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2 years ago

A large insulating sheet has a surface charge density σ. what is the electric field strength near the insulating sheet?

1 answer:

3 0

Free insulative sheets and insulative sheets backed by a grounded conductor are the only cases in which it is possible to extract reliable information from a noncontacting measurement of the charged state of an insulator. In both cases, the electric field from the charge is the deciding factor.<span>

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Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

2 years ago

A certain resistor dissipates 0.5 W when connected to a 3 V potential difference. When connected to a 1 V potential difference,

Stels [109]

Answer:

Explanation:

$Power = IV$

From ohms law we know that

$V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}$

Given data

P1 = 0.5 Watt

P2 = ?

V1= 3 Volts

V2= 1 Volt

Thus we can solve for the power dissipated as follows

$P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}$

$\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W$

<em>The resistor will dissipate 0.056 Watt</em>

7 0

2 years ago

What is the de Broglie wavelength of a bullet of mass 0.075 kg traveling at 350 m/s? Planck's Constant is 6.63 × 10−34 J⋅s.

jasenka [17]

Answer:

De Broglie wavelength of the bullet is given as

$\lambda = 2.5 \times 10^{-35} m$

Explanation:

As per De Broglie hypothesis we know that the wavelength of De Broglie waves is the ratio of Plank's constant and momentum of the particle

here we know that the bullet mass is

m = 0.075 kg

speed of the bullet is given as

v = 350 m/s

Now we have

$P = (0.075)(350)$

$P = 26.25 kg m/s$

Now we know that

$\lambda = \frac{h}{P}$

$\lambda = \frac{6.63 \times 10^{-34}}{26.25}$

$\lambda = 2.5 \times 10^{-35} m$

3 0

2 years ago

1) An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and

Lady_Fox [76]

Answer: Capacitor

Explanation:

Given the following ;

At time, t = T/4 ; I = 0

At time, t = T/2 ; I = - Imax

Where T = period

We can confirm what the unknown element is by using the relation, such that the parameters satisfy the stated condition ;

Recall;

I = Iosin(wt + π/2) [for CAPACITOR]

w = 2π/T

At t = T/4

I = Iosin(2π/T(T/4) + π/2)

I = Iosin(π/2 + π/2)

I = Iosin(0)

I = 0

At t = T/2

I = Iosin(2π/T(T/2) + π/2)

I = Iosin(π + π/2)

I = Iosin(0)

I = Iosin(3π/2)

I = Iosin(540/2)

I = Iosin(270)

I = -Io

Note : Io = Imax

Both conditions are met, Hence, the unknown element is a CAPACITOR.

6 0

2 years ago

An athlete prepares to throw a 2.0-kilogram discus. His arm is 0.75 meters long. He spins around several times with the discus a

wlad13 [49]

Centripetal Force (Fcp) = ?

His arm length = Radius (R) = 0.75 m

Discus velocity = Linear Velocity (V) = 5 m/s

Discus mass (m) = 2 kg

Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).

$fcp = m \times acp \\ fcp = m \times {v}^{2} \div r \\ fcp = 2 \times {5}^{2} \div 0.75 \\ fcp = 2 \times 25 \div 0.75 \\ fcp = 50 \div 0.75$
$fcp = 66.666... = 66 \: newtons$

Answer: Last option, 66 N.

7 0

2 years ago

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