If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be required to stretch it an additional 4.0 cm

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Answer 1
16 additional J of work

Related Questions

Test your prediction through calculation for the situations of the clay bob and the bouncy ball. Assume each has a mass of 100 grams, and each has an initial velocity of 20 m/s straight at the door. Ignore the effects of gravity. Calculate the change in momentum of

Answers

Answer:

a) Δp = -2.0 kgm / s,  b)   Δp = -4 kg m / s

Explanation:

In this exercise the change in moment of a ball is asked in two different cases

a) clay ball, in this case the ball sticks to the door and we have an inelastic collision where the final velocity of the ball is zero

         Δp = p_f - p₀

         Δp = 0 - m v₀

         Δp = - 0.100 20

         Δp = -2.0 kgm / s

b) in this case we have a bouncing ball, this is an elastic collision, as the gate is fixed it can be considered an object of infinite mass, therefore the final speed of the ball has the same modulus of the initial velocity, but address would count

         v_f = - v₀

        Δp = p_f -p₀

        Δp = m v_f - m v₀

        Δp = m (v_f -v₀)

        Δp = 0.100 (-20 - 20)

        Δp = -4 kg m / s

Two electrons are passing 20.0 mm apart. What is the electric repulsive force that they exert on each other

Answers

15.0 I’m pretty sure that’s the answer to your question

Answer:

0.5766422350752*10^-24 N

Explanation:

Couloumb's law states that states that there is an electrical force acting on 2 static charges. The magnitude is directly proportional to the product of the 2 charges.

Strength of electrons : q1 = q2 = 1.602 x 10-19. C

Substitute and solve:

F = (9*10^9)(1.602 x 10-19)(1.602 x 10-19) / (0.02)^2

Done.

Question 9 of 10
According to the law of conservation of momentum, the total initial
momentum equals the total final momentum in a(n)
A. Interacting system
B. System interacting with one other system
C. Isolated system
D. System of balanced forces

Answers

Answer:

The answer is C. Isolated System

Answer:

C. Isolated system

Explanation :

∵According to law of  conservation of momentum ,In an isolated system ,the total momentum remains conserved.

why is unit of power is called derived unit?​

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Distance travelled by a body in unit time is called speed. it is a scalar quantity because it can be specified only by magnitude.

Explore the Prisms screen to see how your understanding of refraction applies when light travels through a medium like glass. Give specific examples and images from the simulation to explain how your understanding applies

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Explanation:

https://tse2.mm.bing.net/th?id=OGC.b52c959ac810db1177599a161631c917&pid=Api&rurl=https%3a%2f%2fupload.wikimedia.org%2fwikipedia%2fcommons%2fthumb%2ff%2ff5%2fLight_dispersion_conceptual_waves.gif%2f266px-Light_dispersion_conceptual_waves.gif&ehk=TdcWPzr5xGP8xUOSOqZXauGOS1jHDMu7WnxPzkl7esw%3d

State whether plastic is biodegradable or non-biodegradable ? Give reasons for your answer.​

Answers

Answer:

non biodegradable

Explanation:

It is non biodegradable because plastic cannot dispose off easily ..

A 1,760 W toaster, a 1,420 W electric frying pan, and an 85 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (The three devices are in parallel when plugged into the same socket.) (a) What current (in A) is drawn by each device

Answers

Answer:

Toaster = I = 14.67 A

Frying Pan = 11.83 A

Lamp = 0.71 A

Explanation:

The electric power is given as:

[tex]P = VI\\\\I = \frac{P}{V}[/tex]

where,

I = current

P = Power

V = Voltage = 120 V

FOR TOASTER:

P = 1760 W

Therefore,

[tex]I = \frac{1760\ W}{120\ V}[/tex]

I = 14.67 A

FOR FRYING PAN:

P = 1420 W

Therefore,

[tex]I = \frac{1420\ W}{120\ V}[/tex]

I = 11.83 A

FOR LAMP:

