A spring stretches from 10cm to 22cm when a force of 4n is applied, if it obeys Hooke’s law,its total length in cm when a force of 6N is applied is?

Answers

Answer 1

Answer:

Explanation:

Hooke's Law is a linear representation that, in equation form, looks like this:

F = -kΔx where F is force, k is the spring constant, and Δx is the displacement of the spring when a Force is applied to it in the form of a mass hanging on the end of it. We need to use the equation to solve for the spring constant, k. Doing this by filling in the values we were given:

4 = -k(-12) the 12 is negative because the mass hangs below the point of equilibrium. This gives us that

k = 1/3. Now we'll use that in the next equation where we need to solve for displacement:

6 = -1/3(Δx) and

Δx = -18 cm


Related Questions

A bat is flying at 0.6 m/s when it spots an appetizing insect. The bat accelerated at a rate of 1.2m/s2 for 1.9
seconds. How fast is the bat traveling at the time?

Answers

Answer:

v_f = 2.9m/s

Explanation:

v_f = v_I + a(t)

v_f = 0.6m/s + 1.2m/s²(1.9s)

v_f = 2.88m/s

v_f = 2.9m/s

A child has a mass of 30 kg on Earth. If th gravity on Moon is one sixth that of the Earth what is the weight of the child on the Moon? Gravity on Earth = 10 N^-1​

Answers

Answer:

the baby would still be 30 kg on the moon because that’s mass, not weight.

However, it will feel like 5 kg, because the acceleration is 1/6 and the force is 1/6 and the actual weight (product of mass and acceleration will be 1/6).

The child has a mass of 30 kg on earth and has one-sixth gravity on the moon will have a weight of 50 kg on the moon.

What is Gravity?

Gravity is a fundamental interaction that causes all objects with mass or energy to attract one another. The weakest of the four fundamental interactions is gravity, by far. The Moon's gravity causes sublunar tides in the oceans, just as gravity on Earth imparts weight to physical objects.

Given:

The mass of the child (m) = 30 kg,

The gravity on the moon (G) = 1/6 of the earth.

To calculate the weight of the child on the moon, use the formula given below,

F = m × a

Here

[tex]F[/tex] is the weight, [tex]m[/tex] is the mass and [tex]a[/tex] is the gravity on earth, gravity on earth [tex]10\ m/s^2[/tex].

Substitute the values

[tex]F=30*10/6[/tex]

[tex]F = 50\ kg[/tex]

Therefore, the weight of a child on the moon is 50 kg.

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What is moment? Write down the law of moment.A long spanner is used to unscrew the tight nut.Why?​

Answers

Answer:

Moment is the product of force and its perpendicular distance from a point along its line of action.

The net moments on clockwise and anticlockwise is zero at a point.

The net force applied on a spanner handle is equal to the moment of the rotation

A carpenter applies a force of 60N horizontally to push a plane 40 cm along a piece of wood, how much work does she do?

Answers

Answer:

W = 24 J

Explanation:

Given that,

Applied force, F = 60 N

Distance moved, d = 40 cm = 0.4 m

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

Work done, W = Fd

Put all the values,

W = 60 N × 0.4 m

= 24 J

Hence, the required work done is equal to 24 J.

An 84% efficient single pulley is used to lift a 230 kg piano 3.5 m. How much work must be input?

Answers

Answer:

35%

Explanation:

win

The amount of work required to lift the 230 kg to a height of 3.5 m at  84% efficiency is 9391.67 J

What is efficiency?

This can be defined as the capacity to achieve a given task with little or no waste. Mathematically, it can be expressed as

Efficiency = (output / input) × 100

How to determine the work output Mass (m) = 230 KgHeight (h) = 3.5 mAcceleration due to gravity (g) = 9.8 m/s²Work output =?

Work output = mgh

Work output = 230 × 9.8 × 3.5

Work output = 7889 J

How to determine the work input Output = 7889 JEfficiency = 84% Input =?

