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3 years ago

A spring is characterized by a spring constant of 60 N/m. How much potential energy does it store, when stretched by 1.0 cm?

Physics

2 answers:

Anestetic [448]3 years ago

8 0

Answer:

3.0 x10^-3 J

Explanation:

The potential energy of a spring is given by PE = (0.5)k*x^2

Where

K: Spring Constant = 60 N/m

x: displacement of the spring from its equilibrium position = 1cm = 0.01m

Then PE = 0.5(60)(.01)^2 = 0.003J = 3.0 x10^-3 J

fiasKO [112]3 years ago

5 0

Answer:

 The potential Energy stored in the spring = 0.003 J

Explanation:

Potential Energy: This is the energy possessed by a body by virtue of its position. The ability to stretched or compressed an elastic material to do work is called the elastic potential energy. The unit of elastic potential energy is Joules (J).

Elastic potential energy can be represented mathematically as

W = 1/2ke² ............................... Equation 1

Where W = Elastic potential Energy, k = force constant of the spring. e = extension.

Given: k = 60 N/m, e = 1.0 cm = (1/100) m = 0.01 m.

Substituting these values into equation 1,

W = 1/2(60)(0.01)²

W = 30×0.0001

W = 0.003 J.

Thus potential Energy stored in the spring = 0.003 J

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Sonja [21]

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2 years ago

Read 2 more answers

How do you convert seconds to minutes, seconds to hours , minutes to seconds and hours to seconds

Bess [88]

Answer:

You could memorize conversions, or use conversion charts, or do a quick internet search for help.

Explanation:

I'll do a few of the conversions for you

Seconds to minutes:

there are 60 seconds in one minute. So if you are wondering how many seconds are in 3 and 1/2 minutes, you would do this conversion:

60 sec/min times x sec/3.5 min

which can be written as 60 x 3.5 = 210. so "x" seconds would be 210 seconds in 3 1/2 minutes.

you could do the same thing in the opposite direction for minutes to seconds:

1 min has 60 seconds. So in 7.25 minutes, how many seconds are there?

1 min/60 sec times 7.25 min/x sec

which can be written as 7.25 x 60 = 435. so "x" seconds would be 435 seconds in 7.25 minutes.

hours to seconds:

this one is slightly more complicated

In one hour there are 60 minutes, and in one minute there are 60 seconds.

so to convert from hours to seconds you would do this conversion:

1 hr/60 min times 1 min/60 sec. then the "min" would cancel out, and you would be left with the label "hr/sec". to do the math, it would be 1 hr / 60 x 60.

60x60 = 3600. so you would have 1 hr/3600 sec. So in one hour there are 3600 seconds.

so if you want to know how many seconds are in 6.75 hours:

6.75 hr/x sec times 3600 sec/1 hr

6.75 x 3600 = 24,300 so there are 24,300 seconds in 6.75 hours.

I hope this helps :)

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3 years ago

One wire carries a current of I1= 2 Amps, and the other carries a parallel current (same direction, wires are side by side) of I

Yuki888 [10]

Answer: $F=3\times 10^{-5} N$

Explanation:

Given

$I_1=2 Amps$

$I_2=0.75 Amps$

Length of each wires $\left ( L\right )=5 m$

Distance between wires $\left ( r\right )=5 cm$

Force per unit length = $\frac{\mu_0I_1I_2}{2\pi r}$

$F'=\frac{2\times 10^{-7}\times 2\times 0.75}{2\pi \times 0.05}$

$F'=6\times 10^{-6}$

Force for $L=5 m$

$F=3\times 10^{-5} N$

6 0

3 years ago

A man wishes to lift a stone weighing 1440 N, using a first-class lever that measures 5 meters. What force should it perform if

Crazy boy [7]

Answer:

3360 N

Explanation:

In a first-class lever, the effort force and load force are on opposite sides of the fulcrum.

The lever is 5 m long. The load force is 1.50 m from the fulcrum, so the effort force must be 3.50 m from the fulcrum.

The torques are equal:

Fr = Fr

(1440 N) (3.5 m) = F (1.5 m)

F = 3360 N

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3 years ago

Which statements are true for two oppositely charged, isolated parallel plates: C=capacitance, U=stored energy (Q and -Q = charg

vlada-n [284]

Answer:

Explanation:

1) True. The stored energy (U) is proportional to the electric field strength (E). The electric field strength decreases when a dielectric is introduced hence inserting a dielectric decreases U.

2) False. From the formula $C=\frac{Q}{V}=\frac{Q}{Vd}$ , capacitance is inversely proportional to distance hence if the distance is doubled, capacitance decreases.

3) False. As the distance between the electric field and the object increases, its electric field decreases.

4) False. If a dielectric is inserted, the plates are further separated. Q stays the same.

5) True. The electric field strength decreases when a dielectric is introduced and capacitance is inversely proportional to electric field hence Inserting a dielectric increases C

6) True. If a dielectric is inserted, the plates are further separated. Q stays the same.

7) True. When the distance is doubled, U increases

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3 years ago