What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

Answers

Answer 1

Answer:

[tex]I=2.71\times 10^{-5}\ A[/tex]

Explanation:

A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.  

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

Let given is,

The diameter of a parallel plate capacitor is 6 cm or 0.06 m

Separation between plates, d = 0.046 mm

The potential difference across the capacitor is increasing at 500,000 V/s

We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :

[tex]C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}[/tex], r is radius

Let I is the displacement current. It is given by :

[tex]I=C\dfrac{dV}{dt}[/tex]

Here, [tex]\dfrac{dV}{dt}[/tex] is rate of increasing potential difference

So

[tex]I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A[/tex]

So, the value of displacement current is [tex]2.71\times 10^{-5}\ A[/tex].


Related Questions

In a physics laboratory experiment, a coil with 250 turns enclosing an area of 14 cm2 is rotated in a time interval of 0.030 s from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is 5.0×10^−5 T.Required:a. What is the total magnetic flux through the coil before it is rotated? After it is rotated? b. What is the average emf induced in the coil?

Answers

Explanation:

Consider a loop of wire, which has an area of [tex]A=14 \mathrm{cm}^{2}[/tex] and [tex]N=250[/tex] turns, it is initially placed perpendicularly in the earth magnetic field. Then it is rotated from this position to a position where its plane is parallel to the field as shown in the following figure in [tex]\Delta t=0.030[/tex] s. Given that the earth's magnetic field at the position of the loop is [tex]B=5.0 \times 10^{-5} \mathrm{T}[/tex], the flux through the loop before it is rotated is,

[tex]\Phi_{B, i} &=B A \cos \left(\phi_{i}\right)=B A \cos \left(0^{\circ}\right[/tex]

[tex]=\left(5.0 \times 10^{-5} \mathrm{T}\right)\left(14 \times 10^{-4} \mathrm{m}^{2}\right)(1)[/tex]

[tex]=7.0 \times 10^{-8} \mathrm{Wb}[/tex]

[tex]\quad\left[\Phi_{B, i}=7.0 \times 10^{-8} \mathrm{Wb}\right[/tex]

after it is rotated, the angle between the area and the magnetic field is [tex]\phi=90^{\circ}[/tex] thus,

[tex]\Phi_{B, f}=B A \cos \left(\phi_{f}\right)=B A \cos \left(90^{\circ}\right)=0[/tex]

[tex]\qquad \Phi_{B, f}=0[/tex]

(b) The average magnitude of the emf induced in the coil equals the change in the flux divided by the time of this change, and multiplied by the number of turns, that is,

[tex]{\left|\mathcal{E}_{\mathrm{av}}\right|=N\left|\frac{\Phi_{B, f}-\Phi_{B, i}}{\Delta t}\right|}{=} & \frac{1.40 \times 10^{-5} \mathrm{Wb}}{0.030 \mathrm{s}}[/tex]

[tex]& 3.6 \times 10^{-4} \mathrm{V}=0.36 \mathrm{mV}[/tex]

[tex]\mathbb{E}=0.36 \mathrm{mV}[/tex]

(a) The initial and final flux through the coil is 1.75 × 10⁻⁵ Wb and 0 Wb

(b) The induced EMF in the coil is 0.583 mV

Flux and induced EMF:

Given that the coil has N = 250 turns

and an area of A = 14cm² = 1.4×10⁻³m².

It is rotated for a time period of Δt = 0.030s such that it is parallel with the earth's magnetic field that is B = 5×10⁻⁵T

(a) The flux passing through the coil is given by:

Ф = NBAcosθ

where θ is the angle between area vector and the magnetic field

The area vector is perpendicular to the plane of the coil.

