How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. the line density of the grating were doubled?

Answer:

Check your router connections then restart your router.

Explanation:

Answer:

Check your router connections then restart your router.

Explanation:

Most internet access comes from routers so the problem is most likely the router.

Answer:

(a) 1.47 x 10⁴ V/m

(b) 1.28 x 10⁻⁷C/m²

(c) 3.9 x 10⁻¹²F

(d) 9.75 x 10⁻¹¹C

Explanation:

(a) For a parallel plate capacitor, the electric field E between the plates is given by;

E = V / d               -----------(i)

Where;

V = potential difference applied to the plates

d = distance between these plates

From the question;

V = 25.0V

d = 1.70mm = 0.0017m

Substitute these values into equation (i) as follows;

E = 25.0 / 0.0017

E = 1.47 x 10⁴ V/m

(c) The capacitance of the capacitor is given by

C = Aε₀ / d

Where

C = capacitance

A = Area of the plates = 7.60cm² = 0.00076m²

ε₀ = permittivity of free space =  8.85 x 10⁻¹²F/m

d = 1.70mm = 0.0017m

C = 0.00076 x  8.85 x 10⁻¹² / 0.0017

C = 3.9 x 10⁻¹²F

(d) The charge, Q, on each plate can be found as follows;

Q = C V

Q =  3.9 x 10⁻¹² x 25.0

Q = 9.75 x 10⁻¹¹C

Now since we have found other quantities, it is way easier to find the surface charge density.

(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e

σ = Q / A

σ = 9.75 x 10⁻¹¹ /  0.00076

σ = 1.28 x 10⁻⁷C/m²

Answer:

The acceleration of the refrigerator together with the objects decreases.

Explanation:

If the mass of the refrigerator is increased by stacking more masses (objects) on it,

and the force applied remains constant, then we know from

F = ma

where

F is the applied force

m is the total mass of the refrigerator and the objects

a is the acceleration of the masses.

If F is constant, and m is increased, the acceleration will decrease

Answer:

The acceleration decreases.

Explanation:

its right

Answer:

8.1 m

Explanation:

Convert km/h to m/s.

45 km/h × (1000 m/km) × (1 h / 3600 s) = 12.5 m/s

Distance = speed × time

d = (12.5 m/s) (0.65 s)

d = 8.125 m

Answer:

Torque = 0.012 N.m

Explanation:

We are given;

Mass of wheel;m = 750 g = 0.75 kg

Radius of wheel;r = 25 cm = 0.25 m

Final angular velocity; ω_f = 0

Initial angular velocity; ω_i = 220 rpm

Time taken;t = 45 seconds

Converting 220 rpm to rad/s we have;

220 × 2π/60 = 22π/3 rad/s

Equation of rotational motion is;

ω_f = ω_i + αt

Where α is angular acceleration

Making α the subject, we have;

α = (ω_f - ω_i)/t

α = (0 - 22π/3)/45

α = -0.512 rad/s²

The formula for the Moment of inertia is given as;

I = ½mr²

I = (1/2) × 0.75 × 0.25²

I = 0.0234375 kg.m²

Formula for torque is;

Torque = Iα

For α, we will take the absolute value as the negative sign denotes decrease in acceleration.

Thus;

Torque = 0.0234375 × 0.512

Torque = 0.012 N.m

Answer:

Everything in the Universe is made up of matter and energy. Matter is anything that has mass and occupies space. ... Energy is the ability to cause change or do work. Some forms of energy include light, heat, chemical, nuclear, electrical energy and mechanical energy.

Explanation:

Answer: Speed = [tex]3.10^{-31}[/tex] m/s

Explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:

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

Relativistic momentum is calculated as:

p = [tex]\frac{mu}{\sqrt{1-\frac{u^{2}}{c^{2}} } }[/tex]

where:

m is rest mass

u is velocity relative to an observer

c is light speed, which is constant (c=[tex]3.10^{8}[/tex]m/s)

Initial momentum is zero, then:

[tex]p_{f}[/tex] = 0

[tex]p_{1}-p_{2}[/tex] = 0

[tex]p_{1} = p_{2}[/tex]

To find speed of the heavier fragment:

[tex]\frac{mu_{1}}{\sqrt{1-\frac{u^{2}_{1}}{c^{2}} } }=\frac{mu_{2}}{\sqrt{1-\frac{u^{2}_{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=\frac{3.10^{-28}.0.844.3.10^{8}}{\sqrt{1-\frac{(0.844c)^{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=1.42.10^{-19}[/tex]

[tex]1.86.10^{-27}u_{1} = 1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }[/tex]

[tex](1.86.10^{-27}u_{1})^{2} = (1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } })^{2}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38}.(1-\frac{u_{1}^{2}}{9.10^{16}} )[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -[2.02.10^{-38}(\frac{u_{1}^{2}}{9.10^{16}} )][/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -2.24.10^{-23}.u^{2}_{1}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2}+2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]u^{2}_{1} = \frac{2.02.10^{-38}}{2.24.10^{-23}}[/tex]

[tex]u_{1} = \sqrt{9.02.10^{-62}}[/tex]

[tex]u_{1} = 3.10^{-31}[/tex]

The speed of the heavier fragment is [tex]u_{1} = 3.10^{-31}[/tex]m/s.

