An experiment is set up as follows:
A mass m = 9.4kg is sent down a frictionless ramp with an initial velocity of 2.5m/s. The ramp is 87cm
long and has an angle of 36°above the horizontal. At the bottom of the ramp is a horizontal surface with a
kinetic friction coefficient of μk = 0.27. At the far side of the horizontal surface, 48cm away, is a spring with
k-constant ks = 3,413.7N
m that will be compressed as the mass collides with the spring. The experiment
ends as the spring is fully compressed and the mass is at rest.
Note: The distance that the spring is compressed is in addition to the 48cm.
7) Find the initial energy of the mass. (10pts)
8) How far will the spring compress if there is no surface friction under the spring? (10pts)
9) How far will the spring compress if the surface friction continues under the spring? (20pts)

Answers

Answer 1

(a) The initial energy of the mass is 76.485 J.

(b) The compression of the spring in the absence of friction is 21.2 cm.

(c)  The compression of the spring in the presence of friction is 19.4 cm.

Initial energy of the mass

The initial energy of the mass is determined as follows;

E = K.E + P.E

E = ¹/₂mv² + mgh

where;

h is the height of the ramp

E = ¹/₂mv²  + mg x Lsinθ

P.E = ¹/₂(9.4)(2.5)² +  (9.4)(9.8)(0.87)(sin36)

P.E = 76.485 J

Compression of the spring when there is no surface tension

The compression of the spring in the absence of friction is calculated as follows;

Ux = E

¹/₂kx² = 76.485

kx² = 2(76.485)

x² = (2 x 76.485)/k

x = √(2 x 76.485)/k

x = √(2 x 76.485 / 3413.7)

x = 0.212 m

x = 21.2 cm

Compression of the spring in presence of friction

The compression of the spring in the presence of friction is determined by applying the principle of conservation of energy.

E - Fd = Ux

E - μmgd = ¹/₂kx²

76.11 - (0.27 x 9.4 x 9.8 x 0.48) = ¹/₂(3413)x²

64.17 = 1706.5x²

x² = 0.0376

x = √0.0376

x = 0.194

x = 19.4 cm

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Related Questions

A student in the Biomechanics class has decided that she would like to make her arms
stronger. She has a mass of 63 kg, She chooses to complete some elbow flexion exercises
using a kettlebell. For this problem, consider the hand and forearm to be a single segment.
The distance from her elbow to her wrist is 22.86 cm.
The force from the kettlebell is applied to her hand, which is 30.48 cm from her elbow joint.
She knows that the moment arm of the elbow extensor muscles about the elbow axis is

Answers

Answer:

what is heat and transfer

PLEASE HELP!! A TOTAL OF 50 POINTS WHEN GIVEN BRAINLIEST!!

Part A: Find the work done in lifting 1 L of blood (mass 1 kg ) from the foot to the head of a 1.9 m -tall person.

Part B: If blood circulates through the body at the rate of 5.0 L/min , estimate the heart's power output. (Your answer underestimates the power by a factor of about 5 because it neglects fluid friction and other factors.)

Answers

Explanation:

for part A:

work done is equal to change in potential energy which is

MgH

so the answer is 1.9 × 1 × 9.81

ANSWER FOR PART A: 18.639

FOR PART B:

power = flow rate x gh

= 5×9.81×1.9

93.195 watts


23) What is the magnitude of the electric field intensity at a point where a proton experiences an
electrostatic force of magnitude 2.30 x 10-25 newton?

Answers

Answer: I am so sorry i hadn't learned this Yet~!

Explanation:

An elevator cabin has a mass of 358.1 ݇݃, and the combined mass of the people inside the cabin is 169.2 ݇݃. The cabin is puled upward by a cable, with a constant acceleration of 4.11 ݉/ݏଶ
. What is the tension in the cable?

Answers

Explanation: First fix your units. acceleration is in m/s/s.

T-Mg=Ma

M=348.1+169.2 =517.3 kg

T= 517.3 (9.81+4.11)

I hope this helps

How do kinetic and potential energy transfer to one throughout a roller coaster ride?

Answers

Answer:

As the cars ascend the next hill, some kinetic energy is transformed back into potential energy. Then, when the cars descend this hill, potential energy is again changed to kinetic energy. This conversion between potential and kinetic energy continues throughout the ride.

