Suppose that 2 J of work is needed to stretch a spring from its natural length of 28 cm to a length of 43 cm. (a) How much work is needed to stretch the spring from 33 cm to 35 cm

Answers

Answer 1

Answer:

0.035 J

Explanation:

Applying,

W = ke²/2.............. Equation 1

Where W = workdone by the stretching the spring, k = spring constant, e = extension.

make k the subject of the equation

k = 2W/e²............... Equation 2

From the question

Given: W = 2 J, e = (43-28) = 15 cm = 0.15 m

Substitute these values into equation 2

k = (2×2)/(0.15²)

k = 177.78 N/m

Hence, work need to stretch the spring from 33 cm to 35 cm

therefore,

e = 35-33 = 2 cm = 0.02 m

Substitute into equation 1

W = 177.78(0.02²)/2

W = 0.035 J


Related Questions

The tires of a car make 60 revolutions as the car reduces its speed uniformly from 95.0 km/h to 60.0 km/h. The tires have a diameter of 0.88 m. If the car continues to decelerate at this rate, how far does it go

Answers

Answer:

-2.869 rad/s2

Explanation:

Data given:

speed, vi at 95.0 km/h = 95 X (1 hour /3600 seconds) X (1000m / 1km)

Note that, for every 1 hour, there will be 60sec X 60sec = 3600 seconds

And for every 1km, there will be 1000m.

So, speed of 95.0 km/h = 26.389 m/s

speed, vi  =  r ω (radius X angular velocity)

 angular velocity, ωi = v/r

ωi = 26.389  m/s ÷ half of 0.88 m diameter

= 59.975 rad/s

decelerating to speed, vf at 60.0 km/h = 60 X X (1 hour /3600 seconds) X (1000m / 1km)

= 16.667m/s

The angular velocity for this speed = 16.667m/s ÷ half of 0.88 m diameter

 = 37.879rad/s

 How far the car goes is equivalent to the angular acceleration which equals to (ωf^2 - ωi^2) ÷ 2θ

= (37.879rad/s)^2 - (59.975 rad/s)^2 ÷ 2 (60 rev X 2π rad/rev)

= -2.869 rad/s2

all
What is Velocity

Answers

Answer:

noun

the speed of something in a given direction.

"the velocities of the emitted particles"

(in general use) speed.

"the tank shot backwards at an incredible velocity"

Similar:

speed

pace

rate

tempo

momentum

impetus

swiftness

swift/fast pace

fastness

quickness

speediness

rapidity

briskness

expeditiousness

expedition

dispatch

acceleration

clip

fair old rate

fair lick

steam

nippiness

fleetness

celerity

ECONOMICS

the rate at which money changes hands within an economy.

noun: velocity of circulation; plural noun: velocities of circulation

Answer:

i hope this helps you

Explanation:

hii

You throw a glob of putty straight up toward the ceiling, which is 3.50 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.10 m/s.
1. What is the speed of the putty just before it strikes the ceiling? Express your answer with the appropriate units.
2. How much time from when it leaves your hand does it take the putty to reach the ceiling? Express your answer with the appropriate units.

Answers

Answer:

Explanation:

Given that:

the putty initial speed (u) = 9.10 m/s

distance (s) between hand and the ceiling = 3.50 m

the speed of the putty prior to the time it hits the ceiling can be determined by considering the second equation of motion.

v² - u² = 2as

Since the putty is moving in a vertical motion(i.e. in an upward direction)

v² - u² = -2gs

v² = u² - 2gs

[tex]v = \sqrt{u^2 - 2gs}[/tex]

[tex]v = \sqrt{(9.10)^2 -( 2* 9.8) (3.50 -0)}[/tex]

[tex]v = \sqrt{82.81 -19.6 (3.50)}[/tex]

[tex]v = \sqrt{82.81 -68.6}[/tex]

[tex]v = \sqrt{14.21}[/tex]

v = 3.77 m/s

2.

The time it takes to reach the ceiling from the moment it leaves your hand can be calculated by using the first equation of motion:

v = u + at

In an upward direction

v = u - gt

making time t the subject;

v - u = -gt

[tex]t = \dfrac{v-u}{-g}[/tex]

[tex]t = \dfrac{3.77 - 9.10}{-9.8}[/tex]

[tex]t = \dfrac{-5.33}{-9.8}[/tex]

t = 0.54s

79. If the takeoff velocity of an airplane on a runway is 300 km/hr with an acceleration of 2 m/s. What is the takeoff time
of the airplane.


