A cart of mass M is on a level, horizontal track. The cart is attached to one end of an ideal spring, and the other end of the spring is attached to a vertical support. The spring has a spring constant k. The cart is pulled to the right, stretching the spring a distance d, as shown above, and released from rest at time t = 0. The cart-spring system begins to oscillate, and the position x of the cart as a function of time t is given by the equation x(t) = d cos(ωt) , where x is in meters, ω is in rad/s , and t is in seconds. Assume the mass of the cart’s wheels to be negligible.

(a) Derive an equation for the velocity of the cart v as a function of time t .

(b) On the axes below, sketch a graph of the velocity v of the cart as a function of time t for two complete oscillations. On the graph, T represents the time for one complete oscillation. Explicitly label asymptotes, maxima, or minima with algebraic expressions, as appropriate.
(c) Express all answers in part (c) in terms of ω, t, M, d, k, and physical constants, as appropriate.

i. Determine an expression for the kinetic energy K of the system as a function of time t.

ii. Determine an expression for the potential energy U of the system as a function of time t.

iii. Using the equations from parts (c)i and (c)ii, show that the rate of change of the total energy of the system dE/dt is zero.
Two springs identical to the spring above are attached to each side of the cart of unknown mass M, as shown in Figure 1 above. The cart is on a level, horizontal track of negligible friction. A piece of clay is added to the top of the cart so that it will stick to the cart. The cart is displaced to the left a distance d and released from rest, as shown in Figure 2 above. The period of oscillation is recorded. This procedure is repeated for several different pieces of clay of different masses. In each trial, the cart is displaced the same distance d. The data are shown below on the linear graph of T2 as a function of mC , where T is the period of oscillation and mC represents the mass of the clay.
(d) From the graph, determine the following.

k, the spring constant of each spring

M, the mass of the cart

(e) The experiment is repeated, but in a second set of trials, the cart is pulled back a distance D, where D > d . Describe any changes that will occur for the straight line on the graph in part (d).

Justify your answer.

If the thrust was cut in half, the new acceleration of the spare shuttle would be half of its original acceleration.

Assuming that the mass of the space shuttle remains constant, the new acceleration of the space shuttle would be half of its original acceleration.

The formula for acceleration is:

a = F/m

where,

a = acceleration

F = force applied

m = mass of the object

If the force is cut in half, the new acceleration can be calculated as follows:

a' = (F/2)/m

where,

a' = new acceleration

Simplifying this equation, we get:

a' = F/2m

Since we know that a = F/m, we can substitute this expression into the equation above to obtain:

a' = a/2

So, the new acceleration of the space shuttle would be half of its original acceleration.

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The bicyclist is applying more force on the pedals at Time 2. Option C

If we look at the image that have been shown, we can be able to see that the force that is acting have been shows by the arrows that have been used to label the movement of the cyclist in the image that is shown here.

We can see that the forward arrow at the time 2 is seen to be larger than the forward arrow that is shown for time 1. The implication of this is that the cyclist is cycling harder and applying more force at time 2 than at time 1.

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An electromagnet is a type of magnet in which the magnetic field is produced by an electric current. Electromagnets usually consist of wire wound into a coil. A current through the wire creates a magnetic field which is concentrated in the hole in the center of the coil.

The volume occupied by 1.80 mol of steam at 200°C and atmospheric pressure is approximately 40.7 L.

We can use the ideal gas law, PV = nRT, to solve for the volume of the steam. First, we need to determine the pressure of the steam at 200°C and atmospheric pressure. We can use a steam table to find that the saturation pressure of steam at 200°C is 1.013 bar.

Next, we can convert this pressure to Pascals,

1.013 bar = 101,325 Pa

We also need to convert the temperature to Kelvin 200°C + 273.15 = 473.15 K.

Now we can plug in our values,

PV = nRT

V = nRT/P

V = (1.80 mol)(8.314 J/mol•K)(473.15 K)/(101,325 Pa)

V ≈ 67.6 L

Therefore, the volume occupied by 1.80 mol of steam at 200°C and atmospheric pressure is approximately 67.6 liters.

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The work-energy theorem states that the change in the kinetic energy of an object will be equal to the net work done on the object.

Mathematically, it can be expressed as;

ΔKE = W

Where; ΔKE represents the change in kinetic energy of the object,

W represents the net work done on the object.

This theorem states that when work is done on an object, it results in a change in its kinetic energy. If work is done on an object, its kinetic energy increases, and if work is done by an object, its kinetic energy decreases.