P = 85 W

Therefore,

[tex]I = \frac{85\ W}{120\ V}[/tex]

I = 0.71 A

a baseball is thrown vertically upward with an initial velocity of 20m/s.
A,what maximum height will it attain? B,what time will elapse before it strike the ground?
C,what is the velocity just before it strike the ground?​

Answers

Answer:

Look at explanation

Explanation:

a)Only force acting on the object is gravity, so a=-g (consider up to be positive)

use: v^2=v0^2+2a(y-y0)

plug in givens, at max height v=0

0=400-19.6(H)

Solve for H

H= 20.41m

b) Use: y=y0+v0t+1/2at^2

Plug in givens

0=0+20t-4.9t^2

solve for t

t=4.08 seconds

c) v=v0+at

v=20-39.984= -19.984m/s

If Katie swims from one end of the pool, to the other side, and then swims back to her original spot, her average velocity is half of her average speed when she swam to the other side.a) trueb) false

Answers

Answer:

false.

Explanation:

Ok, we define average velocity as the sum of the initial and final velocity divided by two.

Remember that the velocity is a vector, so it has a direction.

Then when she goes from the 1st end to the other, the velocity is positive

When she goes back, the velocity is negative

if both cases the magnitude of the velocity, the speed, is the same, then the average velocity is:

AV = (V + (-V))/2  = 0

While the average speed is the quotient between the total distance traveled (twice the length of the pool) and the time it took to travel it.

So we already can see that the average velocity will not be equal to half of the average speed.

The statement is false

A random sample of 22 lunch orders at Noodles & Company showed a mean bill of $10.26
with a standard deviation of $5.21. Find the 99 percent confidence interval for the mean bill of
all lunch orders.

Answers

Answer:

(7.115 ; 13.405)

Explanation:

Given :

Sample size, n = 22

Mean bill, μ = 10.26

Standard deviation, s = 5.21

To obtain the 99% confidence interval for the mean bill of all orders ;

Mean ± margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 99%, df = n-1, 22 - 1 = 21

Tcritical = 2.831

Margin of Error = 2.831 * (5.21/√22) = 3.145

Confidence interval = 10.26 ± 3.145

Lower boundary = 10.26 - 3.145 = 7.115

Upper boundary = 10.26 + 3.145 = 13.405

Confidence interval :

(7.115 ; 13.405)

A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q

Answers

Answer:

0.3 N

Explanation:

Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.

The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N

Coulomb's law equation

F = Kq₁q₂ / r²

Where

F is the force of attraction K is the electrical constant q₁ and q₂ are two point charges r is the distance apart

Data obtained from the question Initial distance apart (r₁) =  rInitial force (F₁) = 1.2 NFinal distance apart (r₂) = 2rFinal force (F₂) =?

How to determine the final force

From Coulomb's law,

F = Kq₁q₂ / r²

Cross multiply

Fr² = Kq₁q₂

Kq₁q₂ = constant

F₁r₁² = F₂r₂²

With the above formula, we can obtain the final force as follow:

F₁r₁² = F₂r₂²

1.2 × r² = F₂ × (2r)²

1.2r² = F₂ × 4r²

Divide both side by 4r²

F₂ = 1.2r² / 4r²

F₂ = 0.3 N

Learn more about Coulomb's law:

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A flag pole 18m high casts a shadow 9.6m long . Find the distance of top of pole from the far of end of Shadow. ​

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Answer:

[tex]{ \bf{pythogras \: theorem :}} \\ \\ { \tt{ = \sqrt{ {9.6}^{2} + {18}^{2} } }} \\ = 20.4 \: cm[/tex]

A baseball of mass 0.145 kg is thrown at a speed of 40.0 m/s. The batter strikes the ball with a force of 15,000 N; the bat and ball are in contact for 0.500 ms. The force is exactly opposite to the original direction of the ball. Determine the final speed of the ball.

Answers

The final speed of the ball is 91.72 m/s.