So, initially, θ = 0°, as area vector and earth's magnetic field both are perpendicular to the plane of the coil

So the initial flux is:

Φ = NABcos0° = NAB

Ф = 250×1.4×10⁻³×5×10⁻⁵ Wb

Ф = 1.75 × 10⁻⁵ Wb

Finally, θ = 90°, and since cos90°, the final flux through the coil is 0

(b) The EMF induced is given by:

E = -ΔФ/Δt

E = -(0 - 1.75 × 10⁻⁵)/0.030

E = 0.583 × 10⁻³ V

E = 0.583 mV

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Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by a clock on board the spacecraft, and then return home at the same speed. Upon return, the people on earth will have advanced exactly 120 years into the future. According to special relativity, how fast must you travel

Answers

Answer:

I must travel with a speed of 2.97 x 10^8 m/s

Explanation:

Sine the spacecraft flies at the same speed in the to and fro distance of the journey, then the time taken will be 6 months plus 6 months

Time that elapses on the spacecraft = 1 year

On earth the people have advanced 120 yrs

According to relativity, the time contraction on the spacecraft is gotten from

[tex]t[/tex] = [tex]t_{0} /\sqrt{1 - \beta ^{2} }[/tex]

where

[tex]t[/tex] is the time that elapses on the spacecraft = 120 years

[tex]t_{0}[/tex] = time here on Earth = 1 year

[tex]\beta[/tex] is the ratio v/c

where

v is the speed of the spacecraft = ?

c is the speed of light = 3 x 10^8 m/s

substituting values, we have

120 = 1/[tex]\sqrt{1 - \beta ^{2} }[/tex]

squaring both sides of the equation, we have

14400 = 1/[tex](1 - \beta ^{2} )[/tex]

14400 - 14400[tex]\beta ^{2}[/tex] = 1

14400 - 1 = 14400[tex]\beta ^{2}[/tex]

14399 = 14400[tex]\beta ^{2}[/tex]

[tex]\beta ^{2}[/tex] =  14399/14400 = 0.99

[tex]\beta = \sqrt{0.99}[/tex] = 0.99

substitute β = v/c

v/c = 0.99

but c = 3 x 10^8 m/s

v = 0.99c = 0.99 x 3 x 10^8 = 2.97 x 10^8 m/s

A long, straight wire carries a 13.0-A current. An electron is fired parallel to this wire with a velocity of 250 km/s in the same direction as the current, 2.00 cm from the wire.
(a) Find the magnitude and direction of the electron’s initial acceleration.
(b) What should be the magnitude and direction of a uniform electric field that will allow the electron to continue to travel parallel to the wire?
(c) Is it necessary to include the effects of gravity? Justify your answer.

Answers

Answer:

A.5.7 x 10^-12m/s²

B. 32.5V/m

C. Pls. See attached file for explanations

Explanation:

14 A particle of mass m strikes a vertical rigid wall perpendicularly from the left with velocity v.
rigid wall
If the collision is perfectly elastic, the total change in momentum of the particle that occurs as a
result of the collision is
A. 2mv to the right.
B. 2my to the left.
C. my to the right.
D. my to the left.​

Answers

Answer:

C. mv to the right

Explanation:

momentum of thr particle=m1v1

momentum of the wall=m2v2

m1v1+ m2v2 =m1u1+ m2u2 since the wall doesn't move it's momentum is zero.

m1v1 =m1u1

therefore change in that occurs as result of the collision is C. mv to the right

The total change in momentum of the particle of mass m that collides elastically with a vertical rigid wall perpendicularly from the left with velocity v is 2mv to the left (option B).  

The total change in momentum is given by:

[tex] \Delta p = p_{f} - p_{i} [/tex]  

In the initial state, the particle is moving to the right until it collides with the rigid wall, so:

[tex] p_{i} = mv [/tex]

In the final state, the particle moves backward after the collision with the wall, so:

[tex] p_{f} = -mv [/tex]  

The minus sign is because it is moving in the negative x-direction (to the left)

Hence, the total change in momentum is:

[tex] \Delta p = -mv - mv = -2mv [/tex]

Therefore, the total change in momentum of the particle is 2mv to the left (option B).

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Water flows through a cylindrical pipe of varying cross-section. The velocity is 5.00 m/s at a point where the pipe diameter is 1.50 cm. At a point where the pipe diameter is 3.00 cm, the velocity is

Answers

Explanation:

We know that rate of flow through a cross section :

[tex]v1 \times a1 = v2 \times a2[/tex]

5 m/s * 1.76cm^2 = v2 * 7.06cm^2

[tex]v2 = 1.24 \: m {s}^{ - 1} [/tex]

At a point where the pipe diameter is 3.00 cm, the velocity is 1.25 m/s.