Answers

Answer:

The take off time is equal to 41.66s.

Explanation:

Given that,

The initial speed of the airplane, u = 0

Acceleration, a = 2 m/s²

Let v is the take off time. Using first equation of motion,

v = u +at

Put v = 300 km/hr = 83.3 m/s

So,

[tex]t=\dfrac{v}{a}\\\\t=\dfrac{83.33}{2}\\\\t=41.66\ s[/tex]

So, the take off time is equal to 41.66s.

Does the same battery always deliver the same amount of flow to any circuit? Mention two observations of any circuits in this lab that support your answer. Explain.

Answers

Answer:

Yes

Explanation:

Given that the battery is the same the PD ( potential difference ) in the circuit will also be the same likewise the flow of charge in the circuit,

Hence the same amount of charge flow is delivered to any circuit.

attached below are examples

The launching catapult of the aircraft carrier gives the jet fighter a constant acceleration of 59 m/s2 from rest relative to the flight deck and launches the aircraft in a distance of 97 m measured along the angled takeoff ramp. If the carrier is moving at a steady 26 knots (1 knot = 1.852 km/h), determine the magnitude v of the actual velocity of the fighter when it is launched.

Answers

Answer:

Explanation:

We shall find the final velocity of  aircraft with respect to aircraft carrier using the following relation.

v² = u² + 2as

v² = 0² + 2 x 59 x 97

v² = 11446

v = 107 m /s

velocity of aircraft carrier = 1.852 x 26 = 48.152 km/h

= 48.152 x 1000 / (60 x 60) m/s

= 13.37 m /s

This velocity of aircraft carrier will be added to the velocity of aircraft .

So absolute velocity of aircraft = 107 m /s + 13.37 m/s

= 120.37 m/s

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery.

Answers

Complete question:

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.

Answer:

The value of its capacitance is 1.027 x 10⁻¹² F

Explanation:

Given;

area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²

separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m

voltage of the battery, V = 18 V

The value of its capacitance is calculated as;

[tex]C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F[/tex]

Therefore, the value of its capacitance is 1.027 x 10⁻¹² F

A circle has a radius of 20 cm and a central angle that measures 216° find the length of the arc defined by the central angle

Answers

Answer:

4320cm

Explanation:

(length of the arc)/(length of the radius)=central angle

lengthof the arc/20=216

lengthof the arc=216×20

length of the arc=4320cm

hope this helps, let me know if I'm right!

The length measured along the arc is called length of the arc. The length of the arc defined by the central angle is 4320cm.

What is central angle?

Central angle is the angle subtended by the arc at the center of circle.

central angle = length of the arc / length of the radius

216 = length of the arc/20

length of the arc = 216×20

length of the arc = 4320cm

Thus, the length of the arc defined by the central angle is 4320cm.

Learn more about central angle.

https://brainly.com/question/15698342

#SPJ2

Find the weight of a man whose mass is 40 kg on earth.


(also
write complete data plus proper formula).



Answers

I am sure it help you with that much ☺️

Explanation:

pleasae give me some thanks please good morning sister

An air-filled pipe is found to have successive harmonics at 700 Hz , 900 Hz , and 1100 Hz . It is unknown whether harmonics below 700 Hz and above 1100 Hz exist in the pipe. What is the length of the pipe

Answers

Answer:

the length of the pipe is 0.85 m or 85 cm

Explanation:

Given the data in the question;

The successive harmonics are;  700 Hz , 900 Hz , and 1100 H

Now, for a closed pipe,

length of pipe (L) = λ/4

Harmonics; 1x, 3x, 5x, 7x, 9x, 11x

1100Hz - 900Hz = 200Hz

⇒ 2x = 200Hz

x = 100Hz ( fundamental frequency )

