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A glass plate 2.95 mmmm thick, with an index of refraction of 1.60, is placed between a point source of light with wavelength 600 nmnm (in vacuum) and a screen. The distance from source to screen is 1.25 cm. How many wavelengths are there between the source and the screen?

Answers

Answer 1

Answer:

Answer:

$N_T=2086285.67$

Explanation:

Given;

Thickness of the glass plate, $x=2.95* 10^(-3)\ m$

refractive index of the glass plate, $n=1.6$

wavelength of light source in vacuum, $\lambda=600* 10^(-9)\ m$

distance between the source and the screen, $d=1.25\ m$

Distance travelled by the light from source to screen in vacuum:

$d_v=d-x$

$d_v=1.25-0.00295$

$d_v=1.24705\ m$

So the no. of wavelengths in the vacuum:

$N=(d_v)/(\lambda)$

$N=(1.24705)/(6* 10^(-7))$

$N\approx2.0784* 10^(6)$ .......................(1)

Now we find the wavelength of the light wave in the glass:

$n=(\lambda)/(\lambda')$

where:

$\lambda'=$ wavelength of light in the medium of glass.

$1.6=(600* 10^(-9))/(\lambda')$

$\lambda'=375* 10^(-9)\ m=375\ nm$

Now the no. of wavelengths in the glass:

$N'=(x)/(\lambda')$

$N'=(2.95* 10^(-3))/(375* 10^(-9))$

$N'=7.8667* 10^(3)$ ............................(2)

From (1) & (2):

total no. of wavelengths are there between the source and the screen:

$N_T=N+N'$

$N_T=2086285.67$

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A 1.5v battery stores 4.5KJ of energy. How long can it light a flashlight bulb that draws 0.60A

Answers

Answer:

The 1.5V battery can power the flashlight bulb drawing 0.60A for 83.33 minutes before it is depleted.

Explanation:

To determine how long a 1.5V battery can power a flashlight bulb drawing 0.60A, you can use the formula for calculating the energy (in joules) consumed by an electrical device over time:

Energy (Joules) = Power (Watts) × Time (Seconds)

In this case, the power (P) is given by the product of the voltage (V) and current (I):

Power (Watts) = Voltage (Volts) × Current (Amperes)

So, first, calculate the power consumption of the flashlight bulb:

Power (Watts) = 1.5V × 0.60A = 0.90 Watts

Now, you want to find out how long the battery can power the bulb, so rearrange the energy formula to solve for time:

Time (Seconds) = Energy (Joules) / Power (Watts)

Given that the battery stores 4.5 kJ (kilojoules), which is equivalent to 4,500 joules, and the power consumption is 0.90 watts:

Time (Seconds) = 4,500 J / 0.90 W = 5,000 seconds

Now, to express the time in more practical units, convert seconds to minutes:

Time (Minutes) = 5,000 seconds / 60 seconds/minute ≈ 83.33 minutes

So, the 1.5V battery can power the flashlight bulb drawing 0.60A for approximately 83.33 minutes before it is depleted.

A charge +Q is located at the origin and a second charge, +4Q, is at distance (d) on the x-axis.a. Where should a third charge, Q , be placed, so that all three charges will be in equilibrium? Express your answer in terms of d.
b. What should be its sign, so that all three charges will be in equilibrium?
c. What should be its magnitude, so that all three charges will be in equilibrium? Express your answer in terms of Q.

Answers

Answer:

a) x = ⅔ d, b) the charge must be negative, c) Q

Explanation:

a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing

∑ F = 0

-F₁₂ + F₂₃ = 0

F₁₂ = F₂₃

let's replace the values

k Q Q / r₁₂² = k Q 4Q / r₂₃²

Q² / r₁₂² = 4 Q² / r₂₃²

suppose charge 3 is placed at point x

r₁₂ = x

r₂₃ = d-x

we substitute

1 / x² = 4 / (d-x) 2

1 / x = 2 / (d-x)

x = 2 (x-d)

x = 2x -2d

3x = 2d

x = ⅔ d

b) The sign of the charge must be negative, to have an attractive charge on the two initial charges

c) Q

Final answer:

The third charge, -Q, can either be placed at a distance of d/4 to the left of origin or at a distance d/2 to the right of +4Q charge to create an equilibrium among the three charges. Its sign should be negative and its magnitude should be equal to Q.

Explanation:

To create equilibrium among all three charges, our third charge can be placed in two potential positions on the x-axis. One position is a distance of d/4 to the left of the origin (negative x-axis), and the other is a distance of d/2 to the right of the +4Q charge (positive x-axis).