Given the following data:

Mass of baseball = 0.145 kgInitial speed = 40.0 m/sForce = 15,000 NewtonTime = 0.500 milliseconds (ms) to seconds = 0.0005 seconds.

To find the final speed of the ball, we would use the following formula:

[tex]F = \frac{M(V - U)}{t}[/tex]

Where:

F is the force applied. u is the initial speed. v is the final speed. t is the time measured in seconds.

Substituting the parameters into the formula, we have;

[tex]15000 = \frac{0.145(V \;- \;40)}{0.0005}\\\\15000(0.0005) = 0.145(V \;- \;40)\\\\7.5 = 0.145V - 5.8\\\\0.145V = 7.5 + 5.8\\\\0.145V = 13.3\\\\V = \frac{13.3}{0.145}[/tex]

Final speed, V = 91.72 m/s

Therefore, the final speed of the ball is 91.72 m/s.

Read more here: https://brainly.com/question/24029674

A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 243 km and a direction 30.0o north of east. The displacement vector for the second segment has a magnitude of 178 km and a direction due west. The resultant displacement vector is R = A + B and makes an angle ? with the direction due east. Using the component method, find (a) the magnitude of R and (b) the directional angle ?.
(a) R = km
(b) ? = degrees

Answers

Answer:

a) [tex]R=126Km[/tex]

b) [tex]\theta=74.6\textdegree[/tex]

Explanation:

From the question we are told that:

1st segment

243km at Angle=30

2nd segment

178km West

Resolving to the X axis

[tex]F_x=243cos30+178[/tex]

[tex]F_x=33.44Km[/tex]

Resolving to the Y axis

[tex]F_y=243sin30+178sin0[/tex]

[tex]R=\sqrt{F_y^2+F_x^2}[/tex]

[tex]F_y=121.5Km[/tex]

Therefore

Generally the equation for Directional Angle is mathematically given by

[tex]\theta=tan^{-1}\frac{F_y}{F_x}[/tex]

[tex]\theta=tan^{-1}\frac{121.5}{33.44}[/tex]

[tex]\theta=74.6\textdegree[/tex]

Generally the equation for Magnitude is mathematically given by

[tex]R=\sqrt{F_y^2+F_x^2}[/tex]

[tex]R=\sqrt{33.44^2+121.5^2}[/tex]

[tex]R=126Km[/tex]

What is the effect on range and maximum height of a projectile as the launch height, launch speed, and launch angle are increased?

Answers

Answer:

The highest point in the trajectory occurs at the midpoint of the path. This highest point increases as the angle increases. At a 75° launch angle, the maximum height is approximately 76 meters. However, a further increase in launch angle beyond this 75° angle will increase the peak height even more.

Differences between angle of twist and angle of shear

Answers

Answer:

idek

Explanation:

a vehicle start moving at 15m/s. How long will it take to stop at a distance of 15m?​

Answers

Speed= distance/time

Or time = distance/speed

According to your question

Speed=15m/s

and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s

D= 1.2km = 1.2×1000m =1200meter

Time = distance/ speed

1200/15 =80second

Or. 1min and 20 sec will be your answer.

When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C where C is a constant. Suppose that at a certain instant the volume is 420 cubic centimeters and the pressure is 99 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?

Answers

Answer:

[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]

Explanation:

We are given that

[tex]PV^{1.4}=C[/tex]

Where C=Constant

[tex]\frac{dP}{dt}=-7KPa/minute[/tex]

V=420 cubic cm and P=99KPa

We have to find the rate at which the  volume increasing at this instant.

Differentiate w.r.t t

[tex]V^{1.4}\frac{dP}{dt}+1.4V^{0.4}P\frac{dV}{dt}=0[/tex]

Substitute the values

[tex](420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0[/tex]

[tex]1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)[/tex]

[tex]\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}[/tex]

[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]

Answer:

[tex]\dot V=2786.52~cm^3/min[/tex]

Explanation:

Given:

initial pressure during adiabatic expansion of air, [tex]P_1=99~kPa[/tex]

initial volume during the process, [tex]V_1=420~cm^3[/tex]

The adiabatic process is governed by the relation [tex]PV^{1.4}=C[/tex] ; where C is a constant.

