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2 years ago

A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm.

Physics

1 answer:

mote1985 [20]2 years ago

7 0

Answer:

a) When its length is 23 cm, the elastic potential energy of the spring is

0.18 J

b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

Explanation:

Hi there!

a) The elastic potential energy (EPE) is calculated using the following equation:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretched lenght.

Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).

First, let´s convert the spring constant units into N/m:

4 N/cm · 100 cm/m = 400 N/m

EPE = 1/2 · 400 N/m · (0.03 m)²

EPE = 0.18 J

When its length is 23 cm, the elastic potential energy of the spring is 0.18 J

b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:

EPE = 1/2 · 400 N/m · (0.06 m)²

EPE = 0.72 J

When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

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A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The

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Answer:

The horizontal distance d does the ball travel before landing is 1.72 m.

Explanation:

Given that,

Height of ramp $h_{1}=2.30\ m$

Height of bottom of ramp $h_{2}=1.69\ m$

Diameter = 0.17 m

Suppose we need to calculate the horizontal distance d does the ball travel before landing?

We need to calculate the time

Using equation of motion

$h_{2}=ut+\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2h_{2}}{g}}$

$t=\sqrt{\dfrac{2\times1.69}{9.8}}$

$t=0.587\ sec$

We need to calculate the velocity of the ball

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2$

$K.E=\dfrac{7}{10}mv^2$

Using conservation of energy

$K.E=mg(h_{1}-h_{2})$

$\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})$

$v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})$

Put the value into the formula

$v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}$

$v=2.922\ m/s$

We need to calculate the horizontal distance d does the ball travel before landing

Using formula of distance

$d =vt$

Where. d = distance

t = time

v = velocity

Put the value into the formula

$d=2.922\times 0.587$

$d=1.72\ m$

Hence, The horizontal distance d does the ball travel before landing is 1.72 m.

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2 years ago

Does the surrounding air become warm or cool when vapour phase of H2O condenses? Explain

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The surrounding air will become warm when water vapor condenses. The vapors when become water will give away latent heat they have, we know that latent heat is required for the object to change states, so, the latent heat the water vapor had when it became water vapor from water will be given out when it again becomes water.

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2 years ago

Pistons are fitted to two cylindrical chambers connected through a horizontal tube to form a hydraulic system. The piston chambe

Ivenika [448]

Answer:

order d> a = e> c> b = f

Explanation:

Pascal's law states that a change in pressure is transmitted by a liquid, all points are transmitted regardless of the form

P₁ = P₂

Using the definition of pressure

F₁ / A₁ = F₂ / A₂

F₂ = A₂ /A₁ F₁

Now we can examine the results

a) F1 = 4.0 N A1 = 0.9 m2 A2 = 1.8 m2

F₂ = 1.8 / 0.9 4

F₂a = 8 N

b) F1 = 2.0 N A1 = 0.9 m2 A2 = 0.45 m2

F₂b = 0.45 / 0.9 2

F₂b = 1 N

c) F1 2.0 N A1 = 1.8 m2 A2 = 3.6 m2

F₂c = 3.6 / 1.8 2

F₂c = 4 N

d) F1 = 4.0N A1 = 0.45 m2 A2 = 1.8 m2

F₂d = 1.8 / 0.45 4.0

F₂d = 16 m2

e) F1 = 4.0 N A1 = 0.45 m2 A2 = 0.9 m2

F₂e = 0.9 / 0.45 4

F₂e = 8 N

f) F1 = 2.0N A1 = 1.8 m2 A2 = 0.9 m2

F₂f = 0.9 / 1.8 2.0

F₂f = 1 N

Let's classify the structure from highest to lowest

F₂d> F₂a = F₂e> F₂c> F₂b = F₂f

I mean the combinations are

d> a = e> c> b = f

6 0

2 years ago

A variable-length air column is placed just below a vibrating wire that is fixed at both ends. The length of the air column, ope

pogonyaev

Answer:

The speed of transverse waves in the wire is 234.26 m/s.

Explanation:

Given that,

Length of wire = 129 cm

Speed of sound in air = 340 m/s

First position of resonance = 31.2 cm

We need to calculate the wavelength

For pipe open at one end and closed at other, there is node at closed end and an anti node at open end for 1st resonance.

At 1st resonance,

$L=\dfrac{\lambda}{4}$

$\lambda=4\times L$

Put the value into the formula

$\lambda=4\times31.2\times10^{-2}$

$\lambda=1.248\ m$

We need to calculate the frequency of sound in pipe

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{340}{1.248}$

$f=272.4\ Hz$

We need to calculate the node distance

For the wire, there are 3 segments, so 4 nodes

Node-node distance ,

$L=\dfrac{l}{3}$

$L = \dfrac{129}{3}$

$L=43\ cm$

We need to calculate the wavelength of the wire

Using formula of length

$L=\dfrac{\lambda}{2}$

$\lambda=L\times 2$

Put the value into the formula

$\lambda=43\times2$

$\lambda=86\ cm$

We need to calculate the speed of the wave

Using formula of frequency

$f=\dfrac{v}{\lambda}$

$v=f\times\lambda$

Put the value into the formula

$v=272.4\times86\times10^{-2}$

$v=234.26\ m/s$

Hence, The speed of transverse waves in the wire is 234.26 m/s.

4 0

2 years ago

What is the work done by the 200.-N tension shown if it is used to drag the 150-N crate 25 m across the floor at a constant spee

WINSTONCH [101]

Answer:

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Explanation:

Work equals force times distance, but the force is zero because the crate being dragged will have zero acceleration. Force equals mass times acceleration and since acceleration is zero, force has to equal zero as well. Since the force is zero, the work required also has to be zero.

8 0

1 year ago