λ = V/f = 340 /100 = 3.4 m

Now

Length L = λ / 4

L = 3.4 / 4

L = 0.85 m or 85 cm

Therefore, the length of the pipe is 0.85 m or 85 cm

15. Muous produced by snails help them move along the ground How does the mucus help maits
move?
A Mucus is cold
C. Mucus reduces friction
B. Mucus leaves a trail D. Mucus makes the snail lighter
16. Which of the following statements does NOT tell about the effects of pravitational force
on objects?
A. The more distant the body from the earth, the lesser is the gravitational force of attraction
B. Gravitational force pushes the objects upward
C. Gravitational force is greater when the objects are closer together
D. The farther you are from earth, the lesser your weight
17. Why is it difficult for us to go up in a mountain than go down?
A We are moving against gravity
B. We are pulled by the wind
C. We are moving towards gravity
D. None of the above
18. Which of these will have the strongest gravitational pull on Earth'?
A a baby
C. a drinking glass
B. a half sack of nice
D. an elephant
19. Without air resistance, all objects fall at the same rate.
D. Neither True or False
C Maybe
B. False
20. The gravity between two objects increases as the distance between them
D stabilizes
C neutralizes
B. increases
A. decreases
A True​

Answers

Answer:

15.C Mucus reduces friction

sometimes balance point may not be obtained on the potentiometer wire why​

Answers

Solution
Let a cell of emf E be connected across the entire length L of a potentiometer wire . Now , if the balance point is obtained at a length l during measurement of an unknown voltage

.
The balance point is not on the potentiometer wire - this statement means that

. In that case ,
l > L
V > E

2 Lights slows down when it enters water from air.
a What happens to its speed?
b What happens to its wavelength?
c What happens to its frequency?​

Answers

Light travels as waves, with the wavefronts perpendicular to the direction of motion. In the animation shown here, the wavefronts are represented by the green parallel lines. The red arrow represents the direction of motion. As light moves from air into water, it not only slows, but the wavelength changes. wavelength becomes shorter in the denser medium of water.


Light will travel in straight line unless it meets any object (medium) in its path. Light changes its speed when it passes from one medium to other and the direction of light changes. This bending of light is called refraction.

This will happen in both when light passes through lighter to denser medium and also while comes from denser medium to lighter medium.

The speed of the light is increases when it comes from denser to less dense medium and decreases when it passes through lighter to denser. Light travels faster in vacuum compared with any other medium.

Refractive index defines the speed of light in a medium.

Refractive index (n)= v1/v2, v1 and v2 are the speed of light in vacuum and medium, respectively.

More the refractive index, slower the speed of light in that medium.

Coming to the wavelength of light, λ=v / f and here the frequency (f) of incident will not change when light passes through different medium.

Wavelength is directly proportional to the speed (v), then the wavelength increases if speed of light increases and vice versa.

The wavelength of light decreases when it goes from air to water because the speed of light decreases.

Mark brainliest

what are the symptoms of hepatitis 'b'​

Answers

Fever might be one of the signs of acute hepatitis B.
Fatigue is a common occurrence.
Appetite loss is common.
Nausea is a feeling of nausea.
Pain in the abdomen.
Urine that is dark in color.
Bowel motions that are clay-colored.

the velocity of a ship in the unit of m/s moving initially along west is given by V(t) = 5-2t. It’s acceleration at t=1 s is given by:
1. 0 m/s^2
2. 2m/s^2 along west
3. 2m/s^2 along east
4. None
Whis one is correct?

Answers

Answer:

4. None

Explanation:

Applying,

a(t) = dV(t)/dt

Where a(t) = Acceleration of the ship at a given time.

From the question,

Given: V(t) = 5-2t

Therefore,

a(t) = dV(t)/dt = 5 m/s²

Hence it's acceleration is 5 m/s²

From the question,

The right option is 4. None

Bola bermassa 200 gram dilempar
ke bawah dari ketinggian 20 m
dengan kecepatan 2 m/s. Jika
percepatan gravitasi bumi 10
m/s2 energi kinetik pada
ketinggian 8 m adalah ......​

Answers

Answer:

0.4

Explanation:

[tex] \frac{1}{2} mv ^{2} [/tex]

kinetic energy formula , potential energy is not considered

0.5×0.2×2×2

what is the suitable way of using social media​

Answers

Answer:

not using it too much and getting addicted

Explanation:

Don’t used everyday or you will be addicted to it, just used when it’s necessary

What are 3 artificial and 2 natural sources of electromagnetic radiation?