Rate of decrease in pressure, [tex]\dot P=7~kPa/min[/tex]

Then the rate of change in volume, [tex]\dot V[/tex] can be determined as:

[tex]P_1.V_1^{1.4}=\dot P.\dot V^{1.4}[/tex]

[tex]99\times 420^{1.4}=7\times V^{1.4}[/tex]

[tex]\dot V=2786.52~cm^3/min[/tex]

[tex]\because P\propto\frac{1}{V}[/tex]

[tex]\therefore[/tex] The rate of change in volume will be increasing.

Puck B has twice the mass of puck A. Starting from rest, both pucks are pulled the same distance across frictionless ice by strings with the same tension.a. Compare the final kinetic energies of pucks A and B. b. Compare the final speeds of pucks A and B.

Answers

Answer:

(a) 1 : 2

(b) same

Explanation:

Let the mass of puck A is m and the mass of puck B is 2 m.

initial speed for both the pucks is same as u and the distance is same for both is s.

let the tension is T for same.

The kinetic energy is given by

[tex]K = 0.5 mv^2[/tex]

(a) As the speed is same, so the kinetic energy depends on the mass.

So, kinetic energy of A : Kinetic energy of B = m : 2m  = 1 : 2

(b) A the distance s same so the final velocities are also same.

(a)  The kinetic energy of puck B is 2 times the kinetic energy of puck A.

(b)  The final speed of both the puck A and B are same.

Let the mass of puck A is m and the mass of puck B is 2 m.

Initial speed for both the pucks is same as u and the distance is same for both is s.

Let the tension is T for same.

Then, the kinetic energy is given as,

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

(a)

As the speed is same, so the kinetic energy depends on the mass.

Then,

[tex]\dfrac{KE_{A}}{KE_{B}} = \dfrac{1/2 \times mv^{2}}{1/2 \times (2m)v^{2}}\\\\\\\dfrac{KE_{A}}{KE_{B}} =\dfrac{1}{2}[/tex]

So, kinetic energy of A : Kinetic energy of B = 1 : 2.

Thus, we can conclude that the kinetic energy of puck B is 2 times the kinetic energy of puck A.

(b)

The final speed for the puck is given as,

v = s/t

here, s is the distance covered.

Since, both pucks are pulled the same distance across frictionless ice. Then, the final speed of each puck is also same.

Thus, we can conclude that the final speed of both the puck A and B are same.

learn more about the kinetic energy here:

https://brainly.com/question/12669551

Một mặt phẳng vô hạn tích điện đều, mật độ σ = 4.10-9 C/cm2 , đặt thẳng đứng trong không khí. Một quả cầu nhỏ có khối lượng 8 g, mang điện tích q = 10-8 C treo gần vào mặt phẳng, sao cho dây treo lúc đầu song song với mặt phẳng. Lấy g = 9,8m/s2 . Khi cân bằng, dây treo quả cầu hợp với mặt phẳng 1 góc bằng bao nhiêu?

Answers

Answer:

The angle is 16 degree.  

Explanation:

A uniformly charged infinite plane, density σ = 4.10-9 C/cm2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2 . In equilibrium, by what angle does the string hanging from the ball make an angle with the plane?

Surface charge density, σ = 4 x 10^-5 C/m^2

charge, q = 10^-8 C

mass, m = 0.008 kg

The electric field due to the plate is

[tex]E= \frac{\sigma }{2\varepsilon 0}[/tex]

Let the angle make with the vertical is A and T is the tension in the string.