This theorem is a fundamental principle in physics that relates the concepts of work and energy, and it is often used to analyze the motion and behavior of objects in various physical systems.

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The change in length of mercury is +0.0441 cm.

With the atomic number 80 and the symbol Hg, mercury is a metallic element. Because of its silvery appearance and capacity to flow smoothly like a liquid, it is also known as quicksilver. The only metal that is liquid at normal pressure and room temperature is mercury.

The following formula describes how a change in temperature causes a material's length to alter:

ΔL = αLΔT

where L is the length change, is the thermal expansion coefficient, L is the starting length, and T is the temperature change.

If we substitute the values provided, we get:

ΔT = 57.0°C - 32.5°C = 24.5°C

L = 3.00 cm

  = 60 x 10⁻⁶ °C (provided)

Thus, the length difference is as follows:

ΔL = (60 × 10⁻⁶/°C) (3.00 cm) (24.5°C)

ΔL = 0.0441 cm

The mercury column will enlarge due to the rise in temperature, making the length change positive.

There will be an estimated +0.0441 cm, an increase in the mercury column's length.

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The compressive stress in the steel column is found to be approximately 397.6 MPa.

The formula for calculating the area of a circle can be used to determine the steel column's cross-sectional area (A),

A = π*(d/2)², diameter of the column is d,

A = π*(0.4/2)²

A = 0.1257m²

The compressive stress (σ) in the column can be calculated using the formula, σ = F/A, F is the load carried by the column is F.

σ = 50 MN/0.1257m²

σ = 397.6 MPa

Therefore, the compressive stress in the steel column is approximately 397.6 MPa.

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According to the question the magnitude of the acceleration of the two objects is 0 and the tension in the cord is also 0.

Acceleration is the rate of change of velocity. It is a vector quantity, meaning it has both magnitude and direction. Acceleration can be caused by a force, such as gravity or friction, or by a change in the velocity itself. In the case of a force, the acceleration is proportional to the force and inversely proportional to the mass of the object.

Let a be the acceleration of the ball and b be the acceleration of the block.

Applying Newton's second law to the ball,

T − mg sin30o = 5a

Applying Newton's second law to the block,

T + mg sin30o − 12b = 0

Solving the two equations,

T = 10a

b = 10a/12

Substituting b in the first equation,

T − mg sin30o = 5a

10a − 5(10) sin30o = 5a

5a(1 − 10 sin30o) = 0

a = 0

b = 0

T = 0

Therefore, the magnitude of the acceleration of the two objects is 0 and the tension in the cord is also 0.

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Sound waves are longitudinal waves.

Since, the particles of the medium through which the sound is transmitted vibrate parallel to the direction that the sound wave moves, they are longitudinal waves.

A sound wave is a pressure wave; as a result of the vibrations of the sound source, zones of high pressure (compressions) and low pressure (rarefactions) are created. These rarefactions and compressions are caused by the sound.

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Answer:

A.

Position-time graph:

The position-time graph for the ball punted vertically will be a parabolic curve, with the vertex at the highest point of the ball's trajectory. Since the initial position is zero, the curve will go through the origin.

Velocity-time graph:

The velocity-time graph will be a straight line that starts at the initial velocity and decreases linearly until it reaches zero at the highest point of the ball's trajectory. After that, the velocity increases linearly in the negative direction until the ball hits the ground.

Acceleration-time graph:

The acceleration-time graph will be a constant negative value, representing the acceleration due to gravity.

Motion map:

A motion map is a diagram that shows the position of an object at several specific times during its motion. For the ball punted vertically, the motion map would look like this:

|O|------|-------|------|-------|------|H|

O represents the initial position, and H represents the highest point of the ball's trajectory. The "|" symbols represent the position of the ball at regular intervals of time.

B.

To find the initial velocity of the ball, we can use the hang time and the acceleration due to gravity.

Hang time = 3.8 seconds

Acceleration due to gravity = -9.8 m/s^2

At the highest point of the ball's trajectory, the velocity is zero. Therefore, we can use the following kinematic equation to find the initial velocity:

hang time = (final velocity - initial velocity) / acceleration

Solving for initial velocity:

initial velocity = final velocity - (hang time x acceleration)

final velocity = 0 m/s

hang time = 3.8 s

acceleration = -9.8 m/s^2

initial velocity = 0 - (3.8 x -9.8)

initial velocity = 37.64 m/s

Therefore, the initial velocity of the ball was 37.64 m/s.

Rotational inertia of the construction about an axis perpendicular to the picture is ml²/6.

According to the perpendicular axis theorem, the sum of any two perpendicular axes of the body that meet the first axis determines the moment of inertia for any axis that is perpendicular to the plane.