Answers

Answer: its b bro

Explanation:

ajafa'jfbA'FJ

A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed.
(a) What is the mass of the other body?
(b) What is the speed of the two-body center of mass if the initial speed of the 2.0 kg body was 4.0 m/s?

Answers

Answer:

(a) the mass of the second body is 1.2 kg

(b)  the speed of the two-body center of mass 2.5 m/s

Explanation:

Given;

mass of the body, m₁ = 2 kg

let the mass of the second body = m₂

let the initial speed of the first body, = u₁ = 4 m/s

then, the final speed of the first body, v₁= ¹/₄u₁ = 0.25u₁

initial speed of the second body, u₂ = 0

let the final speed of the second body = v₂

Apply principle of conservation of linear momentum to determine the mass of the second body;

m₁u₁  +  m₂u₂ = m₁v₁  +  m₂v₂

2u₁  +  0(m₂)  =  2(0.25u₁)  +  m₂v₂

2u₁   =  0.5u₁  +  m₂v₂

1.5(4) = m₂v₂

6 = m₂v₂

Apply one-directional velocity

u₁ + v₁ = u₂ + v₂

u₁ + (0.25u₁) = 0 + v₂

1.25u₁ = v₂

1.25(4) = v₂

5 = v₂

Then, the mass of the second body is calculated as;

m₂v₂ = 6

5m₂ = 6

m₂ = 6/5

m₂ = 1.2 kg

(b)  the speed of the two-body center of mass after collision;

[tex]V_c_m = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} \\\\V_c_m = \frac{2(0.25\times 4) \ + \ 1.2(5)}{2\ + \ 1.2} \\\\V_c_m = 2. 5 \ m/s[/tex]

A photon with a frequency of 5.02 × 1014 hertz is absorbed by an excited hydrogen atom. This causes the electron to be ejected from the atom, forming an ion. Calculate the energy of this photon in joules. [Show all work, including the equation and substitution with units.] Determine the energy of this photon in electron-volts. What is the number of the lowest energy level (closest to the ground state) of a hydrogen atom that contains an electron that would be ejected by the absorption of this photon?

Answers

Answer:

Explanation:

An atom emits a photon (particle of light) when transitioning from a ground state to its excited state. To obey conservation of energy, the energy gained by the atom when an electron moves to a lower energy level is equal to the energy it loses in emitting the photon. (The energy of a photon is E = hf, where E is the energy, h is Planck's constant, and f is the frequency of the photon.) Conversely, when an atom absorbs a photon (as is the case in absorption spectra), the electron absorbing the photon moves to a higher energy level.

an object is 70 um long and 47.66um wide. how long and wide is the object in km?​

Answers

Answer:

length =  7*10^(-8)km

width = 4.666*10^(-8) km

Explanation:

We know that:

1 μm = 1*10^(-6) m

and

1km = 1*10^3 m

or

1m = 1*10^(-3) km

if we replace the meter in the first equation, we get:

1 μm = 1*10^(-6)*1*10^(-3) km

1 μm = 1*10^(-6 - 3)km

1 μm = 1*10^(-9)km

Now with this relationship we can transform our measures:

Length: 70 μm is 70 times 1*10^(-9)km, or:

L = 70*1*10^(-9)km = 7*10^(-8)km

And for width, we have 47.66um, this is 46.66 times 1*10^(-9)km, or:

W = 46.66*1*10^(-9)km = 4.666*10^(-8) km

quick answer please

.
.
.Five identical capacitor plates each of area A are arranged such that adjacent plates are at a distance d apart . The plates are connected to a potential difference of EMF V as shown in the figure. The capacitance :

Answers

Answer:

tq1=EoA/D

-2q1=-2EoAV/d

Option C.

