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3 years ago

What does the elastic region of a force vs. depth plot represents? describe

Physics

1 answer:

PSYCHO15rus [73]3 years ago

4 0

Answer:

The force vs depth plot represent the elastic region and the force per unit depth represent the spring constants

Explanation:

The elastic region is explained by Hooke's law and which states that the that the force needed to extend or compress a spring by some distance is proportional to that distance.

F∝e

F= ke

Where

K is spring constant N/m

e is extension caused by forced (m)

F is the force applied to spring (N)

From the formula, the force per unit depth (F/e) represents the spring constant (k), which is the gradient or the slope of the force vs depth plot. And this is the region where Hooke's is obeyed I.e the elastic region.

F/e = k

Check attachment for proper understanding

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Can anyone help me on question 13

Leona [35]

Given that in a parallel circuit:

R1 = 12 ohms

R2= 15 ohms

I = 12 A

I2 = 4 A

V=?

R=?

R3 =?

P=?

Since,

V= IR

or,

V2 = I2 * R2

V2= 4* 15

V2 = 60V

Since in a parallel circuit voltage remain same in all component of the circuit and is equal to the source voltage.

Therefore,

V= V1 = V2 = V3 = 60V

Since,

V= IR

R= V/I

R= 60/12

R= 5 ohm

That is total resistance is equal to 5 ohms.

Since for parallel circuit,

1/R= 1/R1 + 1/R2 + 1/R3

1/5= 1/12+ 1/15 + 1/R3

1/R3= 1/5- 1/12- 1/15

1/R3= 1/20

R3= 20 ohms

Since,

V=IR

I= V/R

I1= V1/ R1

I1= 60/12

I1= 5 A

I3= V3/R3

I3= 60/20

I3= 3A

Since,

P=VI

P= 60*12

P= 720 watt

P1= V1* I1

P1= 60* 5

P1= 300 watt

P2= V2* I2

P2= 60* 4

P2= 240watt

P3= V3*I3

P3= 60*3

P3= 180 watt

Hence we have,

R1= 12 ohms , R2= 15 ohms, R3= 20 ohms, R= 5 ohms

I1= 5A, I2= 4A, I3= 3A, I= 12 A

V1= V2= V3= V= 60V

P1= 300 watt, P2= 240 watt, P3 = 180 watt, P= 720 watt

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ExtremeBDS [4]

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3 years ago

Given the initial wavefunction Ψ (x, 0) = Axexp (-k x) withx> 0 andk> 0, and Ψ (x, 0) = 0 forx <0, what value must A ta

madam [21]

Answer with explanation:

The Normalization Principle states that

$\int_{-\infty }^{+\infty }f(x)dx=1$

Given

$f(x)=xe^{-kx}(x>0\\\\0(x$

Thus solving the integral we get

$\int_{0 }^{+\infty }A\cdot xe^{-kx}dx=1\\\\A\int_{0 }^{+\infty }\cdot xe^{-kx}dx=1$

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$I=\int xe^{-kx}dx\\\\x\int e^{-kx}dx-\int \frac{d(x)}{dx}\int e^{-kx}dx\\\\-\frac{xe^{-kx}}{k}-\int 1\cdot \frac{-e^{-kx}}{k}\\\\\therefore I=\frac{e^{-kx}}{k}-\frac{xe^{-kx}}{k}$