the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is

Answers

Answer 1

Answer:

The work done on the block by the spring as it accelerates the block is 4kx².

Explanation:

Let initial distance is x.

It was compressed three times farther and then the block is released, new distance is 3x.

The work done in compressing the spring is given by :

[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)[/tex]

[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\W=\dfrac{1}{2}k((3x)^2-x^2)\\\\W=\dfrac{1}{2}k((9x^2-x^2)\\\\W=\dfrac{1}{2}k\times 8x^2\\\\W=4kx^2[/tex]

So, the work done on the block by the spring as it accelerates the block is 4kx².


Related Questions

In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its unstretched length

Answers

Answer: hello  below is the missing part of your question

A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.

answer

x = 0.0962 m

Explanation:

First step :

Determine the length of the rough patch/spot

F = Uₓ (mg)

and  w = F.d = Uₓ (mg)  * d

hence;

d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m

next :

work done on unstretched spring length

Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m

w' = Uₓ (mg)  * d

    = 0.49 * 10 * 9.81 * 0.4847 = 23.27 J

also given that the Elastic energy of spring = work done ( w')

1/2 * kx^2 = 23.27 J

x = [tex]\sqrt{\frac{2*23.27}{5029} }[/tex]  = 0.0962 m

What word chemical equation describes this chemical reaction?

Answer : sodium + chlorine → sodium chloride

Answers

Sodium + chlorine = sodium chloride

how much amount of heat energy is required to convert 5 kg of ice at - 5° c into 100°c steam?​

Answers

Assuming no heat lost to the surrounding,

-5⁰C ice → 0⁰C ice

Specific heat capacity of ice = 2.0 x 10³ J/kg/⁰C

Q = mc∆θ

Q = 5(2.0 x 10³) x (0-(-5))

Q = 50000J

0⁰C ice → 0⁰C water

Specific latent heat of fusion of ice = 3.34 x 10⁵J/kg

Q = mLf

Q = 5(3.34 x 10⁵)

Q = 1670000J

0⁰C water → 100⁰C water

Specific heat capacity of water = 4.2 x 10³ J/kg/⁰C

Q = mc∆θ

Q = 5(4.2 x 10³) x (100-0)

Q = 2100000J

100⁰C water → 100⁰C steam

Specific latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Q = mLv

Q = 5(2.26 x 10⁶)

Q = 11300000J

Total amount of heat required

= 50000 + 1670000 + 2100000 + 11300000

= 15120000J

two point charges with charge q are initially separated by a distance d. if you double the charge on both charges, how far should the charges be separated for the potential energy between them to remain the same

Answers

Answer:

  r ’= 4 r

Explanation:

Electric potential energy is

          U = [tex]k \frac{q_1q_2}{r_{12}}[/tex]k q1q2 / r12

in this exercise

          q₁ = q₂ = q

          U = k q² / r

for when the charge change

           U ’= k q’² / r’

indicate that

      q ’= 2q

      U ’= U

we substitute

           U = k (2q) ² / r ’

           U = 4 k q² / r ’

we substitute

           [tex]k \ \frac{q^2}{r} = 4 k \ \frac{q^2}{r'}[/tex]k q² / r = 4 k q² / r ’

           r ’= 4 r

At which point is the kinetic energy of the pendulum the greatest?

C

A

D

B

Answers

Answer:

Point C

Explanation:

Greatest Kinetic Energy means lowest potential energy since energy is conserved. Lowest potential energy means lowest height which is at Point C.

A hot air balloon is a sphere of volume 2210 m3. The density of the hot air inside is 1.13 kg/m3, while the air outside has a density of 1.29 kg/m3. The balloon itself has a mass of 240 kg. What is the TOTAL NET force acting on the balloon?
[?]N

Answers

The total net force acting on the balloon will be 24498 Newtons

Given that

Volume of the balloon = 2210 cubic meter

Density of the air inside the balloon = 1.13  kg/m3

What will be the net force exerted on the balloon ?