Efficiency = (output / input) × 100

84% = 7889 / input

Cross multiply

84% × input = 7889

Divide both side by 84%

Input = 7889 / 84%

Input = 7889 / 0.84

Input = 9391.67 J

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How do you complete this mixed circuit?

Answers

For the resistor closest to the battery, drawn vertical in the diagram, I = 8 A and P = 320 watts.

Also, the battery voltage is 40 V.

There isn't enough information included in the picture to fill in any of the missing items for the other two resistors.

Vector B has components Bx = 6.18 and By = 4.43. Find the magnitude and direction counter clockwise from the positive x-axis.

Answers

Answer:

Explanation:

The formula for the magnitude of this resultant vector, which is in Q1, btw, is:

[tex]B_{mag}=\sqrt{6.18^2+4.43^2}[/tex] which gives us, to 3 sig fig's:

7.60 m. Since the resultant vector is in Q1, we don't need to add anything to the angle once we find it to 3 sig fig:

[tex]tan^{-1}(\frac{4.43}{6.18})[/tex] gives us an angle of 35.6 degrees

a string 2m long used to whirl a 200gm stone in horizontal circle at a speed of 2m/s . find the tension in the string​

Answers

Answer:

Explanation:

First off, we need the mass in kg and it's in g so we have to convert it. Then we can do the problem. 200 g = .200 kg. Moving on.

The equation used to find the tension in the string is the same one we use to find the centripetal force, because the tension is what is upplying the centripetal force needed to keep the stone moving in a circular manner. The formula for that is

[tex]F_c=T=\frac{mv^2}{r}[/tex] where ma is the mass in kg, v is the velocity in m/s, and r is the radius of the circle about which the stone spins in meters.

Filling in:

[tex]T=\frac{(.200)(2)^2}{2}[/tex] which gives us

T = .4 N

what is the momentum of an object weighing 7.5 kg moving at 1.6 m/s?

Answers

Answer:

[tex]momentum = mass \times velocity \\ = 7.5 \times 1.6 \\ = 12 \: kg {ms}^{ - 1} [/tex]

Mass is 7.5 kg

Velocity is 1.6 m/s

we know that,

momentum = mass × velocity

or; momentum =7. 5 kg × 1. 6 m/s

or; momentum = 12 kgms^1

A car with a mass of 500 kg is moving at a speed of 12 m/s. How much kinetic energy does it have?


someone please help me PLEASEEEE <3

Answers

Answer:

36000

Explanation:

1/2mv²=0.5×500×12²=36000j

A small part of a current loop consisting of a very, very large triangular conductor in the x-y plane is shown below, with the current i=39.30 A.



12 10 4 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 x (cm)



Calculate the magnitude of the B-field at point 'p'.

Answers

Answer: hello your question is poorly written attached below is the well written question

answer: 4.9 * 10^-5 T

Explanation:

Current ( i ) = 39.30 A

Determine the magnitude of the B-field at Point P

Mag of B-field at Point P = μo*I / 2π*r --- ( 1 )

where : I = 39.30 , μo = 4π * 10^-7 , r = 8*10^(-2)

Input values into equation 1

( 10^-7 * 39.30 ) / ( 8*10^-2 )

Mag of B-field = 4.9 * 10^-5 T

a ball of mass 0.2 kg is dropped from a height of 20m on impact to the ground it loses in 30 joule of energy calculate the height it reaches on rebound​

Answers

Answer:

new PE at top=original PE - energy lost

mgh=.2(9.8)20-30

Explanation:

mgh=.2(9.8)20-30 = 9.2

10 points
Calculate the force of attraction between the moon and the earth, if their masses and distance apart are 10^22Kg, 10^24kg and 6.4x10^6m. Take g = 6.67x10^-11Nm^2Kg^-2
a) 0.63x10^24N
b) 1.63x10^22N
c) 2.63x10^20N
d) 3.63x10^18N​

Answers

Answer:

Explanation:

You didn't fill in the proper masses which is why you never got an answer to this. But that's ok...I got you. I happen to know what they are! We will use the universal law of gravitation and the gravitational constant to solve this.