Explanation:

hope it helps U

a diamond sparkles more than a glass cut to similar shapes,why?​

Answers

Answer:

because when you shape glass it's not a shiny as diamond

Why do you think we need evidence to be able to continually test theories about the composition and origin of our solar system and galaxies?

Answers

We need evidence to be able to continually test theories about the composition and origin of our solar system and galaxies in order to verify the theories and to make laws.

What is theory?

A theory is an explanation for observations of the natural world that has been carried by using the scientific method, and which brings together many facts.

So we can conclude that we need evidence to test theories about the composition and origin of our solar system in order to verify the theories and to make laws.

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A proton moves north with a speed of 3 x 10^6 m/s. A 5 Tesla magnetic field is directed west. Determine the magnitude and direction of the magnetic field on the proton.

Answers

The magnitude of magnitude force on the proton is 2.4 x 10⁻¹² N.

Magnitude of magnetic force on the proton

The magnitude of magnitude force on the proton is calculated as follows;

F = qvB sinθ

where;

q is the charge of the protonv is the speed of the protonB is the magnitude of the magnetic filed θ is the angle between the field and speed

Substitute the given parameters and solve for the magnetic force.

F = (1.6 x 10⁻¹⁹) x (3 x 10⁶) x (5) X(sin90)

F = 2.4 x 10⁻¹² N

Thus, the magnitude of magnitude force on the proton is 2.4 x 10⁻¹² N.

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A student is standing at a distance of 45 m from a wall. He gives a loud clap at the echo is heard 0.3a later. Calculate the speed of sound.

Answers

Answer:

300 m/s

Explanation:

2d = vt

v = 2d/t

v = 2×90/.3

v=300 m/s

d = distance

t = time

v = velocity/speed of sound

The SI unit for weight is ________.
A. mass
B. kilogram
C. newton
D. acceleration of gravity
E. weight

Answers

Answer:

Newton

Explanation:

The SI unit for weight is newton.

The lifting force, F, exerted on an airplane wing varies jointly as the area, A, of the wing's surface and the square of the plane's velocity, v. The lift of a wing with an area
of 280 square feet is 27,800 pounds when the plane is going 220 miles per hour. Find the lifting force on the wing if the plane slows down to 130 miles per hour.
(Leave the variation constant in fraction form or round to at least 5 decimal places. Round off your final answer to the nearest pound.)

Answers

Answer:

If the lifting force varies jointly with wing surface area and the square of the plane's velocity, then:

F = kAv²

If the lift is 27,800 pounds when the area is 190 sq. ft and the velocity is 230 mph, then the constant of proportionality is:

27,800 = k(190)(230)²

k = (27,800) / (190)(230)² = 139/50255

The lifting force the plane slows down to 200 mph is:

F = (139/50255)(190)(200)²

F = 21,020.79

Rounded to the nearest pound, the lifting force is 21,021 pounds.

a right angled triangle has 5cm opposite side, 9cm adjacent side. find length of hypotunise

Answers

Perpendicular=P=5cmBase=B=9cm

Hypotenuse be H

Apply Pythagorean theorem

[tex]\\ \sf\Rrightarrow H^2=P^2+B^2[/tex]

[tex]\\ \sf\Rrightarrow H^2=5^2+9^2[/tex]

[tex]\\ \sf\Rrightarrow H^2=25+81[/tex]

[tex]\\ \sf\Rrightarrow H^2=106[/tex]

[tex]\\ \sf\Rrightarrow H=\sqrt{106}cm[/tex]

Why do we tend to attribute others' negative behaviors to dispositions over situational factors?

Answers

Answer:

Because you have a greater understanding of the mentality and conduct of individuals you care about, you are better able to understand their perspective and are more likely to recognize possible situational factors for their behavior.

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Consider the Boeing 787 "Dreamliner", with a mass of 177000 kg. In this particular model, the distance from the front wheels to the rear set of wheels is 21.7 m.
(a) If the center of mass of the airplane is along a line through the center and 3.00 m in front of the rear wheels, how much force, in meganewtons, does the ground exert on each set of rear wheels when the plane is at rest on the runway?
(b) How much force, in meganewtons, does the ground exert on the front set of wheels?

