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3 years ago

A weight of 200 n is hung from a spring with a spring constant of 2500 n/m and lowered slowly. How much will the spring stretch?

2 answers:

8 0

The length by which the spring got stretched will be 0.08 m. The force is directly propotional to the distance by which the spring stretched.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

F is the spring force =200

K is the spring constant= 2500 N/m

x is the length by which spring got stretched =?

The stretch of the spring is found as;

$\rm F=kx \\\\ x = \frac{F}{k} \\\\\ x= \frac{200}{2500} \\\\ x=0.08 \ m$

Hence the length by which the spring got stretched will be 0.08 m.

To learn more about the spring force refer to the link;

brainly.com/question/4291098

#SPJ4

leva [86]3 years ago

3 0

8n/m this is because 200×2500 all over 100 . 0's will cancile 0's living 200×2500×1 ,25 will go into 200 8 times ×1 =8n/m

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Explanation:

Given

Length of Pole is 9 ft

Length of Joe is 3 ft

Joe walks away from Pole at the rate 4 ft/s

Let Joe is x m away from Pole so its shadow length is x'

From Similar triangle concept

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and it is given $\frac{\mathrm{d} x}{\mathrm{d} t}=4 ft/s$

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azamat

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Explanation:

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Here, element C replaces another element B from compound AB.

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A solid sphere of diameter D = 26 cm has a charge of Q = 4 nano-coulombs uniformly distributed on it. Calculate the magnitude of

kozerog [31]

Answer:

E = 4.83 N/ C

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere:

$E= \frac{k*Q}{r^{2} }$ : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

Where:

K: coulomb constant (N*m²/C²)

a: sphere radius (m)

Q: Total sphere charge (C)

r : Distance from the center of the sphere to the region where the electric field is calculated (m)

Equivalences

1nC=10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=4nC=4 *10⁻⁹C

D = 26 cm = 26*10⁻²m = 0.26m

a = D/2 = 0.13m

r= R+a = 2.6 m+ 0.13m = 2.73m

Problem development

Magnitude of the electric field at r = 2.73m from the center of the sphere

r>a , We apply the Formula (1) :

$E= \frac{k*Q}{r^{2} }$

$E= \frac{9*10^{9}*4*10^{-9} }{2.73^{2} }$

E= 4.83 N/ C

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3 years ago