Moment of Inertia of the structure, Iₓ = Ml²/12

By perpendicular axis theorem,

I(cm) = Iₓ + Iy

I(cm) = 2Iₓ = 2 x ml²/12

I(cm) = ml²/6

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Explanation:

Yes, they can....and so can fish.    When a tornado passes ofver water , it forms what is called a 'water spout'  which sucks up water ...and frogs and fish....and drops them elsewhere.

The following is a force that is at work between two objects(A). gravity

is correct option.

Gravity is the force of attraction that exists between any two objects in the universe. It is a fundamental force that affects all objects with mass and is responsible for phenomena like the motion of planets around the sun, the tides in oceans, and the weight of objects on Earth's surface.

Force is a physical quantity that describes the interaction between two or more objects, causing a change in motion or deformation of the objects. It is represented by the symbol F and is measured in units of Newtons (N) in the International System of Units (SI).

There are several types of forces, including gravitational, electromagnetic, nuclear, and frictional forces. Gravitational force is the force of attraction that exists between any two objects with mass, while electromagnetic force is the force of attraction or repulsion between electrically charged particles. Nuclear forces are the forces that hold the atomic nucleus together, and frictional forces are the forces that resist the relative motion of objects in contact.

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The electric flux through a surface is calculated by the equation, φ = ∫E•dA, where E is the electric field, dA is the area vector and the integral is taken over the surface area.

Electric flux is a measure of the amount of electric field passing through a given area. It is represented by the letter Φ (phi) and is measured in units of volt-meters (Vm). Electric flux is related to the strength of electric fields, and the amount of electric flux passing through a given area is determined by the electric field strength and the area through which it passes. Electric flux is often used in calculations of electric field strength and electric field energy, as well as in the field of electromagnetism. Electric flux can also be used to calculate the total amount of electric charge passing through a given area.

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The magnitudes of the two forces are 6 N and 8 N when acting at right angles and their resultant is 10 N.

The problem involves finding two forces, given the magnitudes of their resultants when acting at different angles. Let F1 and F2 be the magnitudes of the two forces. When the two forces are at right angles, the resultant force R is given by:

R = √(F1² + F2²) = 10 N

Squaring both sides, we get:

F1² + F2² = 100

When the two forces act at an angle of 60 degrees, their resultant R is given by:

R = √(F1² + F2² + 2F1F2cos60) = 13 N

Squaring both sides and using cos60 = 0.5, we get:

F1² + F2² + F1F2 = 169

Now, we have two equations with two unknowns. Solving for F1 and F2, we get:

F1 = 6 N

F2 = 8 N

As a result, the magnitudes of the two forces are 6 N and 8 N when operating at right angles, respectively, and the resultant is 10 N. When the two forces act at an angle of 60 degrees, their magnitudes are still 6 N and 8 N, but their resultant is 13 N

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Answer:

D. Mechanical waves in which the vibrations are parallel to the motion of the sound.

Explanation:

gg

Answer: D
Mechanical waves in which the vibrations are parallel to the motion of the sound.

Explanation:
just took

According to the question the velocity and direction of motion of the particle at a height of 130m is 173.20 m/s, 30°.

Velocity is a vector physical quantity that measures the rate of change of an object's position. It is a combination of speed and direction and is measured in units such as meters per second (m/s). Velocity is the rate of change of an object’s position with respect to time. It is the amount of distance covered in a period of time. It is a vector quantity as it has both magnitude as well as direction. It can be calculated by dividing the distance covered by the time taken. Velocity is the rate at which an object is changing its position. It is a measure of how fast an object is traveling in a specific direction.

The velocity components of the particle at a height of 130m will be the same as it was at the time of projection since there is no acceleration, only gravity acting on it.

The velocity of the particle at a height of 130m will be:

Vx = 196 m/s * cos 30° = 173.20 m/s

Vy = 196 m/s * sin 30° = 98 m/s

The direction of motion of the particle at a height of 130m will be the same as the angle of projection i.e. 30°.

Therefore, the velocity and direction of motion of the particle at a height of 130m is 173.20 m/s, 30°.

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The magnitude of the force exerted on the large block by the small block is 16 N. Option A is correct.

The force exerted on the large block by the small block can be found using Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. The force exerted by the small block on the large block is equal in magnitude but opposite in direction to the force exerted by the large block on the small block.