The third charge should be a negative charge, denoted as -Q, in order to balance the positive charges and create equilibrium.

The magnitude of this third charge should be equal to Q, the original charge. The reason is that the forces need to be balanced to create equilibrium and the force is proportional to the charge. Therefore, if -Q is placed at d/4 to the left of the origin or if -Q is placed at d/2 to the right of the +4Q charge, the system will be in equilibrium.

Learn more about Electric Charges and Fields here:

brainly.com/question/18369630

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A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. the bicycle stops in 5.0 s. how far did the bicycle travel during the 5.0 seconds of braking?

Answers

Distance traveled by the bicycle during the 5 seconds of braking is 22m

Explanation:

initial angular velocity= 2 rev/s

final angular velocity= 0 rev/s

Angular displacement Ф= $((wi+wf)/(2) )$ t

Ф= $((0+2)/(2) )5=5$ rev

so the distance travelled= 5(2πr)

distance=5(2π*0.7)

distance=22m

The bicycle traveled about 22 m during the 5.0 seconds of braking

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Further explanation

Centripetal Acceleration can be formulated as follows:

$\large {\boxed {a = \frac{ v^2 } { R } }$

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

$\texttt{ }$

Centripetal Force can be formulated as follows:

$\large {\boxed {F = m \frac{ v^2 } { R } }$

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

$\texttt{ }$

Given:

radius of wheel = R = 0.70 m

initial angular speed = ω = 2.0 rev/s = 4π rad/s

final angular speed = ωo = 0 rad/s

time taken = t = 5.0 s

Asked:

distance covered = d = ?

Solution:

$d = \theta R$

$d = (\omega + \omega_o)(1)/(2)t R$

$d = ( 4 \pi + 0 ) (1)/(2)(5.0)( 0.70 )$

$d = 4\pi (1.75)$

$d = 7\pi \texttt{ m}$

$d \approx 22 \texttt{ m}$

$\texttt{ }$

Learn more

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g glass ball charged to 4.7 nC is shot straight up at 4.8 m/s from the floor level. How high does the ball go if the ceiling voltage is +3.0x10^6V?

Answers

The height at which the ball goes for the given parameters is; 0.827 m

What is the height of the ball?

We are given;

distance between the metal plates; d = 3.1 m

mass of glass; m = 1.1g = 0.0011 kg

charge on the glass; q = 4.7 nC = 4.7 × 10⁻⁹ C

speed of the glass ball; v = 4.8 m/s

voltage of the ceiling; V = +3.0 × 10⁶ V

The repulsive force experienced by the ball is gotten from the formula;

F = qV/d

|F| = (4.7 × 10⁻⁹ × 3 × 10⁶)/3.1

|F| = 4.548 × 10⁻³ N

F = -4.548 × 10⁻³ N (negative because it is repulsive force)

The net horizontal force experienced by the ball is;

F_net = F - mg

F_net = (-4.548 × 10⁻³) - (0.0011 × 9.8)

F_net = -15.328 × 10⁻³ N

To get the height of the ball, we will use the formula;

F_net * h = ¹/₂mv²

h = (¹/₂ * 0.0011 * 4.8²)/(15.328 × 10⁻³)

We took the absolute value of F_net, hence it is not negative

h = 0.827 m

Read more about height of ball at; brainly.com/question/12446886

Answer:

The ball traveled 0.827 m

Explanation:

Given;

distance between the metal plates of the room, d = 3.1 m

mass of the glass, m = 1.1g

charge on the glass, q = 4.7 nC

speed of the glass ball, v = 4.8 m/s

voltage of the ceiling, V = +3.0 x 10⁶ V

The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;

F = qV/d

|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)

|F| = 4.548 x 10⁻³ N

F = - 4.548 x 10⁻³ N

The net horizontal force experienced by this ball is;

$F_(net) = F_c - mg\n\nF_(net) = -4.548 *10^(-3) - (1.1*10^(-3) * 9.8)\n\nF_(net) = -15.328*10^(-3) \ N$

The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.

$W = F_(net) *h\n\nW = 15.328 *10^(-3) * h$

W = K.E

$15.328*10^(-3) *h = (1)/(2)mv^2\n\n 15.328*10^(-3) *h = (1)/(2)(1.1*10^(-3))(4.8)^2\n\n 15.328*10^(-3) *h =0.0127\n\nh = (0.0127)/(15.328*10^(-3))\n\n h = 0.827 \ m$

Therefore, the ball traveled 0.827 m

A 2-C charge experiences a force of 40 N when put at a certain location inspace. The electric field at that location is a. 2 N/C.b. 20 N/C. c. 30 N/C. d.
40 N/C. e. 60 N/C.