What is fluid flow?

Fluid Flow, a branch of fluid dynamics, is concerned with fluids. It involves the movement of a fluid under the influence of uneven forces. As long as unbalanced pressures are applied, this motion will persist.

Given parameters:

Initial velocity of the water: u = 5.00 m/s

Initial diameter of the pipe: d = 1.50 cm.

Final diameter of the pipe: D = 3.00 cm.

Final velocity of the water: v = ?

In fluid motion:

velocity×(diameter)² = constant

Hence, initial velocity × ( initial diameter)² = final velocity × ( final diameter)²

ud² = vD²

v = u (d/D)²

v= 5 × (1.50/3.0)²

v= 5/2²

v= 5/4

v= 1.25 m/s.

Hence, at a point where the pipe diameter is 3.00 cm, the velocity is 1.25 m/s.

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I WILL GIVE BRAINLIEST Identify two types of motion where an object's speed remains the same while it continues to change direction

Answers

Answer:

velocity and acceleration

Answer:

Hey there!

Centripetal (Circular Motion) and Oscillating Motion.

Let me know if this helps :)

how many electrons do calcium have in their outer shell

Answers

Answer:

Calcium has two electrons in its outer shell.

Explanation:

Calcium is defined as a metal due to its physical and chemical traits. The two outer electrons are very reactive. Calcium has a valence of 2.

PLEASEEEEEEEE HELP WILL MARK BRAINLYIST

5. Theories are models that explain but laws __________.​

Answers

Answer:

Theories are models that explain but laws just describes an action under certain circumstances

Explanation:

Evolution is a law that does not explain how and why

But evolution by natural selection is a theory because it explain how it happens

A train on one track moves in the same direction as a second train on the adjacent track. The first train, which is ahead of the second train and moves with a speed of 36.4 m/s , blows a horn whose frequency is 123 Hz .what is its speed?

Answers

Answer:

51. 7m/s

Explanation:

Take speed of sound in air = 340 m/s

fp = fs (V + Vp)/(V + Vs)

128 = 123 (340 + Vp)/(340 + 36.4)

Vp = 51.7m/s

Explanation:

In Young's 2-slit interference experiment, the wavelength of laser light can be determined. The two slits are separated by 0.16 mm. The screen is 1.4 m from the slits. It is observed that the second bright band is located 11 mm from the center of the pattern. Given this information, what is the wavelength of the laser light?

a. 1258 nm
b. 419 nm
c. 500 nm
d. 629 nm

Answers

Answer:

d. 629 nm

Explanation:

slit separation d = .16 x 10⁻³ m

distance of screen D = 1.4 m

distance of second bright band = 11 x 10⁻³

distance of second bright band = 2 x band width

= 2 x λ D /d

Putting the values given ,

11 x 10⁻³  = 2 x λ x 1.4 /  .16 x 10⁻³

λ = 1.76 x 10⁻⁶ / 2.8

= .6285 x 10⁻⁶

= 628.5 x 10⁻⁹

= 629 nm approx .

A viewing screen is separated from a double slit by 5.20 m. The distance between the two slits is 0.0300 mm. Monochromatic light is directed toward the double slit and forms an interference pattern on the screen. The first dark fringe is 3.70 cm from the center line on the screen.

Required:
a. Determine the wavelength of light.
b. Calculate the distance between the adjacent bright fringes.

Answers

Answer:

The wavelength of this light is approximately [tex]427\; \rm nm[/tex] ([tex]4.27\times 10^{-7}\; \rm m[/tex].)The distance between the first and central maxima is approximately [tex]7.40\; \rm cm[/tex] (about twice the distance between the first dark fringe and the central maximum.)  

