s = 260,000 gm/s

The total momentum of the system is the sum of these two momenta:

Ptotal = P1 + P2 = 24,000 gm/s + 260,000 gm/s = 284,000 g*m/s

The third piece has a mass of 2kg - 800g - 500g = 700g. We can use the conservation of momentum equation to find its velocity:

Ptotal = (700g * v3) + (800g * 30 m/s) + (500g * 520 m/s)

v3 = (Ptotal - (800g * 30 m/s) - (500g * 520 m/s)) / 700g

v3 = (284,000 gm/s - (800g * 30 m/s) - (500g * 520 m/s)) / 700g

v3 = (284,000 gm/s - 24,000 gm/s - 260,000 gm/s) / 700g

v3 = (284,000 - 24,000 - 260,000) gm/s / 700g

v3 = 0 gm/s / 700g

The velocity of the third mass is 0 m/s.

Hope this helps.

Determine the speed of the 50-kg cylinder after it has descended a distance of 2 m, starting from rest. Gear A has a mass of 10 kg and a radius of gyration of 125 mm about its center of mass. Gear B and drum C have a combined mass of 30 kg and a radius of gyration about their center of mass of 150 mm

Answers

This question is incomplete, the missing image is uploaded along this answer below.

Answer:

the speed of the 50-kg cylinder after it has descended is 3.67 m/s

Explanation:

 Given the data in the question and the image below;

relation between velocity of cylinder and velocity of the drum is;

V[tex]_D[/tex] = ω[tex]_c[/tex] × r[tex]_c[/tex]  ----- let this be equ 1

where V[tex]_D[/tex] is velocity of cylinder,  ω[tex]_c[/tex] is the angular velocity of drum C and r[tex]_c[/tex] is the radius of drum C

Now, Angular velocity of gear B is;

ω[tex]_B[/tex] = ω[tex]_C[/tex]

ω[tex]_B[/tex] = V[tex]_D[/tex] / r[tex]_c[/tex]  -------- let this equ 2

so;

V[tex]_D[/tex] / 0.1 m = 10V[tex]_D[/tex]

Next, we determine the angular velocity of gear A;

from the diagram;

ω[tex]_A[/tex]( 0.15 m ) = ω[tex]_B[/tex]( 0.2 m )

from equation 2; ω[tex]_B[/tex] = V[tex]_D[/tex] / r[tex]_c[/tex]

so

ω[tex]_A[/tex]( 0.15 m ) = (V[tex]_D[/tex] / r[tex]_c[/tex] ) 0.2 m

substitutive in value of radius r[tex]_c[/tex] (0.1 m)

ω[tex]_A[/tex]( 0.15 m ) = (V[tex]_D[/tex] / 0.1 m ) 0.2 m

ω[tex]_A[/tex]( 0.15 ) = 0.2V[tex]_D[/tex] / 0.1

ω[tex]_A[/tex] =  2V[tex]_D[/tex]  / 0.15

ω[tex]_A[/tex] = 13.333V[tex]_D[/tex]   ----- let this be equation 3

To get the speed of the cylinder, we use energy conversation;

assuming that the final position is;

T₁ + ∑[tex]U_{1-2[/tex] = T₂

0 + m[tex]_D[/tex]gh = [tex]\frac{1}{2}[/tex]m[tex]_D[/tex]V²[tex]_D[/tex] + [tex]\frac{1}{2}I_A[/tex]ω²[tex]_A[/tex] + [tex]\frac{1}{2}I_B[/tex]ω²[tex]_B[/tex]

so

m[tex]_D[/tex]gh = [tex]\frac{1}{2}[/tex]m[tex]_D[/tex]V²[tex]_D[/tex] + [tex]\frac{1}{2}[/tex](m[tex]_A[/tex]k[tex]_A[/tex]²)(13.333V[tex]_D[/tex])² + [tex]\frac{1}{2}[/tex](m[tex]_B[/tex]k[tex]_B[/tex]²)(10V[tex]_D[/tex])²

we given that; m[tex]_D[/tex] = 50 kg, h = 2 m, m[tex]_A[/tex] = 10 kg, k[tex]_A[/tex] 125 mm = 0.125 m, m[tex]_B[/tex] = 30 kg, k[tex]_B[/tex] = 150 mm = 0.15 m.