Answers

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

Center mass of the airplane

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

Some assumptionsThe wheels under the wind do not pass through the center line.The position of the front wheel is constant and it is zero mark (origin).The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

Mass of the plane at the position of the rear wheels

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

[tex]X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}[/tex]

[tex]18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg[/tex]

Force exerted by the ground on each rear wheel

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

[tex]W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN[/tex]

Mass of the plane at the position of the front wheel

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

Force exerted by the ground on the front wheel

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

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A child pushes a toy truck up an inclined plane. The lifting force is...
Select one:
a. the weight of the truck.
b. the length of the inclined plane.
c. the force the child uses to push the truck.
d. equal to the total amount of work done.

Answers

The lifting force when a child pushes a toy truck up an inclined plane is  equal to the total amount of work done.

Work-energy theorem

The work-energy theroem states, the work done on an object is equal to the change in mechanical energy of the object.

The net force applied by the child on the truck, will create a lifting force on the object.

Fd - Ffd = k.E

Where;

Fd is the work done by the childFfd is the work done by frictionK.E is the kinetic energy of the truck

Thus, the lifting force when a child pushes a toy truck up an inclined plane is  equal to the total amount of work done.

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When a penny is dropped, it takes 16 seconds. What is its height

Answers

My guess is 8
16 / 2 = 8

any help here please ???​

Answers

Answer:

16 reflection,incidence

what are the milestones of modern phyiscs?

Answers

Answer:

The articles appearing under "Milestones in Physics" will give an insight into special events or situations that have been decisive for the evolution of Physics

Answer:

We have invented many new ways to get energy, especially renewable means of energy, and we have made many new carbon-neutral devices more widespread and cheaper.

Modern physics milestones are getting harder and harder to get because almost every element has been discovered, many gaseous and liquid substances have been experimented with and dozens of other mid 20th century discoveries.



#1 DESCRIBE how magnets work

Answers

Answer:All magnets have north and south poles. Opposite poles are attracted to each other, while the same poles repel each other. When you rub a piece of iron along a magnet, the north-seeking poles of the atoms in the iron line up in the same direction. The force generated by the aligned atoms creates a magnetic field.

Explanation: I did this exact question before

A popcorn-maker transfers 250J of energy into other energy stores every second. What is its power?

Answers

Work done=250JTime taken=1s

Now

[tex]\\ \sf\rightarrowtail Power=\dfrac{Work\:done}{Time\:taken}[/tex]

[tex]\\ \sf\rightarrowtail Power=\dfrac{250}{1}[/tex]

[tex]\\ \sf\rightarrowtail Power=250W[/tex]

When resting, a person generates about 412005 joules of heat from the body. The person is submerged neck-deep into a tub containing 2124 kg of water at 20.9 °C. If the heat from the person goes only into the water, find the water temperature.

Answers

If a person generates about 412005 joules of heat from the body,  the water temperature is mathematically given as

t=21.6296C

What is the water temperature.?

Question Parameter(s):

The person is submerged neck-deep into a tub containing 2124 kg of water at 20.9 °C

Generally, the equation for the Heat   is mathematically given as

Heat gained =Heat loess

Thereofore

mw*cw*(t-2160)=1.5*10^5

[tex]t=21.60+\frac{1.5*10^5}{mw*Cw}\\\\t=21.60+\frac{1.5*10^5}{1.2*10^3*4186}[/tex]

t=21.6296C

In conclusion, the tempreature

t=21.6296C

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Describe gravitational force in your own words. Which two factors affect the gravitational force between two objects?

Answers

Answer:

Gravitational force is the force that attracts objects towards each other. Two factors that affect the gravitational force between objects are the mass of the two objects and the distance between

Explanation:

Gravity is what pulls us towards the Earth if we were to jump into the air, so it is the force that pulls things towards other things. The bigger the objects are the more gravity they have, so a planet has more gravity than say, an apple. Distance between objects also makes their gravity change, so the Earth's pull on the moon is different than the Earth's pull on the sun.

Hopefully this helps- let me know if you have any questions!

Question :-

Describe gravitational force in your own words. Which two factors affect the gravitational force between two objects?