Explanation:

Wavelength

Convert all lengths to meters:

Separation of the two slits: [tex]0.0300\; \rm mm = 3.00\times 10^{-5}\; \rm m[/tex].Distance between the first dark fringe and the center of the screen: [tex]3.70\; \rm cm = 3.70\times 10^{-2}\; \rm m[/tex].

Refer to the diagram attached (not to scale.) Assuming that the screen is parallel to the line joining the two slits. The following two angles are alternate interior angles and should be equal to each other:

The angle between the filter and the beam of light from the lower slit, andThe angle between the screen and that same beam of light.

These two angles are marked with two grey sectors on the attached diagram. Let the value of these two angles be [tex]\theta[/tex].

The path difference between the two beams is approximately equal to the length of the segment highlighted in green. In order to produce the first dark fringe from the center of the screen (the first minimum,) the length of that segment should be [tex]\lambda / 2[/tex] (one-half the wavelength of the light.)

Therefore:

[tex]\displaystyle \cos \theta \approx \frac{\text{Path difference}}{\text{Slit separation}} = \frac{\lambda / 2}{3.00\times 10^{-5}\; \rm m}[/tex].

On the other hand:

[tex]\begin{aligned} \cot \theta &\approx \frac{\text{Distance between central peak and first minimum}}{\text{Distance between the screen and the slits}} \\ &= \frac{3.70\times 10^{-2}\; \rm m}{5.20\; \rm m} \approx 0.00711538\end{aligned}[/tex].

Because the cotangent of [tex]\theta[/tex] is very close to zero,

[tex]\cos \theta \approx \cot \theta \approx 0.00711538[/tex].

[tex]\displaystyle \frac{\lambda /2}{3.00\times 10^{-5}\; \rm m} \approx \cos\theta\approx 0.00711538[/tex].

[tex]\begin{aligned}\lambda &\approx 2\times 0.00711538 \times \left(3.00\times 10^{-5}\; \rm m\right) \\ &\approx 4.26 \times 10^{-7}\; \rm m = 426\; \rm nm\end{aligned}[/tex].

Distance between two adjacent maxima

If the path difference is increased by one wavelength, then the intersection of the two beams would move from one bright fringe to the next one.

The path difference required for the central maximum is [tex]0[/tex].The path difference required for the first maximum is [tex]\lambda[/tex].The path difference required for the second maximum is [tex]2\,\lambda[/tex].

On the other hand, if the distance between the maximum and the center of the screen is much smaller than the distance between the screen and the filter, then:

[tex]\begin{aligned}&\frac{\text{Distance between image and center of screen}}{\text{Distance between the screen and the slits}} \\ &\approx \cot \theta \\ &\approx \cos \theta \\ &\approx \frac{\text{Path difference}}{\text{Slit separation}}\end{aligned}[/tex].

Under that assumption, the distance between the maximum and the center of the screen is approximately proportional to the path difference. The distance between the image (the first minimum) and the center of the screen is [tex]3.70\; \rm cm[/tex] when the path difference is [tex]\lambda / 2[/tex]. The path difference required for the first maximum is twice as much as that. Therefore, the distance between the first maximum and the center of the screen would be twice the difference between the first minimum and the center of the screen: [tex]2 \times 3.70\; \rm cm = 7.40\; \rm cm[/tex].

If you weigh 685 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 25.0 km

Answers

CHECK COMPLETE QUESTION BELOW

you weigh 685 NN on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 25.0 kmkm ? Take the mass of the sun to be msmsm_s = 1.99×1030 kgkg , the gravitational constant to be GGG = 6.67×10−11 N⋅m2/kg2N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be ggg = 9.8 m/s2m/s2 .