we know that; g = 9.81 m/s²

so we substitute

50 × 9.81 × 2 = ( [tex]\frac{1}{2}[/tex] × 50 × V[tex]_D[/tex]²) + [tex]\frac{1}{2}[/tex]( 10 × (0.125)² )(13.333V[tex]_D[/tex])² + [tex]\frac{1}{2}[/tex]( 30 × (0.15)²)(10V[tex]_D[/tex])²

981 = 25V[tex]_D[/tex]² + 13.888V[tex]_D[/tex]² + 33.75V[tex]_D[/tex]²

981 = 72.638V[tex]_D[/tex]²

V[tex]_D[/tex]² = 981 / 72.638

V[tex]_D[/tex]² = 13.5053

V[tex]_D[/tex] = √13.5053

V[tex]_D[/tex] = 3.674955 ≈ 3.67 m/s

Therefore,  the speed of the 50-kg cylinder after it has descended is 3.67 m/s

A man lifts various loads with the same lever. The distance of the applied force from the fulcrum is 2.00 m and the distance from the fulcrum to the load is 0.500 m. A graph of resistance force vs. effort force is shown. What is the mechanical advantage of the lever? What is the ideal mechanical advantage of the lever? What is the efficiency of the lever? Show your work.

Answers

The mechanical advantage of the lever is 3.5.

The ideal mechanical advantage of the lever is 4

The efficiency of the lever is 87.5 %.

What is the mechanical advantage of the lever?

The mechanical advantage of the lever is the ratio of the load overcame by the lever to the effort applied by the lever.

M.A = Load / Effort

From the graph we will find the change in load to change in effort is calculated as;

M.A = ( 160 N - 20 N ) / ( 50 N - 10 N )

M.A = 3.5

The ideal mechanical advantage of the lever is calculated as follows;

I.M.A = distance travelled by effort / distance travelled by load

I.M.A = ( 2 m ) / ( 0. 5 m )

I.M.A = 4

The efficiency of the lever is calculated as follows;

E = M.A / I.M.A x 100%

E = ( 3.5 / 4 ) x 100%

E = 87.5 %

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Grandma Sue (mass 80 kg) and her grandson James (mass 40 kg) are on a smooth icy surface. As Grandma Sue whizzes around the icy surface at 3 m/s in a straight line, she is suddenly confronted with scared James standing at rest directly in her path. Rather than knock him over, she picks him up and continues her uniform motion in a straight line without braking. Find the speed of Grandma Sue and James after the collision.

Answers

Answer:

v = 2 m/s

Explanation:

Here, we will use the law of conservation of momentum to solve this problem:

[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]

where,

m₁ = mass of grandma = 80 kg

m₂ = mass of James = 40 kg

u₁ = initial speed of grandma = 3 m/s

u₂ = initial speed of James = 0 m/s

v₁ = v₂ = v = final speed of grandm and James = ?

Therefore,

[tex](80\ kg)(3\ m/s)+(40\ kg)(0\ m/s)=(80\ kg)(v)+(40\ kg)(v)\\\\(120\ kg)v = 240\ Ns\\\\v = \frac{240\ N.s}{120\ kg}\\[/tex]

v = 2 m/s

Compare scalar and vector quantities using the definitions of distance and displacement

Answers

Answer:

Distance is a scalar quantity while displacement is a vector quantity

Explanation:

A scalar quantity represents only the magnitude and does not give any detail about the direction of the quantity for example distance. Distance can be any length measured in any direction (no specific direction)

However, a vector quality represents both the magnitude and direction. For instance displacement is a vector quantity. If direction is not defined then displacement becomes equal to distance.

What will be the linear expansion of a piece of wrought iron 8 feet long if it’s temp is raised 70 degrees F?

Answers

The linear expansion of the wrought iron is 0.0038 feet.

What is the linear expansion of the wrought iron?

The linear expansion of the wrought iron is the increase in length of the wrought iron due to change or increase in temperature.