[tex]T sin A = q E....(1)\\\\T cos A = m g .... (2)\\\\Divie (1) by (2)\\\\tan A =\frac{q E}{m g}\\\\tan A = \frac{10^{-8}\times 4\times 10^{-5}}{2\times 8.85\times 10^{-12}\times 0.008\times9.8}\\\\tan A = 0.288\\\\A = 16 degree\\[/tex]

Una pelota se lanza verticalmente hacia arriba desde la azotea de un edificio con una velocidad inicial de 35 m/s. Si se detiene en el aire a 200 m del suelo, ¿Cuál es la altura del edificio?

a. 138,8 m

b. 51.2 m

c. 71,2 m

d. 45,0 m

Answers

the answer is B because i did this in school

I need help with this problem can anybody help me please , it’s physics 2 course

Answers

Answer:

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Explanation:

The 2 capacitors in the middle are connected in parallel so simply add their capacitance together:

[tex]5.0\:\mu\text{F} + 8.0\:\mu\text{F} = 13.0\:\mu \text{F}[/tex]

Now we have 3 capacitors connected in series so their equivalent capacitance [tex]C_{eq}[/tex] is

[tex]\dfrac{1}{C_{eq}} = \dfrac{1}{10.0\:\mu \text{F}} + \dfrac{1}{13.0\:\mu \text{F}} + \dfrac{1}{9.0\:\ mu \text{F}} [/tex]

or

[tex]C_{eq} = 3.5\:\mu \text{F}[/tex]

any one tell me about the earth rotation it rotatining or not with any proof? ​

Answers

The proof that the earth is rotating is the happens of night and day also the seasons, eg. winter, summer, autumn.

What does Boyle's Law state about the relationship between the pressure and volume of an ideal gas at constant temperature?
a) The product of pressure and volume increases as pressure decreases.
b) The sum of pressure and volume is constant.
c) The sum of product and volume decreases as volume increases.
d) The product of pressure and volume is constant.

Answers

Answer:

Option (d).

Explanation:

According to the Boyle's law, for a given mass of a gas, the pressure of the gas is inversely proportional to the volume of the gas keeping the temperature of the gas is constant.

So,

Let the pressure is P, volume is V and T is the absolute temperature of the gas.

Pressure proportional to the reciprocal of the volume.

[tex]P \alpha \frac{1}{V}\\\\P V = constant[/tex]

The correct option is (d).

You are helping your friend move a new refrigerator into his kitchen. You apply a horizontal force of 275 N in the positive x direction to try and move the 61 kg refrigerator. The coefficient of static friction is 0.58. (a) How much static frictional force does the floor exert on the refrigerator

Answers

Answer:

f = 347.08 N

Explanation:

The frictional force exerted by the floor on the refrigerator is given as follows:

[tex]f = \mu R = \mu W[/tex]

where,

f = frictional force = ?

μ = coefficient of static friction = 0.58

W = Weight of refrigerator = mg

m = mass of refrigerator = 61 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]f = \mu mg\\f = (0.58)(61\ kg)(9.81\ m/s^2)\\[/tex]

f = 347.08 N

Two charged particles attract each other with a force of magnitude F acting on each. If the charge of one is doubled and the distance separating the particles is also doubled, the force acting on each of the two particles has magnitude
(a) F/2,
(b) F/4,
(c) F,
(d) 2F,
(e) 4F,
(f) None of the above.

Answers

Answer:

F/2

Explanation:

In the first case, the two charges are Q1 and Q2 and the distance between them is r. K is the Coulomb's constant

Hence;

F= KQ1Q2/r^2 ------(1)

Where the charge on Q1 is doubled and the distance separating the charges is also doubled;

F= K2Q1 Q2/(2r)^2

F2= 2KQ1Q2/4r^2 ----(2)

F2= F/2

Comparing (1) and (2)

The magnitude of force acting on each of the two particles is;

F= F/2

A planet of mass m moves around the Sun of mass M in an elliptical orbit. The maximum and minimum distance of the planet from the Sun are r1 and r2, respectively. Find the relation between the time period of the planet in terms of r1 and r2.

