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1 year ago

a spring is stretched 20 cm by a 30.0 n force. determine the work done in stretching the spring from 0 to 40 cm?

Physics

1 answer:

Alekssandra [29.7K]1 year ago

3 0

The work done when a spring is stretched from 0 to 40cm is 4J.

What is work done?

Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.

The work done on the spring to stretch to 40cm is,

F = kx

where F is force, k is force constant.

k = F / x = 10 N / 20 * 10^-2 m = 50 N/m

W = 0.5 * k * (x)^2

where W = work done, k = force constant.

W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.

Therefore, the work done on the spring when it is stretched to 40cm is 4J.

To learn more about work done click on the given link brainly.com/question/25573309

#SPJ4

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In the International System of Units, the basic unit for measuring mass is what

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Answer:

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Explanation:

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3 years ago

A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold

LUCKY_DIMON [66]

Answer:

(a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

$Q=\dfrac{V}{t}$

Put the value into the formula

$Q=\dfrac{405}{90}$

$Q=4.5\ cm^3/s$

(a). We need to calculate the coefficient of permeability

Using formula of coefficient of permeability

$Q=kiA$

$k=\dfrac{Q}{iA}$

$k=\dfrac{Ql}{Ah}$

Where, Q=discharge

l = length

A = cross section area

h=constant head causing flow

Put the value into the formula

$k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}$

$k=8.6\times10^{-3}\ cm/s$

The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(c). We need to calculate the discharge velocity during the test

Using formula of discharge velocity

$v=ki$

$v=\dfrac{kh}{l}$

Put the value into the formula

$v=\dfrac{8.6\times10^{-3}\times38}{17.5}$

$v=0.0187\ cm/s$

The discharge velocity during the test is 0.0187 cm/s.

(b). We need to calculate the volume of solid in the ample

Using formula of volume

$V_{s}=\dfrac{M_{s}}{V_{s}}$

Put the value into the formula

$V_{s}=\dfrac{4950\times10^{-3}}{2710}$

$V_{s}=1826.56\ cm^3$

We need to calculate the volume of the soil specimen

Using formula of volume

$V=A\times L$

Put the value into the formula

$V=\dfrac{\pi(17.5)^2}{4}\times17.5$

$V=4209.24\ cm^3$

We need to calculate the volume of the voids

$V_{v}=V-V_{s}$

Put the value into the formula

$V_{v}=4209.24-1826.56$

$V_{v}=2382.68\ cm^3$

We need to calculate the seepage velocity

Using formula of velocity

$Av=A_{v}v_{s}$

$v_{s}=\dfrac{Av}{A_{v}}$

$v_{s}=\dfrac{V}{V_{v}}\times v$

Put the value into the formula

$v_{s}=\dfrac{4209.24}{2382.68}\times0.0187$

$v_{s}=0.0330\ cm/s$

The seepage velocity is 0.0330 cm/s.

Hence, (a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

8 0

3 years ago

Please helppp

Y_Kistochka [10]

Answer:

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F = m(g + v²/R ) = 1000(9.8 + 15²/30) = 17,300

6 0

3 years ago

We have three identical metallic spheres A, B, C. Initially sphere A is charged with charge Q, while B and C are neutral. First,

larisa [96]

Answer:

The final charges of each sphere are: q_A = 3/8 Q , q_B = 3/8 Q , q_C = 3/4 Q

Explanation:

This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.

Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point

q_A = Q / 2

q_B = Q / 2

Now sphere A touches sphere C, ending with half the charge

q_A = ½ (Q / 2) = ¼ Q

q_B = ¼ Q

Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge

q = Q / 4 + Q / 2 = ¾ Q

This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q

q_A = 3/8 Q

q_B = 3/8 Q

The final charges of each sphere are:

q_A = 3/8 Q

q_B = 3/8 Q

q_C = 3/4 Q

7 0

3 years ago