Mathematically, the formula for the linear expansion of metals is given as;

ΔL = αLΔθ

where;

ΔL is the linear expansion of the metalα is the coefficient of thermal expansion of the metalL is the original length of the metalΔθ is the change in temperature of the metal

The coefficient of thermal expansion of wrought iron = 0.0000067 / ⁰F

The linear expansion of the wrought iron is calculated as follows;

ΔL = αLΔθ

ΔL = ( 0.0000067/ ⁰F   x   8 ft    x   70 ⁰F )

ΔL = 0.0038 ft

Thus, the linear expansion of the wrought iron is a function of the increase in temperature of the iron and the original length of the iron.

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which of the following changes will increase the frequency of an oscillating pendulum?
a. an increase in the mass of the pendulum.
b. an increase in the initial height of release.
c. an increase in the length of the rope.
d. more than one of the above
e. none of the above
explain your answer.
NO LINKS. ​

Answers

Answer:

b because we apply Hooke's law

Explanation:

Hooke's law

a stage light uses a white lamp wgat color of light will be seen when
a yellow filter is used
an orange filter is used

Answers

Answer:

Explanation:

When a yellow filter is used, the stage light will appear yellow because the filter absorbs all colors of light except for yellow, which it allows through. When an orange filter is used, the stage light will appear orange for the same reason.

It is important to note that the color of the lamp itself will not change, but the light that is seen by the observer will appear to be the color of the filter due to the absorption and transmission properties of the filte

A monkey (mass m) is swinging on a vine of length L while carrying a bunch of bananas (a large bunch, mass m/2). His swinging motion has period T and reaches maximum height during the swing h (measured from the bottom of his arc of motion). He accidentally lets go of his bananas when he is at a height of h/2. What happens to the amplitude and period of his oscillation as a result? Explain.

Answers

Answer:

Explanation:

The period of oscillation will remain unchanged because the period of oscillation of a pendulum does not depend upon the mass of the bob  . Here monkey along with bunch of banana represents bob .

When the monkey and banana were at height h /2 , they have potential energy as well as kinetic energy . banana is separated from the system . It carried its total energy along with it . But the energy of monkey remained intact with it . So it will keep on moving as usual . So it will attain the same maximum height as before .

Hence the amplitude of oscillation too will remain unchanged .

Sarah and Devon are involved in an egg catching contest. They stand a fixed distance apart and toss a raw egg back and forth to each other. The goal is to catch the egg without it breaking. Which of the following would be the best strategy for Devon to employ as he attempts to catch the egg tossed by Sarah? Devon should…

A
Decrease the time it takes to stop the egg once it hits his hand.

B
Increase the time it takes to stop the egg once it hits his hand.

C
Tell Sarah to toss the egg at a very small angle to decrease the time the egg is in the air.

D
Tell Sarah to toss the egg at a rather large angle to increase the time the egg is in the air.

E
Use only one hand to catch the egg so that less mass is involved in stopping the egg.

Answers

Answer:

The correct answer is - B. Increase the time it takes to stop the egg once it hits his hand.

Explanation:

In order to catch the egg the force and time makes an impulse to the egg to change its momentum and bring it to rest. In order to change the momentum, one needs to decrease the force exerted onto the egg and increase the time it takes to make it rest.

Less force can be given to the egg by increasing the time to bring it to rest stop the egg once it hits his hand. Other than this it should be caught to the sides.

in v-belts the contact between the pulley and the belt is at the​

Answers

Answer:

Is at the pivot of the wheel

It’s bungee jumping skydiving and hiking

Answers

That’s a great question

A iguana runs back and forth along the ground. The horizontal position of the iguana in meters over time is shown
What is the displacement of the iguana between 3 s and 6 s?
m
What is the distance traveled by the iguana between 3 s and 6 s?

Answers

The displacement of the iguana between 3 s and 6 s  is 6.71 meters.

The distance traveled by the iguana between 3 s and 6 s is  8.08 meters.

What are distance and displacement?

Distance is the sum of an object's movements, regardless of direction.

The term "displacement" refers to a shift in an object's position.

According to the graph:

The displacement of the iguana between 3 s and 6 s

= √{ (3-6)²+(6-0)²} meters

= 6.71 meters.