Answer:

5.94×10^15N

Explanation:

the weight on the surface of a neutron star can be calculated by below expresion

W= Mg

W= weight of the person

m= mass of the person

g=gravity of the neutron star

But we need the mass which can be calculated as

m= W/g

m= 685/9.81

m= 69.83kg

From the gravitational law equation we have

F= GMm/r^2

G= gravitational constant = 6.67x10⁻¹¹

M= mass of the neutron star = 1.99x10³⁰ kg

r = distance between the person and the surface

Then r can be calculated as = 25/2 = 12.5 km , we divide by two because it's the distance between the person and the surface

g=gravity of the neutron star can be calculated as

g=(6.67×10^-11 ×1.99×10^30)/(12.5×10^3)^2

= 8.50×10^13m/s^2

Then from W= mg we can find our weight

W= 8.50×10^13m/s^2 × 69.83

= 5.94×10^15N

Therefore, weight on the surface of a neutron star is 5.94×10^15N

The following equation is an example of
decay.
181
185
79
Au →
4
2
He+

Answers

Answer:

Alp decay.

Explanation:

From the above equation, the parent nucleus 185 79Au produces a daughter nuclei 181 77 Ir.

A careful observation of the atomic mass of the parent nucleus (185) and the atomic mass of the daughter nuclei (181) shows that the atomic mass of the daughter nuclei decreased by a factor of 4. Also, the atomic number of the daughter nuclei also decreased by a factor of 2 when compared with the parent nucleus as shown in the equation given above.

This simply means that the parent nucleus has undergone alpha decay which is represented with a helium atom as 4 2He.

Therefore, the equation is an example of alpha decay.

Calculate the electromotive force produced by each of the battery combinations shown in the figure, if the emf of each is 1.5 V.

Answers

Answer:

A) 1.5 V

B) 4.5 V

Explanation:

A) Batteries in parallel have the same voltage as an individual battery.

V = 1.5 V

B) Batteries in series have a voltage equal to the sum of the individual batteries.

V = 1.5 V + 1.5 V + 1.5 V

V = 4.5 V

Magnetic resonance imaging needs a magnetic field strength of 1.5 T. The solenoid is 1.8 m long and 75 cm in diameter. It is tightly wound with a single layer of 1.50-mm-diameter superconducting wire.
What current is needed?

Answers

Answer:

The current needed is 1790.26 A

Explanation:

Given;

magnitude of magnetic field, B = 1.5 T

length of the solenoid, L = 1.8 m

diameter of the solenoid, d = 75 cm = 0.75 m

The magnetic field is given by;

[tex]B = \frac{\mu_o NI }{L}[/tex]

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

I is current in the solenoid

N is the number of turns, calculated as;

[tex]N = \frac{Length \ of\ solenoid}{diameter \ of \ wire} \\\\N = \frac{1.8}{1.5*10^{-3}} =1200 \ turns[/tex]

The current needed is calculated as;

[tex]I = \frac{BL}{\mu_o N} \\\\I = \frac{1.5 *1.8}{4\pi *10^{-7} *1200} \\\\I = 1790.26 \ A[/tex]

Therefore, the current needed is 1790.26 A.

Answer:

I = 1790.5 A

Explanation:

The magnetic field due to a solenoid is given by the following formula:

B = μ₀NI/L

where,

B = Magnetic Field Required = 1.5 T

μ₀ = 4π x 10⁻⁷ T/A.m

L = length of Solenoid = 1.8 m

I = Current needed = ?

N = No. of turns = L/diameter of wire = 1.8 m/1.5 x 10⁻³ m = 1200

Therefore,

1.5 T = (4π x 10⁻⁷ T/A.m)(1200)(I)/1.8 m

I = (1.5 T)(1.8 m)/(1200)(4π x 10⁻⁷ T/A.m)

I = 1790.5 A

To protect her new two-wheeler, Iroda Bike
buys a length of chain. She finds that its
linear density is 0.68 lb/ft.
If she wants to keep its weight below 1.4 lb,
what length of chain is she allowed?
Answer in units of ft.

Answers

Answer:

1.8/0.61 =2.95 ft

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

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A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled

Answers

Answer:

y ’= y / 2

thus when the slit width is doubled the pattern width is halved

Explanation:

The diffraction of a slit is given by the expressions

          a sin θ = m λ

where a is the width of the slit, λ is the wavelength and m is an integer that determines the order of diffraction.

          sin θ = m λ / a

If this equation

          a ’= 2 a

we substitute

          2 a sin θ'= m λ

          sin θ'= (m λ / a)  1/2

          sin θ ’= sin θ / 2

           

We can use trigonometry to find the width

         tan θ = y / L

as the angle is small

         tan θ = sin θ / cos θ = sin θ

         sin θ = y / L  

         

we substitute

        y ’/ L = y/L   1/2

        y ’= y / 2

thus when the slit width is doubled the pattern width is halved

What is the minimum thickness of coating which should be placed on a lens in order to minimize reflection of 566 nm light? The index of refraction of the coating material is 1.46 and the index of the glass is 1.71.