Answer:

t₀ = 1.55 s

Explanation:

According to Einstein's Theory of Relativity, when an object moves with a speed comparable to speed of light, the time interval measured for the event, by an observer in  motion relative to the event is not the same as measured by an observer at rest.

It is given as:

t = t₀/[√(1 - v²/c²)]

where,

t = time measured by astronaut in motion = 3.1 s

t₀ = time required according to observer on earth = ?

v = relative velocity = 0.87 c

c = speed of light

3.1 s = t₀/[√(1 - 0.87²c²/c²)]

(3.1 s)(0.5) = t₀

t₀ = 1.55 s

Answer:

Explanation:

Time interval required for this rotation according to an observer on the Earth is given as [tex]\delta t[/tex]

where,

[tex]t_o = 3.1\\\\v = 0.87[/tex]

[tex]\delta t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\\\\\delta t = \frac{3.1}{\sqrt{1-(\frac{0.87c}{c})^2}}\\\\\delta t = 6.29sec[/tex]

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Complete question:

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:

a) the midpoint between the two rings?

b) the center of the left ring?

Answer:

a) the electric field strength at the midpoint between the two rings is 0

b) the electric field strength at the center of the left ring is 2712.44 N/C

Explanation:

Given;

distance between the two rings, d = 25 cm = 0.25 m

diameter of each ring, d = 10 cm = 0.1 m

radius of each ring, r = [tex]\frac{0.1}{2} = 0.05 \ m[/tex]

the charge on each ring, q = 20 nC

Electric field strength for a ring with radius r and distance x from the center of the ring is given as;

[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}}[/tex]

The electric field strength at the midpoint;

the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m

[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9}*0.125*20*10^{-9}}{(0.125^2 + 0.05^2)^{3/2}} \\\\E = 9210.5 \ N/C[/tex]

[tex]E_{left} = 9210.5 \ N/C[/tex]

The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;

[tex]E_{right} = -9210.5 \ N/C[/tex]

The electric field strength at the midpoint;

[tex]E_{mid} = E_{left} + E_{right}\\\\E_{mid} = 9210.5 \ N/C - 9210.5 \ N/C\\\\E_{mid} = 0[/tex]

(b)

The distance from the right ring to center of the left ring, x = 0.25 m.

[tex]E = \frac{KxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9} *0.25*20*10^{-9}}{(0.25^2 + 0.05^2)^{3/2}} \\\\E = 2712.44 \ N/C[/tex]

Answer:

e. It is neither attracted nor repelled.

Explanation:

Electrostatic attraction or repulsion occurs between two or more charged particles or conductors. In this case, if the negatively charged ball is brought close to the neutral end A of the rod, there would be no attraction or repulsion between the rod end A and the negatively charged ball. This is because a charged particle or conductor has no attraction or repulsion to a neutral particle or conductor.

Answer:gamma ray

Explanation:

Answer:

Explanation:

volume of water being lifted

= π r² h , where r is radius of cylinder and h is height of cylinder

= 3.14 x5² x 10

= 785 m³

mass of water = 785 x 10³ kg

mass of this much of water is lifted so that its centre of mass is lifted by height

10 / 2 = 5m .

So work done = mgh , m is mass of water , h is displacement of centre of mass and g is acceleration due to gravity

= 785 x 10³ x 9.8 x 5

= 38.465 x 10⁶ J  

Answer:

The average density of the rod is 1.605 kg/m.

Explanation:

The average density of the rod is given by:

[tex] \rho = \frac{m}{l} [/tex]    

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:

[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]

[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex]   (1)

Using u = x+1  →  du = dx  → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]

[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]

Therefore, the average density of the rod is 1.605 kg/m.  

I hope it helps you!    

The average density of the rod is  [tex]1.605 \;\rm kg/m^{3}[/tex].

Given data:

The length of rod is, L = 3 m.

The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3,  The expression for the average density is given as,

[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)

Using u = x+1  

du = dx

u₁= x₁+1 = 0+1 = 1

and

u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]

Thus, we can conclude that the average density of the rod is  [tex]1.605 \;\rm kg/m^{3}[/tex].

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The correct answer is C. The inner planets are also called terrestrial planets.