Here force on the balloon will be equal to the weight of the air displaced by balloon

[tex]F= mass of air displaced\times gravity[/tex]

[tex]F= Density \times volume \times gravity[/tex]

[tex]F=1.13 \times 2210 \times 9.81[/tex]

[tex]F=24498 N[/tex]

The total net force acting on the balloon will be 24498 Newtons

To know more about buoyancy force follow

https://brainly.com/question/117714

A TV satellite dish is designed to receive radio waves of wavelength
0.0644 meters. What is the frequency of the waves it receives? _______GHz

Give your answer in gigahertz (GHz). 1 GHz = 10^9 Hz.

Give your answer to the nearest tenth of a GHz (one place after the decimal). Just enter the number; do NOT use scientific notation.

Answers

Answer:

4.7 GHz

Explanation:

Applying,

v = λf................. Equation 1

Where v = velocity of the radio wave, λ = wavelength, f = frequency

make f the subject of the equation

f = v/λ.............. Equation 2

Note: A radio wave is an electromagnetic wave, as such it moves with a velocity of 3.00 x 10⁸ m/s

From the question,

Given: λ = 0.0644 meters

Constant: v = 3.00 x 10⁸ m/s

Substitute these values into equation 2

f = (3.00 x 10⁸)/0.0644

f = 4.66×10⁹ Hz

f = 4.7 GHz

Explain the following observations:
a) A balloon filled with hydrogen gas floats in air;
B) A ship made of steel floats on water.

Answers

Answer and Explanation:

a. An oxygen-filled balloon is not able to float in the air, because the oxygen inside the balloon is of the same density, that is, the same "weight" as the oxygen outside the balloon and present in the atmosphere. The balloon can only float if the gas inside it is less dense than atmospheric oxygen. Helium gas is less dense than atmospheric gas, so if a balloon is filled with helium gas, that balloon will be able to float because of the difference in density.

b. The ship is able to float in the water because its steel construction is hollow and full of air. This makes the average density of this ship less than the density of water, which makes the ship lighter than water and for this reason, this ship is able to float. In addition, the ship is partially immersed, allowing the weight of the ship on the water to counteract the buoyant force that the water promotes on the ship. Weight and buoyant are two opposing forces that keep the ship afloat.

High-speed stroboscopic photographs show that the head of a -g golf club is traveling at m/s just before it strikes a -g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at m/s. Find the speed of the golf ball just after impact.

Answers

The question is incomplete. The complete question is :

High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 60 m/s just before it strikes a  50 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after impact.

Solution :

We know that momentum = mass x velocity

The momentum of the golf club before impact = 0.200 x 60

                                                                             = 12 kg m/s

The momentum of the ball before impact is zero. So the total momentum before he impact is 12 kg m/s. Therefore, due to the conservation of momentum of the two bodies after the impact is 12 kg m/s.

Now the momentum of the club after the impact is = 0.2 x 40

                                                                                    = 8 kg m/s

Therefore the momentum of the ball is = 12 - 8

                                                                = 4 kg m/s

We know momentum of the ball, p = mass x velocity

                                                     4 = 0.050 x velocity

∴ Velocity =  [tex]$\frac{4}{0.050}$[/tex]

                 = 80 m/s

Hence the speed of the golf ball after the impact is 80 m/s.

A 0.495-kg hockey puck, moving east with a speed of 4.50 m/s , has a head-on collision with a 0.720-kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed (magnitude of the velocity) of each object after the collision?

Answers

Answer:

a)  [tex]v_1=-0.833m/s[/tex]

b)  [tex]V_2=12.5m/s[/tex]

Explanation:

From the question we are told that:

Hockey puck Mass [tex]m_1=0.495kg[/tex]

Hockey puck Speed [tex]u_1=4.50m/s[/tex]

Puck Mass [tex]m_2=0.720kg[/tex]

Assuming

Initial speed of Puck [tex]u_2=0[/tex]

Generally the equation for Speed of First Puck is mathematically given by

 [tex]v_1=(\frac{m_1-m_2}{m_1+m_2})*v_1+(\frac{2m_2}{m_1+m_2})u_2[/tex]

 [tex]v_1=(\frac{0.495-0.720}{0.495+0.720})*4.50+0[/tex]

 [tex]v_1=-0.833m/s[/tex]

