All categories

3 years ago

PLEASEEEEE HELLLLLPPPPPPP

Physics

2 answers:

torisob [31]3 years ago

7 0

Answer:

I'm pretty sure it's b

Explanation:

masya89 [10]3 years ago

6 0

Answer:

A or B(the answers)

Explanation:

they seem like the most right

You might be interested in

Need help with this question.

mixas84 [53]

I would say the correct answer would be light travels faster in medium 3 then medium 2.

8 0

3 years ago

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

A small 16 kilogram canoe is floating downriver at a speed of 4 m/s. What is the canoe's kinetic energy?

seraphim [82]

Kinetic Energy,K.E=1/2MV²

mass,m=16kg

velocity,v=4m/s

K.E=1/2×16×4²

=128kgm²/s²

=128 Joules

5 0

3 years ago

Read 2 more answers

A car traveling at 60km/h undergoes uniform acceleration at a rate of 2/ms^2 until is velocity reached 120km/h determine the dis

jenyasd209 [6]

Explanation:

Given that,

Initial speed of a car, u = 60 km/h = 16.67 m/s

Acceleration, a = 2m/s²

Final speed, v = 120 km/h = 33.33 m/s

We need to find the distance traveled and the time taken to make the distance.

acceleration = rate of change of velocity

$a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{33.33 -16.67 }{2}\\\\t=8.33\ s$

let the distance be d.

$d=\dfrac{v^2-u^2}{2a}\\\\d=\frac{33.33^{2}-16.67^{2}}{2(2)}\\\\d=208.25\ m$

Hence, the distance traveled and the time taken to make the distance is 208.25 m and 8.33 seconds respectively.

3 0

3 years ago

Who clarified the photoelectric effect?

sergiy2304 [10]

When visible light, X rays, gamma rays, or other forms of electromagnetic radiation are shined on certain kinds of matter, electrons are ejected. That phenomenon is known as the photoelectric effect. The photoelectric effect was discovered by German physicist Heinrich Hertz (1857–1894) in 1887. You can imagine the effect as follows: Suppose that a metal plate is attached by two wires to a galvanometer. (A galvanometer is an instrument for measuring the flow of electric current.) If light of the correct color is shined on the metal plate, the galvanometer may register a current. That reading indicates that electrons have been ejected from the metal plate. Those electrons then flow through the external wires and the galvanometer. HOPE THIS HELPED

7 0

4 years ago