Answer :-

Gravitational Force :-

It is the attractive force between the two body having some mass.Example of gravitational force :- attraction between the human and earth . earth pulls the body towards it with the help of this This force helps the celestial bodies to revolve in there orbits

Factors effecting Gravitational force:-

Mass :- gravitational force directly proportional on the masses of the body between which it is acting [tex]gravitational \: force ∝m{ \tiny1}m{ \tiny2}[/tex]Distance between the masses :- Gravitational force is inversely proportional to the square of the distance between the masses [tex]gravitational \: force∝ \frac{1}{ {r}^{2} } [/tex]

Formula of the Gravitational force :-

[tex]F = \frac{G m{ \tiny1}m{ \tiny2} }{ {r}^{2} } [/tex]

where

F is gravitational force G is gravitational constant ( universal constant ) m 1 is mass of first body m2 is mass of second body r is the distance between the two bodies

How can an athlete participating in a 40m sprint modify and improve their performance based on the kinematic variable of speed and acceleration?

Answers

The athlete can improve performance by building strength, coordination and balance.

Who is an Athlete?

This is an individual who is proficient in sports and other forms of physical exercise.

Improvement of performance based on the kinematic variable of speed and acceleration can be achieved by building strength, coordination and balance by performing plyometric exercises etc.

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A thin 1.5 mm coating of glycerine has been placed between two microscope slides of width 0.8 cm and length 3.9 cm . Find the force required to pull one of the microscope slides at a constant speed of 0.28 m/s relative to the other.

Answers

The  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

Force required to pull one end at a constant speed

The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is determined by applying Newton's second law of motion as shown below;

F = ma

where;

m is massa is acceleration

At a constant speed, the acceleration of the object will be zero.

F = m x 0

F = 0

Thus, the  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

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A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in the xy plane.
What is the electric potential V(z) on the z axis as a function of z , for z>0 ?
What is the magnitude E of the electric field on the z axis, as a function of z , for z>0 ?

Answers

The electric potential V(z) on the z-axis is :  V = [tex](\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z[/tex]

The magnitude of the electric field on the z axis is : E = kб 2[tex]\pi[/tex]( 1 - [z / √(z² + a² ) ] )

Given data :

V(z) =2kQ / a²(v(a² + z²) ) -z  

Determine the electric potential V(z) on the z axis and magnitude of the electric field

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq =  б( 2πR ) dr

dV  [tex]\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }[/tex]  ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = [tex]\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,[/tex]

 V = [tex]\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ][/tex]

     = [tex]\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ][/tex]

Therefore the electric potential V(z) = [tex](\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z[/tex]

Also

The magnitude of the electric field on the z axis is : E = kб 2[tex]\pi[/tex]( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

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Which of the following can only be a situation of increasing temperature?

1. The phase change of the material is going from water to solid
2. The height of the object is increasing
3. The objects speed in increasing at a constant rate
4. The average kinetic energy of the particles of a material is increasing

Answers

Increasing the temperature causes an increase in the average kinetic energy of the particles of a material.

What is average kinetic energy of particles?

The average kinetic energy of particles is the energy possessed by particles due to their constant motion.

The constant motion of particles occurs due to the energy acquired by the particles, when the temperature of the particles increases, the average kinetic energy increases which in turn increases the speed of the particles.

Thus, we can conclude that, increasing the temperature causes an increase in the average kinetic energy of the particles of a material.

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The hot exhaust from a rocket travels in one direction, and the rocket travels in the opposite direction. This is an example of?
A. Equal and opposite forces.
B. Friction.
C. Balanced forces.
D. Inertia.

Answers

Answer:

Explanation:

The hot exhaust from a rocket travels in one direction, and the rocket travels in the opposite direction. This is an example of?

A. Equal and opposite forces.

B. Friction.

C. Balanced forces.

D. Inertia.

2. A tennis ball machine launches balls horizontally with an initial speed of 5.3 m/s, from a height of 1.2 m.
a) What will the time of flight be for a tennis ball launched by the ball machine? (3)
b) What will the range of the tennis ball be? (2)
c) What will be the final velocity of the ball with which it reaches the ground? (3)

Answers

(a) The time of flight be for a tennis ball launched by the ball machine is 0.19 s.

(b) The range of the tennis ball be is 1.01 m.

(c)  The final velocity of the ball with which it reaches the ground is 7.16 m/s.