Generally the equation for Speed of Second Puck is mathematically given by

 [tex]V_2=(\frac{2m_1}{m_1+m_2})u_2-(\frac{m_1-m_2}{m_1+m_2})v[/tex]

 [tex]V_2=(\frac{2*0.495}{0.495*0.720})*4.50-0[/tex]

 [tex]V_2=12.5m/s[/tex]

recognizing forms of energy

Answers

Answer:

hi the question isn't obvious and need a photo I guess

The question is missing

Preocupada com o aumento da tarifa na conta de luz, uma pessoa resolve economizar diminuindo o tempo de banho de 20 para 15 minutos. Seu chuveiro possui as seguintes especificações: 4200 W e 220V. Sabendo que o kWh custa R$0,30, a economia feita em 10 dias foi de aproximadamente​

Answers

Answer:

The mount saved is $ 0.105.

Explanation:

Concerned about the increase in the electricity bill, a person decides to save by reducing bathing time from 20 to 15 minutes. Your shower has the following specifications: 4200 W and 220V. Knowing that the kWh costs R$0.30, the savings made in 10 days were approximately​.

The electrical energy is given by

E = P x t

where, P is the electrical power and t is the time.

When he is using the shower for 20 minutes, the energy consumed is

E = 4200 x 20 x 60 = 5040,000 J = 1.4 kWh

When he is using the shower for 15 minutes, the energy consumed is

E' = 4200 x 15 x 60 = 3780000 J = 1.05 kWh

The difference in energy is

E'' = E - E' = 1.4 - 1.05 = 0.35 kWh

The money saved is

= 4 0.3 x 0.35 = $ 0.105

A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.0 in. (50.8 mm) is pulled in tension. Use the load-elongation characteristics tabulated below to complete parts (a) through (f).
a. Plot the data as engineering stress versus engineering strain.
b. Compute the modulus of elasticity.
c. Determine the yield strength at a strain offset of 0.002.
d. Determine the tensile strength of this alloy.
e. What is the approximate ductility, in percent elongation?
f. Compute the strain energy density up to yielding (modulus of resilience).
( Load in N Load in lb Length in mm Length in in. 2.000 2.002 2.004 2.006 2.008 2.010 2.020 2.040 2.080 2.120 2.160 2.200 2.240 2.270 2.300 2.330 Fracture 50.800 7330 15,100 3400 23,100 5200 30,400 6850 34,400 7750 38,400 8650 41,3009300 44,800 10,100 46,200 10,400 53, 47,300 10,650 54.864 47,500 10,700 55.880 46,100 10,400 44,800 10,100 42,600 9600 3,400 8200 Fracture Fracture Fracture 50.851 50.902 50.952 51.003 51.054 1650 51.308 51.816 52.832 848 56.896 57.658 58.420 59.182

Answers

Answer:

A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.0 in. (50.8 mm) is pulled in tension. Use the load-elongation characteristics tabulated below to complete parts (a) through (f).

A uniform metre rule of mass 10g is balanced on a knife edge placed at 45cm mark. Calculate the distance of a mass 25g from the pivot​

Answers

Answer:

The distance of a mass 25g from the pivot​ is 18cm

Explanation:

Given

[tex]m_1 = 10g[/tex]

[tex]d_1 = 45cm[/tex]

[tex]m_2 = 25g[/tex]

Required

Distance of m2 from the pivot

To do this, we make use of:

[tex]m_1 * d_1 = m_2 * d_2[/tex] --- moments of the masses

So, we have:

[tex]10 * 45= 25* d_2[/tex]

[tex]450= 25* d_2[/tex]

Divide both sides by 25

[tex]18= d_2[/tex]

Hence:

[tex]d_2 = 18[/tex]

A particle of mass 1.2 mg is projected vertically upward from the ground with a velocity of 1.62 x 10 cm/h. Use the above information to answer the following four questions: 7. The kinetic energy of the particle at time t = 0 s is A. 1.215 x 10-3 J B. 2.430 J C. 1215 J D. 9.72 x 106 J E. OJ (2)​

Answers

Answer:

K = 0 J

Explanation:

Given that,

The mass of the particle, m = 1.2 mg

The speed of the particle, [tex]v=1.62\times 10\ cm/h[/tex]

We need to find the kinetic energy of the particle at time t = 0 s.

