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3 years ago

The angle between reflected ray and the normal line is

Physics

1 answer:

Alja [10]3 years ago

7 0

Answer:

<h2>Angle(Δ) of Reflection</h2>

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Two fully charged cylindrical capacitors are connected to two identical batteries. The capacitors are identical except that the

Leni [432]

Answer:

Part(a): The relative capacitance is $\dfrac{C_{A}}{C_{B}} = 0.33$

Part(b): The relative energy stored is $\dfrac{U_{A}}{U_{B}} = 0.33$

Part(c): The relative charge stored is $\dfrac{Q_{A}}{Q_{B}} = 0.33$

Explanation:

We know the capacitance ( $C$ ) of a capacitor having charge ( $Q$ ) and subjected to a potential difference of ( $V$ ) is given by

$C = \dfrac{Q}{V}$

Also, the energy ( $U$ ) stored by a capacitor can be written as

$U = \dfrac{1}{2}C~V^{2}$

Let us assume that the inner radius of the Capacitor B, as shown in the figure, be $\textbf{r_{i}^{B}}$ $\bf{r_{i}^{B}}$ , the outer radius be $\bf{r_{o}^{B}}$ , the inner radius of Capacitor A be $\bf{r_{i}^{A}}$ and the outer radius be $\bf{r_{o}^{B}}$ .

Given in the problem,

$&& r_{o}^{B} = 2~r_{B}^{i}\\&& r_{o}^{A} = 4~r_{B}^{i}\\&& and~r_{i}^{B} = 4~r_{o}^{B} = 8~r_{B}^{i}$

Now, the capacitance ( $C$ ) of a cylindrical capacitor is given by,

$\bf{C = \dfrac{2~\pi~\epsilon_{0}~L}{ln(\dfrac{r_{o}}{r_{i}})}}$

where $\epsilon_{o}$ is the permittivity of the free space, $L$ is the length of the cylindrical capacitor.

Part(a):

The capacitance of capacitor A,

$C_{A} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{A}}{r_{i}^{A}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{8~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(8)}$

and the capacitance of capacitor B,

$C_{B} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{2~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(2)}$

giving the relative capacitance of each capacitor to be

$\dfrac{C_{A}}{C_{B}} = \dfrac{ln(2)}{ln(8)} = \dfrac{ln(2)}{3~\ln(2)} = \dfrac{1}{3} = 0.33$

Part(b):

Energy stored by capacitor A,

$U_{A} = \dfrac{1}{2}~C_{A}~V^{2}$

Energy stored by capacitor B,

$U_{B} = \dfrac{1}{2}~C_{B}~V^{2}$

giving the relative energy stored by each capacitor to be

$\dfrac{U_{A}}{U_{B}} = \dfrac{C_{A}}{C_{B}} = 0.33$

Part(c):

The charge stored by capacitor A,

$Q_{A} = C_{A}~V$

The charge stored by capacitor B,

$Q_{B} = C_{B}~V$

giving the relative charge stored by each capacitor to be

$\dfrac{Q_{A}}{Q_{B}} = \dfrac{C_{A}}{C_{B}} = 0.33$

8 0

3 years ago

if a runners power is 400 watts as she runs, how much chemical energy does she convert into other forms in 10 minutes

AnnZ [28]

Answer:

Energy converted = $240000\,Joules = 240\, kJoules$

Explanation:

Recall that Power is the rate at which energy is transferred therefore defined by the mathematical formula: $Power\,=\,\frac{Energy\,transferred}{time}$

Since the information on the power of the runner is given, as well as the time the energy conversion takes place, we can then use this equation to find how much energy is been converted. Notice that we just need to change the given time *10 minutes) into the appropriate units (seconds)to get the answer in SI units of energy (Joules). The conversion of 10 minutes into seconds is done by multiplying : 10 minutes * 60 seconds/minute = 600 seconds.

We use this then to find the energy converted by the runner:

$Power\,=\,\frac{Energy\,transferred}{time}\\400 \,W = \frac{E}{600\,sec} \\400 \,W * 600\,sec=E\\E=240000\,Joules = 240\, kJoules$

3 0

3 years ago

A wire, of length L = 4.1 mm, on a circuit board carries a current of I = 1.96 μA in the j direction. A nearby circuit element g

tamaranim1 [39]

Answer:

The magnitude of the magnetic field B is 5.921 T.

Explanation:

Given that,

Length = 4.1 mm

$B_{x}=4.9\ G$

$B_{y}=2.3\ G$

$B_{z}=2.4\ G$

Current $I = 1.96\ mu A$

We need to calculate the magnetic field

Using formula of magnetic field

$B=\sqrt{B_{x}^2+B_{y}^2+B_{z}^2}$

Put the value into the formula

$B=\sqrt{(4.9)^2+(2.3)^2+(2.4)^2}$

$B=5.921\ T$

Hence, The magnitude of the magnetic field B is 5.921 T.

6 0

3 years ago

What is the acceleration of a softball if it hits the ground with a force 0.50 kg and hits the catchers glove with a force of 25

Rom4ik [11]

Acceleration of the ball is $50 m/s^2$

Explanation:

The acceleration of the ball can be found by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:

$F=ma$

where

F is the net force

m is the mass

a is the acceleration

For the ball in this problem, we have

m = 0.50 kg (mass)

F = 25 N (force)

thereofre, the acceleration of the ball is

$a=\frac{F}{m}=\frac{25}{0.50}=50 m/s^2$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0

3 years ago

What is the distance from one crest to the next crest?

ella [17]

<span>The distance between wave crests is called wavelength. It is a characteristic shared by waves of all kinds, including ocean waves and sound waves. Wavelength is measured from the highest point, or summit, of one wave's crest to the summit of the next wave's <span>crest</span></span>

<span><span>hope this helps</span></span>

3 0

3 years ago