Time of flight of tennis ball

The time of flight of the tennis ball is calculated as follows;

h = vt + ¹/₂gt²

1.2 = 5.3t + 0.5(9.8)t²

1.2 = 5.3t + 4.9t²

4.9t² + 5.3t - 1.2 = 0

a = 4.9, b = 5.3, c = 1.2

solve using quadratic formula

t = 0.19 s

Thus, the time of flight be for a tennis ball launched by the ball machine is 0.19 s.

Range of the tennis ball

The range of the tennis ball is calculated as follows;

R = vt

R = 5.3 x 0.19

R = 1.01 m

Final velocity of the ball

The final velocity of the ball with which it reaches the ground is calculated as follows;

vf = vo + gt

vf = 5.3 + 9.8(0.19)

vf = 7.16 m/s

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Explain why is dressing table mirror may become dirsty if wiped with a cloth on a warm day.​

Answers

Answer:

When you rub a dry cloth across glass, it creates charged static electricity, which attracts little non-charged dust particles.

A bucket of mass m is attached to a rope that is wound around the outside of a solid sphere (I = 2/5 M^2) of radius R. When the bucket is allowed to fall from rest, it falls with an acceleration of a down. What is the mass of the sphere in terms of m, R, a, and g?

Answers

Answer:

[tex]\displaystyle \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{M^{2}\, a}}[/tex], assuming that the tension in the rope is the only tangential force on the sphere ([tex]g[/tex] denote the gravitational acceleration.)

Explanation:

The forces on the bucket are:

Weight of the bucket: [tex]m\, g[/tex] (downward.)Tension in the rope (upward.)

Since the weight of the bucket and the tension from the rope are in opposite directions, the magnitude of the net force would be:

[tex]\begin{aligned} \|\text{Net Force}\| =\; & \|\text{Weight}\| - \|\text{Tension}\| \end{aligned}[/tex].

The upward tension in the rope prevents the bucket from accelerating at [tex]g[/tex] (free fall.) Rather, the bucket is accelerating at an acceleration of only [tex]a[/tex]. The net force on the bucket would be thus [tex]m\, a[/tex].

Rearrange the equation for the net force on the bucket to find the magnitude of the tension in the rope would be:

[tex]\begin{aligned} & \|\text{Tension}\| \\ =\; & \|\text{Weight}\| - \|\text{Net Force}\| \\ =\; & m\, g - m\, a \\ =\; & (g - a)\, m\end{aligned}[/tex].

At a distance of [tex]R[/tex] from the center of the sphere, the tension in the rope [tex](g - a)\, m[/tex] would exert a torque of [tex](g - a)\, m\, R[/tex] on the sphere. If this tension is the only tangential force on this sphere, the net torque on the sphere would be [tex](g - a)\, m\, R\![/tex].

Let [tex]M[/tex] denote the mass of this sphere. The moment of inertia of this filled sphere would be [tex]I = (2/5)\, M^{2}[/tex].

Therefore, the magnitude of the angular acceleration of this sphere would be:

[tex]\begin{aligned}& \|\text{Angular Acceleration}\| \\ =\; & \frac{\|\text{Net Torque}\|}{(\text{Moment of Inertia})} \\ =\; & \frac{(g - a)\, m\, R}{(2/5)\, M^{2}} \end{aligned}[/tex].

The bucket is accelerating at a magnutide of [tex]a[/tex] downwards. The rope around the sphere need to unroll at an acceleration of the same magnitude, [tex]a\![/tex]. The tangential acceleration of the sphere at the surface would also need to be [tex]\! a[/tex].

Since the surface of the sphere is at a distance of [tex]R[/tex] from the center, the angular acceleration of this sphere would be [tex](a / R)[/tex].

Hence the equation:

[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \|\text{Angular Acceleration}\| = \frac{a}{R} \end{aligned}[/tex].

Solve this equation for [tex]M[/tex], the mass of this sphere:

[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \frac{a}{R} \end{aligned}[/tex].

[tex]\begin{aligned}M^{2} &= \frac{(g - a)\, m\, R^{2}}{(2/5)\, a} \\ &= \frac{(5/2)\, (g - a)\, m\, R^{2}}{a}\end{aligned}[/tex].

[tex]\begin{aligned}M&= \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{a}}\end{aligned}[/tex].

Other Questions
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