Answers

Answer:

The  thickness is   [tex]t = 1.415 *10^{-7 } \ m[/tex]

Explanation:

From the question we are told that

    The wavelength is  [tex]\lambda = 566 \ nm = 566 *10^{-9} \ m[/tex]

     The  index of refraction of glass is  [tex]n_g = 1.71[/tex]

     The index of refraction of the coating is  [tex]n= 1.46[/tex]

Generally the condition for destructive interference is  

         [tex]2 t = (m + \frac{1}{2} ) * \frac{\lambda }{n }[/tex]

Here m is the order of the interference pattern and given from the question that we are considering minimizing  reflection  m = 0

t = thickness of the coating

substituting values

         [tex]2 t = (0 + \frac{1}{2} ) * \frac{ 566 *10^{-9}}{ 1.46 }[/tex]

    =>    [tex]t = 1.415 *10^{-7 } \ m[/tex]

           

please help !!!!!!!!!!!!!!!!!! give the answer to the question i. which lighthouse will be warmer during the day time and why ? ii. which lighthouse will be warmer during the night time and why ? please help

Answers

Answer:

I. light house 1 will be warmer during the day ii. light house 2 will be warmer at night.

Explanation:

Because the land conducts heat better than water the light house farthest away from the water will get hotter during as the ground will heat up faster than the water. But this also means that the ground will lose heat faster at night where the water won't making the light house closest to the water hotter at night.

A microwave oven operates at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly
downward toward the base. The base is lined with a material that completely absorbs microwave energy. The total microwave energy content of the cavity is 0.50 mJ.

A)The power output of the microwave oven is?
B) The intensity of the microwave beam is?
C) The electric field amplitude is?
D) The force on the base due to the radiation is?

Answers

Answer:

power = 600000 W

intensity = 6666666.66 W/m²

Em = 70880.18 N/m

F = 2 × [tex]10^{-3}[/tex] N  

Explanation:

given data

frequency f = 2400 MHz

height oven cavity h = 25 cm = 0.25 m

base area measures A =  30 cm by 30 cm

total microwave energy content of cavity E = 0.50 mJ = 0.50 × [tex]10^{-3}[/tex]  

solution

first, we get here total time taken from top to bottom that is express as

Δt = [tex]\frac{h}{c}[/tex]  ...............1

Δt = [tex]\frac{0.25}{3\times 10^8}[/tex]  

Δt = 8.33 × [tex]10^{-10}[/tex] s

and

power output will be

power = [tex]\frac{E}{\Delta t}[/tex]   ..............2

power = [tex]\frac{0.50 \times 10^{-3}}{8.33 \times 10^{-10}}[/tex]

power = 600000 W

and

intensity of the microwave beam is  

intensity = power output ÷ base area    ..............2

intensity =  [tex]\frac{600000}{30 \times 30 \times 10^{-4}}[/tex]  

intensity = 6666666.66 W/m²

and

electric field amplitude is

as we know intensity  I = [tex]\frac{E^2}{c \mu o}[/tex]     ...............3

[tex]E(rms) = \sqrt{Ic\ \mu o} \\E(rms) = \sqrt{6666666.66 \times 3 \times 10^{8} \times 4 \pi \times 10^{-7} }[/tex]

E(rms) = 50119.87 N/m

and we know

[tex]E(rms) = \frac{Em}{\sqrt{2}}\\50119.87 = \frac{Em}{\sqrt{2}}[/tex]  

Em = 70880.18 N/m

and

force on the base due to the radiation is by the radiation pressure

[tex]Pr = \frac{l}{c}[/tex]    ..................4

[tex]\frac{F}{A} = \frac{l}{c}[/tex]  

so

F = [tex]\frac{6666666.66 \times 900 \times 10^{-4}}{3\times 10^8}[/tex]  

F = 2 × [tex]10^{-3}[/tex] N  

A beam of light from a laser illuminates a glass how long will a short pulse of light beam take to travel the length of the glass.