[tex]F_g=\frac{Gm_1m_2}{r^2}[/tex] and filling in:

[tex]F_g=\frac{(6.67*10^{-11})(5.98*10^{24})(7.36*10^{22})}{(3.84*10^8)^2}[/tex] The denominator is the radius of the earth plus the radius of the moon plus the distance between their surfaces, just FYI.

That gives us that

[tex]F_g=1.99*10^{20}N[/tex] Not sure what your choices entail, but I'd have to say, taking into consideration that maybe your problem didn't figure in the distance between the surfaces, you'd be at choice B.

What's scientific notation

Answers

Scientific notation is a way of expressing numbers that are too large or too small to be conveniently written in decimal form. It may be referred to as scientific form or standard index form, or standard form in the UK.

Water is being heated on the stove at 90 ⁰C. What is this temperature on the Fahrenheit degrees?

Answers

Answer:

194 degrees farenheit! hot

Explanation:

(90°C × 9/5) + 32 = 194°F

Answer:

194

Explanation:

to get from celcius to Fahrenheit

multiply temperature by 2 and add 30

what the balance electrons for calcium​

Answers

Answer:

its two valence electrons

Explanation:

Calcium is a group 2 element with two valence electrons. Therefore, it is very reactive and gives up electrons in chemical reactions.

UR WELCOME!! :)

what is electricity?write any two defects of a simple cell.​

Answers

Answer:

Electricity is the set of physical phenomena associated with the presence and motion of matter that has a property of electric charge. Electricity is related to magnetism, both being part of the phenomenon of electromagnetism, as described by Maxwell's equations.

A pendulum is moving 2.0 m/s at the bottom of its swing. How high vertically will it go before it begins to swing back? Group of answer choices 0.4 m 0.1 m 0.2 m 0.8 m 1.0 m

Answers

Answer:

h = 0.2 m

Explanation:

Given that,

A pendulum is moving 2.0 m/s at the bottom of its swing.

We need to find the height high it swing back. Let the height is h.

Using the conservation of energy such that,

[tex]\dfrac{1}{2}mv^2=mgh\\\\h=\dfrac{v^2}{2g}[/tex]

Put all the values,

[tex]h=\dfrac{(2)^2}{2\times 10}\\\\h=0.2\ m[/tex]

So, it will reach to a height of 0.2 m.

The rock falls from the distance of 15 m before it hits the water. Calculate its kinetic energy just before hitting the water. Show your working

Answers

Answer:

[tex]k.e = \frac{1}{2} m {v}^{2} \\ {v}^{2} = {u}^{2} + 2gs \\ {v}^{2} = 0 + (2 \times 9.8 \times 15) \\ v = 17.1 \: m {s}^{ - 1} \\ k.e = \frac{1}{2} \times m \times {17.1}^{2} \\ = 147m \: joules \\ m \: is \: mass[/tex]

Kinetic energy of the rock just before hitting the water is 147m joules where m is the mass of the object.

What is Kinetic Energy?

The kinetic energy of an object is defined as the energy it possesses due to its motion. It is the work required to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains the kinetic energy as long as its momentum does not change.

It is directly proportional to the mass of the object and the square of its velocity:[tex]K.E. = 1/2 mv^2[/tex].

Where, m= mass of the object

v= velocity

So if the units of mass are in kilograms and the units of velocity are meters per second, then kinetic energy has units of kilogram-meter squared per second squared.

For above given information,

[tex]K.E= 1/2 mv^2\\v^2= u^2 + 2gs\\ \\v^2= 0+ (2* 9.8*15)\\v^2= 294 m/s\\So, K.E. = 1/2 m* 294 m/s\\K.E.= 147m joules[/tex]

Thus, Kinetic energy of the rock just before hitting the water is 147m joules where m is the mass of the object.

