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1 year ago

How much work is needed to stretch this spring a distance of 5.0 cm , starting with it unstretched?

Physics

1 answer:

olganol [36]1 year ago

4 0

The spring's work output while it is not extended is 5 J.

<h3>What is The Hooke's law ?</h3>

According to Hooke's law, as long as the elastic limit is not exceeded, the extension of a particular material is directly proportional to the force applied. Hence, we must be aware of that F = Ke

F = Applied force

Force constant is K.

Extension = e

F = applied force

Force constant is K.

Extension = e

Using the graph

K = F/e

F = 200N

E equals 5 cm or 5 * 10-2 m

K = 200N/ 5 * 10^-2 m

K = 4000 N/m

Now;

Work = 1/2Ke^2

Work = 0.5 * 4000 N/m * (5 * 10^-2 m)^2

Work = 5 J

To know more about Hooke's Law visit:

brainly.com/question/13348278

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3 years ago

Una ardilla corre desarrollando una aceleracion de 3.5 m/s2 la fuerza ejecidas por las patas traseras es de 5 N. Cual sera la ma

ziro4ka [17]

Answer:

1.43 kg

Explanation:

We can solve the problem by using Newton's second law:

$F=ma$

where

F is the net force on an object

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3 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

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2 years ago