Answers

Answer:

The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]

Explanation:

Given that,

A beam of light from a laser illuminates a glass.

Suppose, the length of piece is [tex]L=25.21\times10^{-2}\ m[/tex]

Index of refraction is 2.83.

We need to calculate the speed of light pulse in glass

Using formula of speed

[tex]v=\dfrac{c}{\mu}[/tex]

Put the value into the formula

[tex]v=\dfrac{3\times10^{8}}{2.83}[/tex]

[tex]v=1.06\times10^{8}\ m/s[/tex]

We need to calculate the time of short pulse of light beam

Using formula of velocity

[tex]v=\dfrac{d}{t}[/tex]

[tex]t=\dfrac{d}{v}[/tex]

Put the value into the formula

[tex]t=\dfrac{25.21\times10^{-2}}{1.06\times10^{8}}[/tex]

[tex]t=2.37\times10^{-9}\ sec[/tex]

Hence, The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]

How long will it take a spacecraft travelling at 99% the speed of light (gamma = 7) to reach

the star Sirius which is 8.6 light-years away according to people on Earth ? How long will it

take according to the crew of the ship?

Answers

Answer:

The time taken is  [tex]t = 2.739 *10^{8} \ s[/tex]

Explanation:

From the question we are told that

    The speed of the spacecraft is [tex]v = 0.99c[/tex]

    where c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]

    =>   [tex]v = 0.99 * 3.0 *10^{8 } = 2.97*10^{8}\ m/s[/tex]

    The distance of Sirius is [tex]d = 8.6 \ light-years = 8.6 * 9.461*10^{15}= 8.135*10^{16} \ m[/tex]

   

Generally the time taken is mathematically represented as

       [tex]t = \frac{d}{v}[/tex]

substituting values

      [tex]t = \frac{8.136 *10^{16}}{2.97 *10^{8}}[/tex]

      [tex]t = 2.739 *10^{8} \ s[/tex]

A 70 kg human body typically contains 140 g of potassium. Potassium has a chemical atomic mass of 39.1 u and has three naturally occurring isotopes. One of those isotopes, 40K,is radioactive with a half-life of 1.3 billion years and a natural abundance of 0.012%. Each 40K decay deposits, on average, 1.0 MeV of energy into the body. What yearly dose in Gy does the typical person receive from the decay of 40K in the body?

Answers

Answer:

0.03143 Gy

Explanation:

Mass of the human body = 70 kg

Mass of potassium in the human body = 140 g

chemical atomic mass of potassium = 39.1

From avogadros number, we know that 1 atomic mass of an element contains 6.023 × 10^(23) atoms

Thus,

140g of potassium will contain;

(140 × 6.023 × 10^(23))/(39.1) = 2.1566 × 10^(24) atoms

We are told that the natural abundance of one of the 40K isotopes is 0.012%.

Thus;

Number of atoms of this isotope = 0.012% × 6.023 × 10^(23) = 7.2276 × 10^(19) K-40 atoms

Formula for activity of K-40 is given as;

Activity = (0.693 × number of K-40 atoms)/half life

Activity = (0.693 × 7.2276 × 10^(19))/1300000000

Activity = 3.85 × 10^(10)

We are told that each decay deposits 1.0 MeV of energy into the body.

Thus;

Total energy absorbed by the body in a year = 3.85 × 10^(10) × 1 × 365 = 1405.25 × 10^(10) MeV

Now, 1 MeV = 1.602 × 10^(-13) joules

Thus;

Total energy absorbed by the body in a year = 1405.25 × 10^(10) × 1.602 × 10^(-13) = 2.25 J

1 Gy = 1 J/kg

Thus;

Yearly dose = 2.25/70 = 0.03143 Gy

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?

