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adelina 88 [10]
4 years ago
13

A spring is stretched from x=0 to x=d, where x=0 is the equilibrium position of the spring. It is then compressed from x=0 to x=

−d. What can be said about the energy required to stretch or compress the spring? View Available Hint(s) A spring is stretched from to , where is the equilibrium position of the spring. It is then compressed from to . What can be said about the energy required to stretch or compress the spring? More energy is required to stretch the spring than to compress it. The same amount of energy is required to either stretch or compress the spring. Less energy is required to stretch the spring than to compress it.
Physics
1 answer:
VARVARA [1.3K]4 years ago
7 0

Answer:

The same amount of energy is required to either stretch or compress the spring.

Explanation:

The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:

U=\frac{1}{2}k (\Delta x)^2

where

k is the spring constant

\Delta x is the stretch/compression of the spring

In the first case, the spring is stretched from x=0 to x=d, so

\Delta x = d-0=d

and the amount of energy required is

U=\frac{1}{2}k d^2

In the second case, the spring is compressed from x=0 to x=-d, so

\Delta x = -d -0 = -d

and the amount of energy required is

U=\frac{1}{2}k (-d)^2= \frac{1}{2}kd^2

so we see that the amount of energy required is the same.

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A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to
White raven [17]

Answer:

1) \Delta s=1000\ ft

2)  \Delta s'=998.11\ ft.s^{-1}

3) t\approx125\ s

t'\approx463.733\ s

Explanation:

Given:

width of river, w=500\ ft

speed of stream with respect to the ground, v_s=8\ ft.s^{-1}

speed of the swimmer with respect to water, v=4\ ft.s^{-1}

<u>Now the resultant of the two velocities perpendicular to each other:</u>

v_r=\sqrt{v^2+v_s^2}

v_r=\sqrt{4^2+8^2}

v_r=8.9442\ ft.s^{-1}

<u>Now the angle of the resultant velocity form the vertical:</u>

\tan\beta=\frac{v_s}{v}

\tan\beta=\frac{8}{4}

\beta=63.43^{\circ}

  • Now the distance swam by the swimmer in this direction be d.

so,

d.\cos\beta=w

d\times \cos\ 63.43=500

d=1118.034\ ft

Now the distance swept downward:

\Delta s=\sqrt{d^2-w^2}

\Delta s=\sqrt{1118.034^2-500^2}

\Delta s=1000\ ft

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

v'=v.\cos37

v'=4\times \cos37

v'=3.1945\ ft.s^{-1}

<u>Now the net effective speed of stream sweeping the swimmer:</u>

v_n=v_s-v'

v_n=8-3.1945

v_n=4.8055\ ft.s^{-1}

<u>The  component of swimmer's velocity heading directly towards the opposite bank:</u>

v'_r=v.\sin37

v'_r=4\sin37

v'_r=2.4073\ ft.s^{-1}

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

\tan\phi=\frac{v_n}{v'_r}

\tan\phi=\frac{4.8055}{2.4073}

\phi=63.39^{\circ}

  • Now let the distance swam in this direction be d'.

d'\times \cos\phi=w

d'=\frac{500}{\cos63.39}

d'=1116.344\ ft

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\Delta s'=\sqrt{d'^2-w^2}

\Delta s'=\sqrt{1116.344^2-500^2}

\Delta s'=998.11\ ft.s^{-1}

3)

Time taken in crossing the rive in case 1:

t=\frac{d}{v_r}

t=\frac{1118.034}{8.9442}

t\approx125\ s

Time taken in crossing the rive in case 2:

t'=\frac{d'}{v'_r}

t'=\frac{1116.344}{2.4073}

t'\approx463.733\ s

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4 years ago
L. What does it mean to "convert" from one unit to another? Give an example.
TiliK225 [7]

These are four questions and four answers


Question 1. What does it mean to "convert" from one unit to another? Give an example.


Convert from one unit to another is transforming or changing the magnitude to an equivalent measure according to a different pattern of comparison or unit of measure.


For example, converting from meter to centimeters, means to transform the unit of measure (the pattern of comparison) from meter to centimeter.


When your measure is 3 meters, means the length is 3 times the standard pattern named meter.


Therefore, converting it to centimeters means that you want to know how many times is 3 meters relative to the a new pattern named centimeter.


Since, 1 meter equals 100 cm, you build a conversion factor which you can use to make the conversion:


1 m = 100 cm ⇒ 1 = 100 cm / 1 m


∴ 3 m = 3 m × 100 cm / 1 m = 300 cm.


Question 2. What are the general mathematical operations you use when converting from one SI unit to another?


The most common mathematical operations are the basic ones: addition, substraction, multiplication and division.


See one example of each one:


a) Addition:


Transform 20°C to K: 20 + 273.15 K = 293.15 K


b) Subtraction:


Convert 300 K to celsius degrees: 300 - 273.15°C = 26.85°C


c) Multiplication


Convert 1000 Km to meter


1000 Km × 1000 m/Km = 1,000,000 m


d) Division


Convert 1540 mm Hg to atm


1 atm = 760 mmHg ⇒ 1 = 760 mmHg / atm

1540 mmHg / (760 mmHg / atm) = 2.026 atm


Question 3. How many meters do you cover in a 10 km (10-K) race?


This requires to convert 10 km to the equivalent number of meters.


Conversion factor: 1 km = 1000 m ⇒ 1 = 1000 m/Km


∴ 10 Km = 10 Km × 1000 m/km = 10,000 m ← answer.


Quesiton 4. An Olympic swimming pool is 50 meters long. You swim from one end to the other four times.


a. How many meters do you swim?

4 laps × 50 m/lap = 200 m ← answer


b. How many kilometers do you swim?


1 = 1 km / 1000m

200 m = 200 × 1 km / 1000 m = 0.200 km ← answer


c. How many centimeters do you swim?


1 m = 100 cm ⇒ 1 = 100 cm / m

200 m = 200 m × 100 cm/m = 20,000 cm ← answer

5 0
3 years ago
Read 2 more answers
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