The distance traveled by the iguana between 3 s and 6 s

= [ √{ (3-5)²+(6-6)²} +√{ (5-6)²+(6-0)²}] meters

= [2+ 6.08] meters

= 8.08 meters.

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Part A
What is the radius of the hydrogen-atom Bohr orbit shown in the figure? (Figure 1)
r = ____ nm

Answers

The radius of the hydrogen-atom Bohr orbit shown in the figure is 5.3 nm.

What is  Bohr orbit?

The path that hypothetical electrons take around the nucleus is known as Bohr's orbit.

These orbits are described by Bohr in his hypothesis of the structure of an atom as energy levels or shells where electrons move in a fixed circle around the nucleus.

These orbits resemble solar system orbits, with the exception that they are attracted by electrical forces rather than gravity. The term "ground state" refers to the amount of energy that an electron typically occupies.

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A 6 kg bowling ball is lifted 1.2 m into a storage rack. The acceleration of gravity is 9.8 m/s² Calculate the increase in the ball's potential energy. Answer in units of J.​

Answers

Answer:

70.56 J

Explanation:

Gravitational Potential Energy= mass×gravitational pull× height

= 6×9.8×1.2= 70.56 J

A 2kg block is attached to a spring for which K=200N/m it is held at an extension of 5 cm and then released at t=0.
A, the displacement as a function of time?
B, the acceleration when X=+A/2
C, the total energy when X=+A/2
D, the velocity when X=+A/2

Answers

The displacement, acceleration, energy and velocity of the simple harmonic motion of the mass attached to the spring are as follows;

A) x(t) = 0.05·sin(10·t + π/2)

B) The acceleration is; a(t) = -2.5 m/s²

C) The total energy is 0.0625 J

D) The velocity is ±√3/4 m/s

What is a simple harmonic motion?

The restoring force of a body in simple harmonic motion is directly proportional to the displacement of the body from its mean or central position.

Mass of the block, m = 2 kg

The spring constant, k = 200 N/m

The extension of the spring = 5 cm

Time at which the spring is released, t = 0

A. The motion of the spring with the mass is a Simple Harmonic Motion

The angular velocity can be obtained using the formula;

ω = √(k/m)

Therefore;
ω = √(200/2) = 10

The angular velocity of the block on the spring is, ω = 10 rad/s

The period, T = The time to complete 2·π rad

Therefore; T = 2·π rad/(10 rad/s) = π/5 s

The amplitude, A, is the cistance of the mass from the at rest position, which is 5 cm = 0.05 m

The equation of the extension of the spring is therefore;

x(t) = 0.05·sin(10·t + c)

At t = 0, x(t) = 0.05, therefore;

sin(10 × 0 + c) = sin(c) = 1

c = π/2

The equation for the displacement as a function of time is therefore;

x(t) = 0.05·sin(10·t + π/2)

B. The acceleration when x(t) = A/2 is obtained as follows;

x(t) = 0.05·sin(10·t + π/2)

A/2 = 0.05·sin(10·t + π/2)

A = 0.05

0.05/2 = 0.05·sin(10·t + π/2)

sin(10·t + π/2) = 1/2

10·t + π/2 = π/6

t = -π/30

cos(10×(-π/30) + π/2) = ±√3/2

v(t) = x'(t) = 0.05 × 10 × cos(10·t + π/2)

a(t) = v'(t) = -5·sin(10·t + π/2)

a(t) = v'(t) = -5·sin(10·t + π/2) = -5 × 1/2 = -2.5

The acceleration when X = + A/2 is -2.5 m/s²

C. The energy in a pring = (1/2)·k·x²

When x = A/2, we get;

E = (1/2) × 200 × (0.05/2)² = 0.0625

The energy in the spring when x = A/2 is 0.0625 J

D) The velocity when x = A/2 is; v(t) = x'(t) = 0.05 × 10 × cos(10·t + π/2)

v(t) = 0.5 × cos(10·t + π/2)

When x = A/2, sin(10·t + π/2) = 1/2, therefore;

cos(10·t + π/2) = ±√3/2

v(t) = 0.5 × ±√3/2 = ±√3/4

When x = A/2, the velocity, v(t) = ±√3/4 m/s

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