Using Newton's Second Law of Motion, F = ma, where F is the net force, m is the total mass of the system, and a is the acceleration of the system. Therefore, the acceleration of the system is:

a = F/m = 16 N / (1 kg + 2 kg) = 5.33 m/s²

The force of friction between the two blocks is equal in magnitude but opposite in direction to the force exerted by the small block on the large block. Therefore, the force of friction is:

f = ma = 3 kg * 5.33 m/s² = 16 N

Therefore, the magnitude of the force exerted on the large block by the small block is 16 N. Option A is correct.

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The package hits the ground after 12.6 seconds.

The package's speed immediately before contact is 83.2 m/s.

To solve this problem using the kinematic equations of motion. First find the time it takes for the package to hit the ground.

(A) Using the vertical motion equation:

y = vi×t + (1/2)gt²

where y = vertical displacement,

vi = initial vertical velocity,

t = time, and

g = acceleration due to gravity.

The initial vertical velocity is given by:

vi = vsinθ = (175 m/s)sin(15°) = 45.12 m/s

Take upward as positive, so g is negative:

g = -9.8 m/s²

Substituting the values into the equation:

543 m = (45.12 m/s)t + (1/2)(-9.8 m/s²)t²

Simplifying and solving for t:

t = 12.6 s

Therefore, it takes 12.6 seconds for the package to hit the ground.

(B) Now, let's find the speed of the package just before the impact. Use the vertical motion equation again:

vf = vi + gt

where vf = final vertical velocity.

At the instant just before the package hits the ground, the final vertical velocity is:

vf = -√(2gh)

where h = height from which the package was released 543 m.

Substituting the values:

vf = -√(2(9.8 m/s²)(543 m)) = -83.2 m/s

Again, taking upward as positive, the speed of the package just before impact is:

v = |vf| = 83.2 m/s

Therefore, the speed of the package just before impact is 83.2 m/s.

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Answer:

the answer is b

Explanation:

The average velocity in the following cases are 1.743m/s and 1.22m/s.

To find the average velocity is given by:

V = [tex]\frac{total distance (d)}{time taken (t)}[/tex]

(a) From the question we are given:

d1 = 73.2m,

d2 = 73.2m,

v1 = 1.22m/s

v2 = 3.05m/s

We need to calculate the time taken to complete each of the distance using the formula:

t = [tex]\frac{total distance}{speed}[/tex]

t1 = 73.2/1.22 = 60s

t2 = 73.2/3.05 = 24s

The average velocity is the total distance divided by the total time:

Average velocity (V) = (d1 + d2)  / (t1 + t2)

V = 146.4 m / 84 s

Therefore, V = 1.743 m/s (to 3 significant figures)

(b) From the question we are given:

t1 = 1min = 60s,

t2 = 1min = 60s,

v1 = 1.22m/s

v2 = 3.05m/s

We need to calculate the distance covered for each of the time using the formula:

d1 = velocity * time = v1 * t1 = 1.22 * 60 = 73.2m

d2 = velocity * time = v2 * t2 = 3.05 * 60 = 183m

The average velocity is the total distance divided by the total time:

Average velocity (V) = (d1 + d2)  / (t1 + t2)

V = (73.2 + 183)/ (60 + 60)

V = 256.2/120 = 2.135m/s

Therefore, the average velocity (V) is 2.135m/s (to 3 significant figures)

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According to the question the image is located 18.00 cm away from the lens and the image is a real, inverted image.

Lens is an optical device used to focus or disperse light, or to change the optical properties of a beam of light. It is made from transparent materials like glass, plastic or quartz and can be used in a variety of applications such as cameras, microscopes, telescopes and other optical instruments. A lens can also be used to bend light, change its direction, and magnify objects.

The image is located 18.00 cm away from the lens and the image is a real, inverted image. This can be determined using the lens equation given by 1/f = 1/d_o + 1/d_i, where f is the focal length, d_o is the object distance, and d_i is the image distance. Plugging in the given values, we get 1/9.00 = 1/27.0 + 1/d_i, which simplifies to 1/d_i = 1/9.00 - 1/27.0, or d_i = 18.00 cm.

Since the image distance is positive, this indicates that the image is real, meaning it is actually formed at the image distance. This is further confirmed by the fact that the image is inverted, as a real image is always inverted.

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Answer:

Incident Ray:- A ray of light that strikes a surface.

Focal point:- Location where parallel rays of radiation meet after being refracted or reflected.

Principal axis:- Line passing through the center of the surface of a lens or spherical mirror.

Focal length:- Distance between the center of a convex lens or mirror and the focal point of the lens or mirror.

Normal:- A line perpendicular to the surface at a particular point.

Convex:- Curving outward with edges pointing away from the viewer .