At t = 0 s, the particle is at rest, v = 0

So,

[tex]K=\dfrac{1}{2}mv^2[/tex]

If v = 0,

[tex]K=0\ J[/tex]

So, the kinetic energy of the particle at time t = 0 s is 0 J.

As a roller coaster car crosses the top of a 48.01-m-diameter loop-the-loop, its apparent weight is the same as its true weight. What is the car's speed at the top?

Answers

Answer:

The speed of the car, v = 21.69 m/s

Explanation:

The diameter is  = 48.01 m

Therefore, the radius of the loop R = 24.005 m

Weight at the top is n = mv^2/R - mg

Since the apparent weight is equal to the real weight.

So, mv^2/R - mg = mg

v = √(2Rg)

v = √[2(24.005 m)(9.8 m/s^2)]

The speed of the car, v = 21.69 m/s

Answer:

The speed is 15.34 m/s.

Explanation:

Diameter, d = 48.01 m

Radius, R = 24.005 m

Let the speed is v and the mass is m.

Here, the weight of the car is balanced by the centripetal force.

According to the question

[tex]m g = \frac{mv^2}{R}\\\\v =\sqrt{24.005\times9.8}\\\\v = 15.34 m/s[/tex]

IS ANYONE THERE..??!​

Answers

Answer:

hmmmmmmmmmmmmmmmmmmmmmmmm y

PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?

Answers

[tex]A=2.01×10^{16}\:\text{nuclei}[/tex]

Explanation:

Given:

[tex]\lambda = 4.96×10^3 s[/tex]

[tex]A_0 = 3.21x10^{17}[/tex] nuclei

t = 1.98×10^4 s

[tex]A=A_02^{-\frac{t}{\lambda}}[/tex]

[tex]A=(3.21×10^{17}\:\text{nuclei}) \left(2^{-\frac{1.98×10^4}{4.96×10^3}} \right)[/tex]

[tex]\:\:\:\:\:\:\:=2.01×10^{16}\:\text{nuclei}[/tex]

what is the dimensional formula of force and torque​

Answers

Answer:

Units. Torque has the dimension of force times distance, symbolically T−2L2M. Although those fundamental dimensions are the same as that for energy or work, official SI literature suggests using the unit newton metre (N⋅m) and never the joule. The unit newton metre is properly denoted N⋅m.

Dimension: M L2T−2

In SI base units: kg⋅m2⋅s−2

Other units: pound-force-feet, lbf⋅inch, ozf⋅in

Answer:

hope it is helpful to you

Three 30 g metal balls, one of aluminum, copper and lead, are placed in a large beaker of hot water for a few minutes. [The specific heats of aluminum, copper, and lead are 903, 385, and 130 J / (kg ° C), respectively].
to. Which of the balls, if any, will reach the highest temperature? Explain.
b. Which of the balls, if any, will have the most heat energy? Explain.

Answers

Answer:

The answer is below

Explanation:

Specific heat capacity is an intensive property of a material. The specific heat of a material is the amount of energy required to raise the temperature of one unit mass m of material by one unit of temperature.

a) Temperature is inversely proportional to specific heat capacity. If the same amount of heat is applied to all three balls, the ball that will reach the highest temperature is the ball with the least specific heat capacity.

Hence lead will have the highest temperature since it has the least specific heat capacity.

b) The quantity of heat is directly proportional to the specific heat capacity. Hence if all balls experience the same temperature change, the ball that have the most energy will be that with the highest specific heat capacity.

Hence aluminum will have the most heat since it has the highest specific heat capacity.

When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not noticeable?

Answers

Answer:

Yes, when an apple falls towards the earth, the apple gets accelerated and comes down due to the gravitational force of attraction used by the earth. The apple also exerts an equal and opposite force on the earth but the earth does not move because the mass of the apple is very small, due to which the gravitational force produces a large acceleration in it (a = F/m) but the mass of the earth is very large, the same gravitational force produces very small acceleration in the earth and we don't see the earth rising towards the apple.