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In April 1974, Steve Prefontaine completed a 10.0 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.71 km mark at a time of 25.0 min . If he accelerated for 60 s and then maintained his increased speed for the remainder of the race, calculate his acceleration over the 60 s interval. Assume his instantaneous speed at the 7.71 km mark was the same as his overall average speed up to that time.

Answers

Answer:

0.18 m/s²

Explanation:

The total time taken to cover a 10 km race is 27 min, 43.6 seconds.

At 25 min, Pre was at 7.71 km mark. Therefore the average speed = 7.71 km / 25 min = 7710 m / (25 * 60) s = 5.14 m/s

The distance remaining = 10 km - 7.71 km = 2.29 km = 2290 m

The remaining time = 27 min, 43.6 seconds - 25 min = 2 min 43.6 second = 163.6 seconds

She accelerates for 60 seconds, therefore the distance covered (S) during the acceleration (a) is:

S₁ = 5.14(60) + 0.5a(60)² = 308.4 + 1800a

She maintains the speed for the remaining distance (S₂). The remaining time = 163.6 seconds - 60 seconds = 103.6 seconds. The final speed after the acceleration = (5.14 + 60a) m/s

S₂ = (5.14 + 60a)* 103.6 = 532.5 + 6216a

S₁ + S₂ = 2290 m

(308.4 + 1800a) + (532.5 + 6216a) = 2290

8016a + 840.9 = 2290

8016a = 1449.1

a = 0.18 m/s²

Un reloj de péndulo de largo L y período T, aumenta su largo en ΔL (ΔL << L). Demuestre que su período aumenta en: ΔT = π ΔL /√(L g)

Answers

Answer:

 ΔT = [tex]\pi \ \frac{\Delta L}{\sqrt{Lg} }[/tex]

Explanation:

In a simple harmonic motion, specifically in the simple pendulum, the angular velocity

          w = [tex]\sqrt{\frac{g}{L} }[/tex]

angular velocity and period are related

          w = 2π / T

we substitute

          2π / T = \sqrt{\frac{g}{L} }

          T = [tex]2\pi \ \sqrt{\frac{L}{g} }[/tex]

In this exercise indicate that for a long Lo the period is To, then and increase the long

          L = L₀ + ΔL

we substitute

           T = [tex]2\pi \ \sqrt{\frac{L + \Delta L}{g} }[/tex]

            T = [tex]2\pi \ \sqrt{\frac{L}{g} } \ \sqrt{1+ \frac{\Delta L}{L} }[/tex]

in general the length increments are small ΔL/L «1, let's use a series expansion

           [tex]\sqrt{1+ \ \frac{\Delta L}{L} } = 1 + \frac{1}{2} \frac{\Delta L}{L} + ...[/tex]  

we keep the linear term, let's substitute

           T = [tex]2\pi \ \sqrt{\frac{L}{g} } \ ( 1 + \frac{1}{2} \frac{\Delta L}{L} )[/tex]  

if we do

           T = T₀ + ΔT

           

           T₀ + ΔT = [tex]2\pi \sqrt{\frac{\Delta L}{g} } + \pi \ \sqrt{\frac{L}{g} } \ \frac{\Delta L}{L}[/tex]

            T₀ + ΔT = T₀ + [tex]\pi \sqrt{\frac{1}{Lg} } \ \Delta L[/tex]

            ΔT = [tex]\pi \ \frac{\Delta L}{\sqrt{Lg} }[/tex]

for the light bulb to produce the aame illuminace with the candle to the screen, the luminous intensity of the bulb is _______ that of the candle?

Answers

Answer:

9 times

Explanation:

luminosity decreases to the square of the distance

A cyclist is taking part in the Tour de France, which is a bicycle race that takes place every year.
a Two forces acting on the cyclist are weight and reaction. Name two other forces acting on him
as he cycles along.

Answers

Solve the following word problems.
1. The ratio of red marbles and blue marbles that Carlo has is 8: 3. When he
exchanged 35 red marbles for 20 blue marbles from his brother, he was left with
equal number of red and blue marbles.
How many red and blue marbles did he have at the beginning
How many red and blue marbles did he have now

A student is in class 400 minutes every day for 5 days each week. How many seconds is this?