Answers

Answer:

the relation between the time period of the planet is

T = 2π √[( r1 + r2 )³ / 8GM ]

Explanation:

Given the data i  the question;

mass of sun = M

minimum and maximum distance = r1 and r2 respectively

Now, using Kepler's third law,

" the square of period T of any planet is proportional to the cube of average distance "

T² ∝ R³

average distance a = ( r1 + r2 ) / 2

we know that

T² = 4π²a³ / GM

T² = 4π² [( ( r1 + r2 ) / 2 )³ / GM ]

T² = 4π² [( ( r1 + r2 )³ / 8 ) / GM ]

T² = 4π² [( r1 + r2 )³ / 8GM ]

T = √[ 4π² [( r1 + r2 )³ / 8GM ] ]

T = 2π √[( r1 + r2 )³ / 8GM ]

Therefore, the relation between the time period of the planet is

T = 2π √[( r1 + r2 )³ / 8GM ]

Which one of the following is not an example of convection? An eagle soars on an updraft of wind. A person gets a suntan on a beach. An electric heater warms a room. Smoke rises above a fire. Spaghetti is cooked in water.

Answers

Answer: The statement that is not an example of convection is (A person gets a suntan on a beach).

Explanation:

There are different modes of heat energy transfer which includes:

--> conduction

--> Radiation and

--> Convection

CONVECTION is a process by which heat energy is transferred in a fluid or air by the actual movement of the heated molecules. The cooler portion of the air surrounding a warmer part exerts a buoyant force on it. As the warmer part of the air moves, it is replaced by cooler air that is subsequently warmed.

Convection in gases is very common and gas expands more than liquid when subjected to high temperature.

--> it is used in bringing about the circulation of fresh air in the room in a process known as ventilation.Here, cool air is constantly being replaced with denser air ( warm air).

-->An electric heater warms a room and Smoke rises above a fire are typical example of convection in gases.

-->Spaghetti is cooked in water: As the water close to the burner warms, it rises to the top and boils. At the same time, cooler water on top moves downward to replace the rising hot water.

--> also the eagle uses convection current to stay afloat in the sky without flapping its wings to conserve energy.

But the option (A person gets a suntan on a beach) is an example of heat transfer through radiation. This is because the sun emits it's rays from the sky down to earth without any material medium unlike others. Therefore, this option is the ODD one out.

A person jumps out of an airplane above the surface of the Earth, and falls a distance h before opening their parachute. Once the prachute is open the person coasts to the ground a distance d at constant velocity.

a. The work done on the person by the Earth is:
b. The change in gravitational potential energy of the person + Earch system is:

Answers

Answer:

a)   W_total = mg (2h + d)   , b)     E_total = - mg (h + d)

Explanation:

a) We must solve this problem in two parts, the first for the accelerated movement and the second for the movement with constant speed

Let's look for work for the part that is in free fall

        y = y₀ + v₀ t - ½ g t²

when he jumps out of a plane his vertical speed is zero

        y =y₀ - ½ g t²

        dy = 0 - ½ g 2t dt

the work in this first part is

        W₁ = ∫ F dy

        W₁ = mg ∫ g t dt

        W₁ = m g² t² / 2

the time it takes to travel the distance y₀-y = h is

         y₀-y = ½ g t²

         

         t =[tex]\sqrt{2h/g}[/tex]

we substitute

          W₁ = m g² 2h / g

          W₁ = m g 2h

now we look for the work for the part with constant speed

since the velocity is constant let's use the uniform motion ratio

          W₂ = F d

           W₂ = mg d

       

the total work is

           W_total = W₁ + W₂

           W_total = 2mgh + m gd

           W_total = mg (2h + d)

b) The change in gravitational potential energy

           U = mg Δy

in the part with accelerated movement

           U₁ = mg h

in the part with uniform movement

            U₂ = mg d

the total potential energy is

           E_total = U₁ + U₂

           E_total = - mg (h + d)

The gravitational force Asteroid A experiences is the gravitational force Asteroid C experiences

Answers

Answer:

The gravitational force Asteroid A experiences is greater than the gravitational force Asteroid C experiences

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