Typhoon signal number 2 is raised. What is the speed of the expected typhoon?​

Answers

the simple answer is from 61kmph to 120kmph

Explanation:

no explanation is needed

Explain what a circuit breaker is and how it helps protect your house?

Answers

Explanation:

A circuit breaker is an electrical switch designed to protect an electrical circuit from damage caused by overcurrent/overload or short circuit. Its basic function is to interrupt current flow after protective relays detect a fault.

Circuit breakers have been designed to detect when there is a fault in the electricity, so it will “trip” and shut down electrical flow. ... This detection is key to preventing surges of electricity that travel to appliances or other outlets, which can cause them to break down

Every object around you is attracted to you. In fact, every object in the galaxy is attracted to every other object in the galaxy.

a. True
b. False

Answers

Answer:

True

Explanation:

With the gravitational pull that our planets have, we are able to remain in orbit. This demonstrates how every object in the galaxy is attracted to every other object. Every object in the universe that has mass exerts a gravitational pull on every other mass. We as humans do it too, but since our force isn't strong, we don't have much of an effect. I hope this helped and please don't hesitate to reach out with more questions!

Water is falling on the blades of a turbine at a rate of 100 kg/s from a certain spring. If the height of spring be 100m, then the power transferred to the turbine will be: a) 100 KW b) 10 KW c) 1 KW d) 100 W​

Answers

Answer:

Natae Si Jordan Kaya Sya Napaihe

Explanation:

haha

In 2009 Usain Bolt set the world record time by running 100 meters in 9.58 s. Assume that during this race he ran in a straight line with constant acceleration a. What would be the required constant acceleration a

Answers

Usain Bolt I think, sorry if it’s not right

In 2009 Usain Bolt set the world record time by running 100 meters in 9.58 seconds, assuming that he ran this race with constant acceleration, then the required constant acceleration would have been

What are the three equations of motion?

There are three equations of motion given by  Newton

v = u + at

S = ut + 1/2×a×t²

v² - u² = 2×a×s

By using the second equation of motion given by Newton,

S = ut + 1/2at²

100= 0 + 0.5*a*9.58²

a = 2.17 meters / second²

Thus,the required constant acceleration of Usain Bolt would have been 2.17 meters / second².

To learn more about equations of motion here, refer to the link;

brainly.com/question/5955789

#SPJ2

write down the following units in the ascending of their value A) mm nm cm um B) 1m 1cm 1km 1mm. convert the following units into SI without changing their values? A)3500g B)2.5km C)2h​

Answers

Answer:

A) nm, um, mm, cm

B) 1mm, 1cm, 1m, 1km

A) 3500g, B) 2500m, C) 7200 seconds

There are two beakers of water on the table. We can compare the average kinetic energy of the water molecules in the two beakers by measuring their

A temperatures.

B volumes.

C densities.

D masses.

Answers

Answer: masses

Explanation:

Trust me

A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg. How far will block m drop in the first seconds after the system is released?
How long will block M move during above time?
At the time, calculate the velocity of block M
Find out the deceleration of the block M, if the connected string is
removal by cutting after the first second. Then, calculate the time
taken to contact block M and pulley.

Answers

Answer:

a)  y = 0.98 t², t=1s y= 0.98 m,  

b) he two blocks must move the same distance

c) v = 1.96 m / s,  d)  a = -1.96 m / s², e)  x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

small block

             w- T = m a

             

we write the system of equations

             T - fr = M a

             mg - T = m a

we add and resolved

             mg-  μ Mg = (M + m) a

             a = [tex]g \ \frac{m - \mu M}{m+M}[/tex]

             a = [tex]9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}[/tex]

             a = 9.8 (6/30)

             a = 1.96 m / s²

a) now we can use the kinematic relations

             y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

           -fr = M a

          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

the time to travel this distance is

        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration

Answers

Answer:

a= -80.357 m/s

Explanation:

use the formula

vf^2=vi^2+2a(xf-xi)

Plug in givens

0=(7.50)^2+2a(0.350m)

solve for acceleration

a= -80.357 m/s

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