Answers

Answer:

6000 counts per second

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample

UVC light used in sterilizers, has wavelengths between 100 to 280 nm. If a certain UVC wave has a wavelength of 142.9 nm, what is the energy of one of its photons in J

Answers

Answer:

The energy of one of its photons is 1.391 x 10⁻¹⁸ J

Explanation:

Given;

wavelength of the UVC light, λ = 142.9 nm = 142.9 x 10⁻⁹ m

The energy of one photon of the UVC light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f is frequency of the light

f = c / λ

where;

c is speed of light = 3 x 10⁸ m/s

λ  is wavelength

substitute in the value of f into the main equation;

E = hf

[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{142.9*10^{-9}} \\\\E = 1.391*10^{-18} \ J[/tex]

Therefore, the energy of one of its photons is 1.391 x 10⁻¹⁸ J

find it ...............​

Answers

Answer:

9 N

Explanation:

Treat X and y as a single book with weight 9 N.

Draw a free body diagram of this book combination.  There are two forces: weight force 9 N pulling down, and normal force F pushing up.

Sum of forces in the y direction:

∑F = ma

F − 9 N = 0

F = 9 N

g To decrease the intensity of the sound you are hearing from your speaker system by a factor of 36, you can

Answers

Answer:

Increase the distance by a factor of 6.

Explanation:

The intensity at a distance r is given by :

[tex]I=\dfrac{P}{4\pi r^2}[/tex]

Here,

P is power emitted

r is distance from source

It means that the intensity is inversely proportional to the distance from the source.

To decrease the intensity of the sound you are hearing from your speaker system by a factor of 36, we can increase the distance by a factor of 6. Hence, this is the required solution.

One of the two slits in a Young’s experiment is painted over so that it transmits only one-half the intensity of the other slit. As a result:
A. the fringe system disappears
B. the bright fringes get brighter and the dark ones get darker
C. the fringes just get dimmer
D. the dark fringes just get brighter
E. the dark fringes get brighter and the bright ones get darker

Answers

Answer:

B. the bright fringes get brighter and the dark ones get darker

Explanation:

Let us consider when the intensities are equal, we use the equation

[tex]I_{max} = I_{1} + I_{2} + 2\sqrt{I*I}[/tex]     for light fringes and,

[tex]I_{min} = I_{1} + I_{2} - 2\sqrt{I*I}[/tex]     for dark fringes  

where [tex]I_{1}[/tex] and [tex]I_{1}[/tex] are the light intensities from the first and second slits respectively.

For the first case where the light from the two slits have the same intensities, we can say both have intensity [tex]I[/tex]

[tex]I_{max} = I + I + 2\sqrt{I*I}[/tex] = [tex]2I + 2I = 4I[/tex]

[tex]I_{min} = I + I - 2\sqrt{I*I} = 2I - 2I = 0[/tex]

For the case where one of the intensities has half the intensity of the other.

one has intensity [tex]I[/tex] and the other one has intensity [tex]\frac{I}{2}[/tex]

inserting, we have

[tex]I_{max} = I + \frac{I}{2} + 2\sqrt{I*\frac{I}{2}} = 2.932I[/tex]

[tex]I_{min} = I + \frac{I}{2} - 2\sqrt{I*\frac{I}{2}} = 0.068I[/tex]

this shows that the bright fringes get brighter and the dark ones get darker.


A load of 1 kW takes a current of 5 A from a 230 V supply. Calculate the power factor.

Answers

Answer:

Power factor = 0.87 (Approx)

Explanation:

Given:

Load = 1 Kw = 1000 watt

Current (I) = 5 A

Supply (V) = 230 V

Find:

Power factor.

Computation:

Power factor = watts / (V)(I)

Power factor = 1,000 / (230)(5)

Power factor = 1,000 / (1,150)

Power factor = 0.8695

Power factor = 0.87 (Approx)

2.) Is it possible to have negative velocity but positive acceleration? If so, what would
this mean?

Answers

Answer:

Yes, yes it would

Explanation:

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