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Lubov Fominskaja [6]

1 year ago

11

A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l

ong line of charge with charge per unit length -6.00 μC/m is parallel to the x-axis at y = 11.0 cm and its midpoint is also at x = 0. At what point on the y -axis is the resultant electric field of the two lines of charge equal to zero?

1 answer:

artcher [175]1 year ago

9 0

Answer:

at y=6.29 cm the charge of the two distribution will be equal.

Explanation:

Given:

linear charge density on the x-axis, $\lambda_1=8\times 10^{-6}\ C$

linear charge density of the other charge distribution, $\lambda_2=-6\times 10^{-6}\ C$

Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.

Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.

<u>we know, the electric field due to linear charge is given as:</u>

$E=\frac{\lambda}{2\pi.r.\epsilon_0}$

where:

$\lambda=$ linear charge density

r = radial distance from the center of wire

$\epsilon_0=$ permittivity of free space

Therefore,

$E_1=E_2$

$\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}$

$\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}$

$\frac{8\times 10^{-6}}{x} =\frac{6\times 10^{-6}}{0.11-x}$

$x=0.0629\ m$

∴at y=6.29 cm the charge of the two distribution will be equal.

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CaHeK987 [17]

Answer:

a.) 10Hz

b.) 0.1 s

c.) 187.4 m/s

d.) -412.6 m/s^2

Explanation:

Given that an object is moving back and forth on the x-axis according to the equation x(t) = 3sin(20πt), t> 0, where x(t) is measured in cm and t in seconds. Give decimal answers below.

(a) How many complete back-and-forth motions (from the origin to the right, back to the origin, to the left and finally back to the origin) does the object make in one second?

from the equation given, the angular speed w = 20π

but w = 2πf

where f = frequency.

substitute w for 20π

20π = 2πf

f = 20π/2π

f = 10 Hz

(b) What is t the first time that the object is at its farthest right?

since F = 1/T

T = 1 / f

T = 1/10

T = 0.1 s

Therefore, the t of first time that the object is at its farthest right is 0.1 s

(c) At the time found in part (b), what is the object's velocity?

The velocity can be found by differentiating the equation;

x(t) = 3sin(20πt)

dx/dt = 60πcos(20πt)

where dx/dt = velocity V

V = 60πcos(20π * 0.1)

V = 187.4 m/s

(d) At the time found in part (b), what is the object's acceleration?

to get the acceleration, differentiate equation V = 60πcos(20πt)

dv/dt = -1200πSin(20πt)

dv/dt = acceleration a

a = -1200πSin(20πt)

substitute t into the equation

a = -1200πSin(20π * 0.1)

a = - 412.6 m/s^2

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Kathmandu lies at high altitude than biratnagar from sea level.Where does an object has more weight between two places?Give reas

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Answer:

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As the altitude get higher, the gravitational pull of the earth on the object increases, therefore, the mass is higher up above.

8 0

2 years ago

A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. Assume the test specimen is 12.8-m

ser-zykov [4K]

To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

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$S_{el} = 414Mpa$

Yield Strain of the Specimen

$\epsilon_{el} = 0.002$

Diameter of the test-specimen

$d_0 = 12.8mm$

Gage length of the Specimen

$L_0 = 50mm$

Modulus of elasticity

$E = \frac{S_{el}}{\epsilon_{el}}$

$E = \frac{414Mpa}{0.002}$

$E = 207Gpa$

Strain energy per unit volume at the elastic limit is

$U'_{el} = \frac{1}{2} S_{el} \cdot \epsilon_{el}$

$U'_{el} = \frac{1}{2} (414)(0.002)$

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Considering that the net strain energy of the sample is

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$U_{el} = U_{el}'(\frac{\pi d_0^2}{4})(L_0)$

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Therefore the net strain energy of the sample is $2.663N\codt m$

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1 year ago

Which of the following statements cannot be supported by Kepler's laws of planetary motion?

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Answer:

The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.

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Kepler's three laws are:

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2) A line connecting the Sun with each planet sweeps out equal areas in equal time intervals

3) The cube of the semi-major axis of the orbit of one planet is proportional to the square of its orbital period

There 3 laws help explaining the following statements:

- <em>A planet's distance from the sun will not be the same in six months. --> </em>using the 1st law. In fact, since the orbit is an ellipse (and not a circle), and the Sun is at one of the focii, the distance of the planet from the Sun keeps changing during the year.

-<em> A planet's speed as it moves around the sun will not be the same in six months. -</em>-> using the 2nd law. In fact, since the line connecting the Sun to the planet must cover equal areas in the same time interval, it follows that the speed of the planet cannot be constant during the year (it will be faster when closer to the sun and slower when far from the sun).

- <em>The average distance of Saturn can be calculated using the average distance of Neptune and the orbital period of both planets. </em>--> using the 3rd law. In fact, the ratio $\frac{a^3}{T^2}$ (where a is the semi-major axis of the orbit and T the orbital period) is constant and it is the same for every planet orbiting the sun, so by knowing the data of Neptune and the orbital period of Saturn, it is possible to calculate Saturn's average distance.

Instead, the following statement:

<em>The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.</em>

Is not supported by any Kepler's law.

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2 years ago

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i

Elis [28]

Four electrons are placed at the corner of a square

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now potential is given as

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Now work done in this process is given as

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2 years ago