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Korolek [52]
1 year ago
5

The electric potential in a region that is within 2.00 mm of the origin of a rectangular coordinate system is given by V=Axl+Bym

+Czn+DV=Axl+Bym+Czn+Dwhere AA, BB, CC, DD, ll, mm, and nn are constants. The units of AA, BB, CC, and DD are such that if xx, yy, and zz are in meters, then VV is in volts. You measure VV and each component of the electric field at four points and obtain these results:Point (x,y,z)(m) V(V) Ex(V/m) Ey(V/m) Ez(V/m) 1 (0, 0, 0) 10.0 0 0 0 2 (1.00, 0, 0) 4.0 16.0 0 0 3 (0, 1.00, 0) 6.0 0 16.0 0 4 (0, 0, 1.00) 8.0 0 0 16.01. Use the data to calculate A.2. Use the data to calculate B3. Use the data to calculate C4. Use the data to calculate D5. Use the data to calculate E6. Use the data to calculate l7. Use the data to calculate m8. Use the data to calculate n
Physics
1 answer:
Fynjy0 [20]1 year ago
4 0

Answer:

Given the potential, V = Ax^l+By^m+Cz^n+D

The components of the electric field are:

E_x = \frac{-dV}{dx} = -Alx^l^-^1

E_y = \frac{-dV}{dy} = - Bmy^m^-^1

E_z = \frac{-dV}{dz} = - nCzn^n^-^1

Let's calculate the potential difference for all given points.

V(0, 0, 0) = 10V => Ax^l+By^m+Cz^n+D = 10

=> D = 10

V(1, 0, 0) = 4V => A + 10 = 4

Solving for A, we have:

A = 4 - 10

A = -6

V(0, 1, 0) = 6V => B + 10 = 6

Solving for B, we have:

B = 6 - 10

B = -4

V(0, 0, 1) = 8V => C + 10 = 4

Solving for C, we have:

C = 8 - 10

C = -2

For all given points, let's calculate the magnitude of electric field as follow:

E_x (1, 0, 0) = 16 => - Alx^l^-^1 = 16

Al = -16

Solving for l, we have:

l = \frac{-16}{A}

From above, A = -6

l = \frac{-16}{-6}

l = \frac{8}{3}

E_y (0, 1, 0) = 16=> Bmy^m^-^1 = 16

Bm = -16

Solving for m, we have:

m = \frac{-16}{A}

From above, B = -4

m = \frac{-16}{-4}

m = 4

E_y (0, 0, 1) = 16=> nCz^n^-^1 = 16

nC = - 16

Solving for n, we have:

n = \frac{-16}{C}

From above, C = -2

n = \frac{-16}{-2}

n = 8

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a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio
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a) factor b=\sqrt{2}

b) factor b=\frac{1}{2}

c) factor b=1

d) factor b=1

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Time period of oscillating spring-mass system is given as:

T=\frac{1}{f}

T={2\pi} \sqrt{\frac{m}{k} }

where:

f= frequency of oscillation

m= mass of the object attached to the spring

k= stiffness constant of the spring

a) <u>On doubling the mass:</u>

  • New mass, m'=2m

<u>Then the new time period:</u>

T'=2\pi\sqrt{\frac{m'}{k} }

T'=2\pi\sqrt{\frac{2m}{k} }

T'=\sqrt{2}\times  2\pi\sqrt{\frac{m}{k} } }

T'=\sqrt{2} \times T

where the factor b=\sqrt{2} as asked in the question.

b) On quadrupling the stiffness constant while other factors are constant:

New stiffness constant, k'=4k

<u>Then the new time period:</u>

T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times  2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T

where the factor  b=\frac{1}{2}  as asked in the question.

c) On quadrupling the stiffness constant as well as mass:

New stiffness constant, k'=4k

New mas, m'=4m

<u>Then the new time period:</u>

T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times  2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T

where factor b=1 as asked in the question.

d) On quadrupling the amplitude there will be no effect on the time period because T is independent of amplitude as we can observe in the equation.

so, factor b=1

7 0
1 year ago
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