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soldier1979 [14.2K]
2 years ago
8

I don't understand this QUESTION​

Physics
1 answer:
alex41 [277]2 years ago
6 0

This electric force calculator will enable you to determine the repulsive or attractive force between two static charged particles. Continue reading to get a better understanding of Coulomb's law, the conditions of its validity, and the physical interpretation of the obtained result.

How to use Coulomb's law

Coulomb's law, otherwise known as Coulomb's inverse-square law, describes the electrostatic force acting between two charges. The force acts along the shortest line that joins the charges. It is repulsive if both charges have the same sign and attractive if they have opposite signs.

Coulomb's law is formulated as follows:

F = keq₁q₂/r²

where:

F is the electrostatic force between charges (in Newtons),

q₁ is the magnitude of the first charge (in Coulombs),

q₂ is the magnitude of the second charge (in Coulombs),

r is the shortest distance between the charges (in m),

ke is the Coulomb's constant. It is equal to 8.98755 × 10⁹ N·m²/C². This value is already embedded in the calculator - you don't have to remember it :)

Simply input any three values

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Answer:

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Explanation:

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force = mass × acceleration

From the question we have

force = 15 × 2

We have the final answer as

<h3>30 N</h3>

Hope this helps you

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Find the mass and center of mass of the solid E with the given density function ρ. E lies under the plane z = 3 + x + y and abov
makvit [3.9K]

Answer:

The mass of the solid is 16 units.

The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

Work:

Density function: ρ(x, y, z) = 8

x-bounds: [0, 1], y-bounds: [0, x], z-bounds: [0, x+y+3]

The mass M of the solid is given by:

M = ∫∫∫ρ(dV) = ∫∫∫ρ(dx)(dy)(dz) = ∫∫∫8(dx)(dy)(dz)

First integrate with respect to z:

∫∫8z(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8x+8y+24](dx)(dy)

Then integrate with respect to y:

∫[8xy+4y²+24y]dx, evaluate y from 0 to x

= ∫[8x²+4x²+24x]dx

Finally integrate with respect to x:

[8x³/3+4x³/3+12x²], evaluate x from 0 to 1

= 8/3+4/3+12

= 16

The mass of the solid is 16 units.

Now we have to find the center of mass of the solid which requires calculating the center of mass in the x, y, and z dimensions.

The z-coordinate of the center of mass Z is given by:

Z = (1/M)∫∫∫ρz(dV) = (1/16)∫∫∫8z(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫4z²(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[4(x+y+3)²](dx)(dy)

= ∫∫[4x²+24x+8xy+4y²+24y+36](dx)(dy)

Then integrate with respect to y:

∫[4x²y+24xy+4xy²+4y³/3+12y²+36y]dx, evaluate y from 0 to x

= ∫[28x³/3+36x²+36x]dx

Finally integrate with respect to x:

[7x⁴/3+12x³+18x²], evaluate x from 0 to 1

= 7/3+12+18

Z = (7/3+12+18)/16 = <u>2.021</u>

The y-coordinate of the center of mass Y is given by:

Y = (1/M)∫∫∫ρy(dV) = (1/16)∫∫∫8y(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫8yz(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8xy+8y²+24y](dx)(dy)

Then integrate with respect to y:

∫[4xy²+8y³/3+12y²]dx, evaluate y from 0 to x

= ∫[20x³/3+12x²]dx

Finally integrate with respect to x:

[5x⁴/3+4x³], evaluate x from 0 to 1

= 5/3+4

Y = (5/3+4)/16 = <u>0.3542</u>

<u />

The x-coordinate of the center of mass X is given by:

X = (1/M)∫∫∫ρx(dV) = (1/16)∫∫∫8x(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫8xz(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8x²+8xy+24x](dx)(dy)

Then integrate with respect to y:

∫[8x²y+4xy²+24xy]dx, evaluate y from 0 to x

= ∫[12x³+24x²]dx

Finally integrate with respect to x:

[3x⁴+8x³], evaluate x from 0 to 1

= 3+8 = 11

X = 11/16 = <u>0.6875</u>

<u />

The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

4 0
3 years ago
A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0
3 years ago
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