When a ray of light is incident normally on an air glass interface what is its angle of refraction

If the distance between two objects decreased, what would happen to the force of gravity between them?

evablogger [386]

Answer: A- It would increase

Explanation:

According to the law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the attraction force exerted between both objects

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both objects

$r$ is the distance between both objects

As we can see, the gravity force is directly proportional to the mass of the bodies or objects and inversely proportional to the square of the distance that separates them.

In other words:

<h2>If we decrease the distance between both objects, the gravitational force between them will increase. </h2>

6 0

3 years ago

Read 2 more answers

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

3 0

3 years ago

A handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with a dia

Bingel [31]

Answer

given,

diameter of the pipe is = (14 ft)4.27 m

minimum speed of the skater must have at very top = ?

At the topmost point of the pipe the normal force will be equal to zero.

F = mg

centripetal force acting on the skateboard

$F = \dfrac{mv^2}{r}$

equating both the force equation

$mg = \dfrac{mv^2}{r}$

$v = \sqrt{gr}$

r = d/2 = 14/ 2 = 7 ft

or

r = 4.27/2 = 2.135 m

g = 32 ft/s² or g = 9.8 m/s²

$v = \sqrt{32 \times 7}$

v = 14.96 ft/s

or

$v = \sqrt{9.8 \times 2.135}$

v = 4.57 m/s

5 0

3 years ago

1mm is equal to .1 what

bulgar [2K]

Answer:

When you have to do an English-Metric (SI) length conversion, and you already know the English units of length (miles, yards, feet, inches, etc.), all you need to remember is one simple relationship, and you can readily convert any length in the SI system, to the equivalent length in the other.

1 foot (ft) = 0.3048 meters (m)

BIn this case you need your answer in inches. You (hopefully) know there are 12 inches in a foot, so you just do the following:

1 inch (in) = 1/12 ft = 0.3048/12 m = 0.0254 m

8 0

2 years ago

Which is a characteristic of a physical change? (1 point)

krek1111 [17]

The answer would be a.

a chemical change is a change to the chemical makeup of a substance so if the bonds are unchanged it would be a characteristic of a physical change

6 0

3 years ago

When a ray of light is incident normally on an air glass interface what is its angle of refraction​

When a ray of light is incident normally on an air glass interface what is its angle of refraction