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kipiarov [429]
1 year ago
15

A potential energy function for system 1 is given by U1(x)=Cx2+Bx3. The potential energy function for system 2 is given by U2(x)

=A+Cx2+Bx3, where A is a positive quantity. How does the force on system 1 relate to the force on system 2 at a given position?
Mathematics
1 answer:
BaLLatris [955]1 year ago
6 0

Answer:

F_1=F_2

Step-by-step explanation:

  Given that

For system 1

U_1(x)=Cx^2+Bx^3

For system 2

U_2(x)=A+Cx^2+Bx^3

A is positive quantity

We know that

Force F

F=\dfrac{dU}{dx}

For system 1

\dfrac{dU}{dx}=2Cx+3Bx^2

F_1=2Cx+3Bx^2

For system 2

\dfrac{dU}{dx}=0+2Cx+3Bx^2

F_2=2Cx+3Bx^2

F_1=2Cx+3Bx^2

F_2=2Cx+3Bx^2

F_1=F_2

So from above equations we can say that force on both system is equal at given value of x.

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One of the vertices of △PQR is P(2, −1). The midpoint of PQ is M(3, 0). The midpoint of QR is N(5, 3). Show that MN || PR and MN
VLD [36.1K]

Answer:

<em>See the proof below</em>

Step-by-step explanation:

Given the following coordinates

P(2, −1)

Midpoint of PQ M(3, 0)

We can get the coordinate point Q using the midpoint formula;

M(X,Y) = (x1+x2/2, y1+y2/2)

X = x1+x2/2

3 = 2+x2/2

6 = 2+x2

x2 = 6-2

x2 = 4

Y = y1+y2/2

0 = -1+y2/2

0 = -1 + y2

y2 = 0+1

y2 = 1

<em>Hence the coordinate of Q is (4, 1)</em>

Next is to get the coordinate of R

Given the midpoint of QR to be N(5, 3)

(5,3) = (4+x2/2, 1+y2/2)

5 = 4+x2/2

10 = 4+x2

x2 = 10-4

x2 = 6

1+y2/2 = 3

1+y2 = 6

y2 = 6-1

y2 = 5

<em>Hence the coordinate of R is (6,5)</em>

<em></em>

Given the coordinates M(3, 0) and N(5, 3)

Slope is expressed as:

m = y2-y1/x2-x1

m = 3-0/5-3

m = 3/2

Slope of MN = 3/2

Get the slope of PR

Given the coordinates P(2, −1) and R (6,5)

Slope of PR = 5-(-1)/6-2

Slope of PR = 5+1/4

Slope of PR = 6/4 = 3/2

<em>Since the slope of MN is equal to that of PR, hence MN is parallel to PR i.e MN || PR</em>

<em></em>

To show that MN = 1/2PR, we will have to take the distance between M and N and also P and R first as shown:

For MN with coordinates  M(3, 0) and N(5, 3)

MN = √(x2-x1)²+(y2-y1)²

MN = √(5-3)²+(3-0)²

MN = √2²+3²

MN = √13

Get the length of PR where P(2, −1) and R (6,5)

PR = √(6-2)²+(5+1)²

PR = √4²+6²

PR = √16+36

PR = √52

PR = √4*13

PR = √4*√13

PR = 2√13

Since MN = √13

PR = 2MN

Divide both sides by 2

PR/2 = 2MN/2

PR/2 = MN

Hence MN = 1/2 PR (Proved!)

8 0
1 year ago
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