Refracted:- Changed direction after passing from one medium to another.

Concave:- lens or mirror curving inward so the edges point toward the viewer.

Reflected:- bounced off of a surface.

1.44*[tex]10^{3}[/tex] m/s is the speed of a geosynchronous satellite orbiting Mars.   17.1* [tex]10^{6}[/tex]m is the altitude of a geosynchronous satellite orbiting Mars

What does geosynchronous satellite mean?

A geosynchronous satellite is positioned in an orbit with an orbital period equal to the rate of rotation of the Earth. It takes these satellites 24 hours to orbit the earth once. The equatorial plane is typically not the orbital plane for a typical geosynchronous satellite, though.

The period of rotation of the Mars is  T = 24.8 hours.

The radius of the Mars, R is 3.3895×[tex]10^{6}[/tex]

i.e. 2.05*[tex]10^{7}[/tex]m

The satellite's orbital speed is determined by; v⇒2πr/T i.e. 1.44* [tex]10^{3}[/tex]m/s

A geosynchronous satellite orbiting Mars has a height of: h⇒r-R i.e. 17.1* [tex]10^{6}[/tex]m

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a. The student's final position is 100.5 meters at an angle of 25.2° north of east from her initial position and b. The student would have to head in the direction of 334.8° south of east to return to her initial position.

a. To find the student's final position, we can use vector addition. We can start by breaking each displacement vector into its x- and y-components. First displacement (70 m at 37° north of east),

x-component = 70 cos(37°) = 56.02 m

y-component = 70 sin(37°) = 42.49 m

Second displacement (82 m at 20° south of east),

x-component = 82 cos(-20°) = 78.72 m

y-component = 82 sin(-20°) = -28.12 m

Third displacement (28 m at 30° west of north),

x-component = 28 sin(30°) = 14 m

y-component = 28 cos(30°) = 24.24 m

We can add the x-components and y-components separately to get the total displacement,

x-displacement = 56.02 m + 78.72 m - 14 m = 120.74 m

y-displacement = 42.49 m - 28.12 m + 24.24 m = 38.61 m

The magnitude of the total displacement is,

|d| = √(x-displacement² + y-displacement²)

|d| = √(120.74² + 38.61²)

|d| = 126.0 m.

The direction of the total displacement can be found using the inverse tangent function,

θ = atan(y-displacement/x-displacement)

θ = atan(38.61/120.74)

θ = 17.4° north of east. Therefore, the student's final position is 126.0 m at 17.4° north of east.

b. To return to her initial position, the student would need to walk the same distance and direction as her total displacement but in the opposite direction. The opposite direction of 17.4° north of east is 162.6° south of west. Therefore, she would need to walk 126.0 m at 162.6° south of west to return to her initial position.

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If one of the forces acts in the negative y-direction with a magnitude of 410 N. Then, the magnitude of the second force (F₂) is 410 N.

We can use Newton's second law of motion, which states that the net force acting on an object is equal to the object's mass times its acceleration:

ΣF = m × a

where ΣF is the net force, m is the mass, and a is the acceleration.

In this problem, we know the mass (m = 32 kg) and acceleration (a = 44 m/s²) of the object, and we also know that one force (F₁) acts in the negative y-direction with a magnitude of 410 N. Let's call the second force F₂, which acts in the positive x-direction.

To find the magnitude of F₂, we need to use vector addition to find the net force in the x-direction;

ΣFx = F₂

And in the y-direction;

ΣFy = -F₁

Since the acceleration is only in the x-direction, the net force in the y-direction must be zero. Therefore, we have;

ΣFy = 0 = F₂sin(θ) - F₁

where θ is the angle between F₂ and the x-axis, which is 90 degrees since F₂ acts in the x-direction.

Solving for F₂, we get;

F₂ = F₁ / sin(θ) = 410 N / sin(90°)

= 410 N

Therefore, the magnitude of the second force (F₂) is 410 N.

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Using Gauss's law, the net electric flux through a closed surface is given by:

Φ = Q_in / ε0

where Φ is the electric flux, Q_in is the total charge enclosed by the surface, and ε0 is the electric constant.

In this case, we can consider the sphere as the closed surface. The electric dipole will produce electric field lines that pass through the surface, but these lines will be symmetric and cancel out each other's contribution to the net flux. Therefore, the only contribution to the flux will be from the point charge Q.

The charge enclosed by the surface is simply Q, so the net electric flux is:

Φ = Q / ε0

This result does not depend on the position of the electric dipole inside the sphere, as long as it is symmetrically placed at the center.