What are the four main elements of matter

A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

Mars2501 [29]

Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

8 0

4 years ago

If m=120kg and a=15m/s2, what is the force

Scorpion4ik [409]

Answer:

F= 1800N

Explanation:

the equation for force is F= ma

so plug in the numbers: F= (120)(15)

solve this to get F= 1800N

tip: don't forget to add the units when writing your answer :)

6 0

3 years ago

Two models of the same compound are shown. In what way is Model A better than Model B?

astra-53 [7]

The correct answer is Model A shows the three-dimensional shape of the molecule, but Model B does not.

Explanation:

Model A and B show the structure of a molecule. In the case of model A, the structure is represented through the use of three-dimensional shapes, while in model B the structure is represented using the letters of each element and showing how each element is connected to others.

In this context, one feature that makes model A better is that this represents the molecule using a 3D model, which is better to understand how the molecule looks like and what is its structure. Moreover, both models are alike because they show the number of atoms of each element, although model A does not show the types of elements.

7 0

3 years ago

Read 2 more answers

A two-dimensional section of the body is known as an

lesya692 [45]

Median plane ..........

7 0

3 years ago

A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens m

kirill [66]

a) 50.8 mm

b) The whole image (1:1)

c) It seems reasonable

Explanation:

a)

To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)

p = 3.00 m is the distance of the person from the lens

Therefore, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.050}-\frac{1}{3.00}=19.667m^{-1}\\q=\frac{1}{19.667}=0.051 m=50.8 mm$

b)

Here we need to find the height of the image first.

This can be done by using the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where:

y' is the height of the image

y = 1.75 m is the height of the real person

q = 50.8 mm = 0.0508 m is the distance of the image from the lens

p = 3.00 m is the distance of the person from the lens

Solving for y', we find:

$y'=-\frac{qy}{p}=-\frac{(0.0508)(1.75)}{3.00}=-0.0296 m=-29.6mm$

(the negative sign means the image is inverted)

Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.

c)

This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.

3 0

3 years ago