Explanation:

Our solar system includes a total of eight planets. Additionally, planets are classified into broad categories including inner planets and outer planets. The inner planets category applies to planets such as Earth, Mercury, or Mars because these are located within the asteroid belt (region of asteroids between Mars and Jupiter). Moreover, inner planets differ from others due to their composition as they are composed of rocks and metals. Also, due to this composition, these are known as terrestrial planets. According to this, the statement that best describes inner planets is "The inner planets are also called terrestrial planets".

Answer:

The answer is c.) The inner planets are also called terrestrial planets.

Explanation:

Answer:

A simple arrangement by means of which e.m.f,s. are compared is known as?

(a)Voltmeter

(b)Potentiometer

(c)Ammeter

(d)None of the above

Explanation:

Answer:

  t = 1.81 min ,     the correct answer is c

Explanation:

This is a missile throwing exercise

The object is thrown horizontally, so its vertical speed is zero (voy = 0), let's use the equation

             y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

the final height is y = 0 and the initial height is y₀ = 22000 m

            0 = y₀ + 0 - ½ g t²

            t = √y 2y₀ / g

let's calculate

           t = √(2  22000 / 3.72)

           t = 108.76 s

let's reduce to minutes

           t = 108.76 s (1 min / 60 s)

           t = 1.81 min

The correct answer is c

Answer:

Matter's resistance to a change in motion is called INERTIA and is directly proportional to the mass of an object.

Explanation:

Answer:

This means that mercury has a higher or faster expansion rate than glass

Explanation:

This is because When a container expands, the reservoir in the glass expands at the same rate as the glass. Thus, if there is something in a glass and both expand at the same rate, they have no change - but if the contents expand faster, they will fill the container to a higher level, and if the contents expand slower, they will fill the container to a lower level (relative to the new size of the container).

Answer:

Explanation:

From the question we are told that

    The radius is  [tex]r = 1.4 \ mm = 1.4 *10^{-3} \ m[/tex]

     The  current is  [tex]I = 4.5 \ A[/tex]

Generally the electric field is mathematically represented as

         [tex]E = \frac{J}{\sigma }[/tex]

Where [tex]\sigma[/tex] is the conductivity of  aluminum with value [tex]\sigma = 3.5 *10^{7} \ s/m[/tex]

J is the current density which mathematically represented as  

      [tex]J = \frac{I}{A}[/tex]

Here A is the cross-sectional area which is mathematically represented as  

       [tex]A = \pi r^2[/tex]

       [tex]A = 3.142 * (1.4*10^{-3})^2[/tex]

       [tex]A = 6.158*10^{-6} \ m^2[/tex]

So

    [tex]J = \frac{ 4.5 }{6.158*10^{-6}}[/tex]

    [tex]J = 730757 A/m^2[/tex]

So

       [tex]E = \frac{ 730757}{3.5*10^{7} }[/tex]

       [tex]E = 0.021 \ N/C[/tex]

Answer:

-0.24diopters

Explanation:

The lens is intended that makes an object at infinity appear to be 4.2 m away, so do=infinity, dI = - 4.2m (minus sign because image is on same side of lens as object)

So 1/do +1/di = 1/f

1/infinity + 1/-4.2 = 1/f

1/f = 1/-4.2 = -0.24diopters

Answer:

Pls seeattached file

Explanation:

A resistor made of Ni chrome wire is used in an application where its resistance cannot be more than 1.35 % so its temperature range will be from 33.75 to -33.75 °C.

Electrical resistance, or resistance to electricity, is a force that opposes the flow of current. Ohms are used to expressing resistance values.

When there is an electron difference between two terminals, electricity will flow from high to low. In opposition to that flow is resistance. As resistance rises, the current declines. On the other side, when the resistance falls, the current rises.

According to the question,

R = R₀ (1 + α ΔT)

(1 + 0.0135)R₀ = R₀(1 + α ΔT)

ΔT = (1 + 0.0135) / α

= 0.0135 / 0.0004

= 33.75 °C.

ΔT = [(1 - 0.0135) -1]/0.004

= -33.75 °C

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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]

Answer:

   v₂ = 9 v

Explanation:

For this exercise in fluid mechanics, let's use the continuity equation

           v₁ A₁ = v₂ A₂

where v is the velocity of the fluid, A the area of ​​the pipe and the subscripts correspond to two places of interest.

The area of ​​a circle is

           A = π R²

let's use the subscript 1 for the starting point and the subscript 2 for the part with the constraint

In this case v₁ = v and the area is

            A₁ = π R²

in the second point

           A₂= π (R / 3)²

we substitute in the continuity equation

           v π R² = v₂ π R² / 9

            v = v₂ / 9

            v₂ = 9 v

Answer:

[tex]v=1.24\times 10^8\ m/s[/tex]

Explanation:

Given that,

The refractive index of benzene is 2.419

We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,

[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s[/tex]

So, the speed of light in bezene is [tex]1.24\times 10^8\ m/s[/tex].