Answers

Answer:

120,000seconds

Explanation:

400 minutes x 5= 2000minutes (2000minutes x 60)seconds =120,000seconds.

btw :

stay safe! :3

Answer:

12000

Explanation:

5 x 400 = 2000, 2000 x 60 = 12000

A container of gas is at a pressure of 3.7 x 10^5 Pa. How much work is done by the gas if its volume expands by 1.6 m^3 ?

Answers

Answer:

592000 J

Explanation:

We'll begin by converting 3.7×10⁵ Pa to Kg/ms². This can be obtained as follow:

1 Pa = 1 Kg/ms²

Therefore,

3.7×10⁵ Pa = 3.7×10⁵ Kg/ms²

Next, we shall determine the workdone.

Workdone is given by the following equation:

Workdone (Wd) = pressure (P) × change in volume (ΔV)

Wd = PΔV

With the above formula, the work done can be obtained as follow:

Pressure (P) = 3.7×10⁵ Kg/ms²

Change in volume (ΔV) = 1.6 m³

Workdone (Wd) =?

Wd = PΔV

Wd = 3.7×10⁵ × 1.6

Wd = 592000 Kgm²/s²

Finally, we shall convert 592000 Kgm²/s² to Joule (J). This can be obtained as follow:

1 Kgm²/s² = 1 J

Therefore,

592000 Kgm²/s² = 592000 J

Therefore, the Workdone is 592000 J.

what are scalar quantities and list their examples​

Answers

Explanation:

these are quantities that have magnitude but no direction

Examples include : speed, mass, volume, density, everything, time

speed, mass, volume, density, time

a 10 kg box is pulled across an ice rink for a distance of 50m. it is pulled with a constant force of 10 N on a rope angled at 60° to the horizontal. How much work is done on the box? (A) 50 ) (B) 100 ) (C) 250 ) (D) 500)​

Answers

Answer:

(C) 250

Explanation:

Given the following data;

Mass = 10 kg

Distance = 50 m

Force applied = 10 N

Angle = 60°

To find the work done on the box, we would first of all find the horizontal component of the force applied.

[tex] Horizontal force, Fx = mgCosd[/tex]

Where;

Fx represents the horizontal force.

m is the mass of an object.

g is the acceleration due to gravity.

d is the angle of inclination (theta).

mg = weight = 10 N

Substituting into the equation, we have;

[tex] Fx = 10 * Cos60 [/tex]

[tex] Fx = 125 * 0.5 [/tex]

Fx = 5 Newton

Next, we find the work done.

Mathematically, workdone is given by the formula;

[tex] Workdone = force * distance[/tex]

Substituting into the formula, we have;

[tex] Workdone = 5 * 50 [/tex]

Workdone = 250 Nm

Can someone please help with part c)

Answers

Energy released ,

[tex] = 3.789 \times {10}^{ - 12} [/tex] J


Which is NOT a characteristic of a solid?

a) All solids have a definite shape.
b) All solids have molecules that are loosely packed together.
C) All solids have mass.
d) All solids take up space

Answers

The correct answer is B

Hope this helps ;)

Answer:

b

Explanation:

9) What is the temperature of 5 moles of nitrogen at 1 atm in a 2 liter container?*


- 4.88 K
- 0.0328 K
- 0.41 K.
- 0.0164 K

Answers

Answer:

4.88 K.

Explanation:

From the question given above, the following data were obtained:

Number of mole (n) = 5 moles

Pressure (P) = 1 atm

Volume (V) = 2 L

Gas constant (R) = 0.082 atm.L/Kmol

Temperature (T) =?

The temperature of the gas can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

1 × 2 = 5 × 0.082 × T

2 = 0.41 × T

Divide both side by 0.41

T = 2 / 0.41

T = 4.88 K

Therefore, the temperature of the gas is 4.88 K.

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