Answer:

tq1=EoA/D

-2q1=-2EoAV/D

Explanation:

option C is correct

A ball is thrown vertically upward at 24.0 ms can reach a height of 28.8m ( neglecting air resistance).The speed,in m/s,when it is halfway to its highest point is (using g= 10 ms ^2)

Answers

Answer:

The answer is "[tex]16.79\ \frac{m}{s}[/tex]"

Explanation:

In this question, the halfway indicates the height that is [tex]\frac{28.8}{2}=14.4 \ m[/tex]

Using formula:

[tex]v^2=u^2+2as\\\\v^2=24^2+2(-10)(14.4)\\\\[/tex]

[tex]v^2=576-288\\\\v^2=288\\\\v=\sqrt{288}\\\\v=16.97 \ \frac{m}{s}[/tex]

A pressure sensor was used to measure the unsteady pressure in a cylinder. The sensor output was acquired for 15 seconds at a rate of 202 Hz and spectral analysis was performed using FFT. What is the maximum frequency that will be displayed on the power spectrum plot

Answers

Answer:

Maximum frequency on power spectrum plot = 101 Hz

Explanation:

Given:

Time taken for output = 15 seconds

Frequency rate = 202 Hz

Find:

Maximum frequency on power spectrum plot

Computation:

Maximum frequency = Given frequency rate / 2

Maximum frequency on power spectrum plot = Frequency rate / 2

Maximum frequency on power spectrum plot = 202 / 2

Maximum frequency on power spectrum plot = 101 Hz

Jacie made a model to show the water cycle. The model she made is shown
below.
Which process in the model represents condensation?
A. As water vapor transfers heat to ice cubes, it forms clear droplets outside the
plastic wrap.
B. As water vapor gains heat from ice cubes, it forms clear droplets outside the
plastic wrap.
C. As water vapor transfers heat to ice cubes, it forms colored droplets inside the
plastic wrap.
D. As water vapor gains heat from ice cubes, it forms colored droplets inside the
plastic wrap

Answers

Answer:

option C

Explanation:

as water vapor transfer heat, colored drops are seen inside the wrap.

You send a traveling wave along a particular string by oscillating one end. If you increase the frequency of oscillations, does the speed of the wave increase, decrease, or remain the same?

Answers

Answer:

The speed of the wave remains the same

Explanation:

Since the speed of the wave v = √(T/μ) where T is the tension in the string and μ is the linear density of the string.

We observed that the speed, v is independent of the frequency of the  wave in the string. So, increasing the frequency of the wave has no effect on the speed of the wave in the string, since the speed of the wave in the string is only dependent on the properties of the string.

So, If you increase the frequency of oscillations, the speed of the wave remains the same.

With neat circuit diagrams where possible show any two (2) ways of direct current motor
excitations,

Answers

Mainly there are two types of DC Motors. One is Separately Excited DC Motor and other is Self-excited DC Motor. The self-excited motors are further classified as Shunt wound or shunt motor, Series wound or series motor and Compound wound or compound motor. The dc motor converts electrical power into mechanical power.

Radiation exerts pressure on surfaces on which it lalls (radintion pressure). Will this pressure be greater on a shiny surface or a dark surface

Answers

Answer:

Shiny surface.

Explanation:

We know that radiation pressure is the pressure over a surface exposed to electromagnetic radiation.

Where if the radiation is absorbed by the material (like in the case of a dark surface), the pressure is the energy density flux divided by the speed of light, while if the radiation is totally reflected (idealized case, but we can suppose that this happens for a shiny surface) the pressure is twice pressure for the absorbed case.

This is a simplification for the radiation pressure but is enough to conclude that the radiation pressure is always greater on reflective surfaces, then for this case, the pressure will be greater on a shiny surface than in a dark surface,

How long does take for a freely falling object to reach 4.0 m/s

Answers

Answer :

Considering initial velocity is 0,

It takes about 0.4 seconds.

Use the equation v-u =at

4-0 = 9.8×t

t = 4/9.8 = 0.4 seconds approximately.

In the Biomedical and Physical Sciences building at MSU there are 135 steps from the ground floor to the sixth floor. Each step is 16.6 cm tall. It takes 5 minutes and 30 seconds for a person with a mass of 73.5 kg to walk all the way up. How much work did the person do?

Answers

Answer:

W = 16.4 kJ

Explanation:

Given that,

There are 135 steps from the ground floor to the sixth floor.

Each step is 16.6 cm tall.

The mass of a person, m = 73.5 kg

We need to find the work done by the person. We know that,

Work done = Fd

Where

d is the displacement, d = 135 × 0.166 = 22.41

So,

W = 73.5 × 10 × 22.41

= 16471.35  J

or

W = 16.4 kJ

So, 16.4 kJ is the work done by the person.

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