Answer: The correct answer for this question is letter (B) The electromagnetic waves reach Earth, while the mechanical waves do not. The sun generates both mechanical and electromagnetic waves. Space, between the sun and the earth is a nearly vacuum. So mechanical wave can not spread out in the vacuum.

Hope this helps!

Answer:

The electromagnetic waves reach Earth, while the mechanical waves do not

Answer:

a) k = 95.54 N / m,   c =   19.55 , b)      m₃ = 0.9078 kg

Explanation:

In a simple harmonic movement with friction, we can assume that this is provided by the speed

          fr = -c v

when solving the system the angular value remains

          w² = w₀² + (c / 2m)²

They give two conditions

1) m₁ = 1 kg

     f₁ = 1.1 Hz

the angular velocity is related to frequency

         w = 2π f₁

Let's find the angular velocity without friction is

         w₂ = k / m₁

we substitute

        (2π f₁)² = k / m₁ + (c / 2m₁)²

2) m₂ = 2 kg

    f₂ = 0.8 Hz

        (2π f₂)² = k / m₂ + (c / 2m₂)²

we have a system of two equations with two unknowns, so we can solve it

we solve (c / 2m)² is we equalize the expression

           (2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁

           k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)

           k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)

a) Let's calculate

           k = 4 π² (0.8² -1.1²) / (½ -1/1)

           k = 39.4784 (1.21) / (-0.5)

           k = 95.54 N / m

now we can find the constant of friction

              (2π f₁) 2 = k / m₁ + (c / 2m₁)²

           c2 = ((2π f₁)² - k / m₁) 4m₁²

           c2 = (4ππ² f₁² - k / m₁) 4 m₁²

let's calculate

           c² = (4π² 1,1² - 95,54 / 1) 4 1²

           c² = (47.768885 - 95.54) 8

           c² = -382.1689

           c =   19.55    

b) f₃ = 0.2 Hz

   m₃ =?

              (2πf₃)² = k / m₃ + (c / 2m₃) 2

we substitute the values

              (4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²

              1.579 = 95.54 / m₃ + 95.542225 / m₃²

let's call

              x = 1 / m₃

              x² = 1 / m₃²

- 1.579 + 95.54 x + 95.542225 x² = 0

              60.5080 x² + 60.5080 x -1 = 0

                x² + x - 1.65 10⁻² = 0

                  x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2

                  x = [1 ± 1.03] / 2

                  x₁ = 1.015 kg

                  x₂ = -0.015 kg

Since the mass must be positive we eliminate the second results

                  x₁ = 1 / m₃

                 m₃ = 1 / x₁

                  m₃ = 1 / 1.1015

Answer:

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

Explanation:

Given:

wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m

Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m

Separation distance (D) = 5.4 cm = 0.054 m

Find:

Maximum altitude to see(L)

Computation:

Resolving power = 1.22(λ / d)

D / L = 1.22(λ / d)

0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]

0.054 / L = 1.22 [0.03 × 10⁻⁶]

L = 0.054 / 1.22 [0.03 × 10⁻⁶]

L = 0.054 / [0.0366 × 10⁻⁶]

L = 1.47 × 10⁶

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

Answer: Ф = 17.2657 ≈ 17°

Explanation:

we simply apply ET =0 about the ending of the rod

so In.g.L/2sinФ - In.a.L/2cosФ = 0

g.sinФ - a.cosФ = 0

g.sinФ = a.cosФ

∴ tanФ = a/g

Ф =  tan⁻¹ a / g

Ф = tan⁻¹ ( 10 / 32.17405)

Ф = tan⁻¹ 0.31080948777

Ф = 17.2657 ≈ 17°

Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°

Answer:

L = 0.8 m

Explanation:

Since, the distance between first and third dark fringes is 4 mm. Therefore, the fringe spacing between consecutive dark fringes will be:

Δx = 4 mm/2 = 2 mm = 2 x 10⁻³ m

but,

Δx = λL/d

λ = wavelength of the light = 500 nm = 5 x 10⁻⁷ m

d = slit spacing = 0.2 mm = 0.2 x 10⁻³ m

L = Distance between slits and screen = ?

Therefore, using the values, we get:

2 x 10⁻³ m = (5 x 10⁻⁷ m)(L)/(0.2 x 10⁻³)

L = (2 x 10⁻³ m)(0.2 x 10⁻³ m)/(5 x 10⁻⁷ m)

L = 0.8 m

Answer:

"Endoscope" is the correct answer.

Explanation: