Can someone please help me? Please answer the question in the screenshot. Thank you! I need this answered!

Answer:

  E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]

Explanation:

In this exercise you are asked to calculate the electric field between two plates, the electric field is a vector

         E_ {total} = E₁ + E₂

         E_ {total} = 2 E

where E₁ and E₂ are the fields of each plate, we have used that for the positively charged plate the field is outgoing and for the negatively charged plate the field is incoming, therefore in the space between the plates for a test charge the two fields point in the same direction

to calculate the field created by a plate let's use Gauss's law

          Ф = ∫ E . dA = q_{int} /ε₀

As a Gaussian surface we use a cylinder with the base parallel to the plate, therefore the direction of the electric field and the normal to the surface are parallel, therefore the scalar product is reduced to the algebraic product.

           E 2A = q_{int} / ε₀

where the 2 is due to the surface has two faces

indicate that the surface has a uniform charge for which we can define a surface density

           σ = q_{int} / A

           q_{int} = σ A

we substitute

           E 2A = σ A /ε₀

           E = σ / 2ε₀  

therefore the total field is

           E_ {total} = σ /ε₀

let's substitute the density for the charge of the whole plate

           σ= Q / L²

            E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]

The index of refraction of the plastic is approximately 1.461

The known values in the question are;

The thickness of the piece of plastic placed on the dot = 1.30 cm

The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm

The unknown values in the question are;

The index of refraction

Strategy;

Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic

[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]

The real depth of the dot below the piece of plastic, d₁ = 1.30 cm

The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised

Therefore;

The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm

[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]

Therefore, n = 1.30/0.89 ≈ 1.461

The refractive index of the plastic block, n ≈ 1.461

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The minimum width of the screen is 34 cm.

For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/5640 lines per cm = 1/5640 cm per line = 1/5640 × 10⁻² m per line, θ = angle between principal maximum and the center axis of the grating, m = order of maxima = 1 (since we require the position of the principal maximum) and λ = wavelength = 455 nm = 455 × 10⁻⁹ m

So, sinθ = mλ/d

Also tanθ = L/D where θ = angle between principal maximum and the center axis of the grating, L = distance between central maximum and principal maximum and D = distance between grating and screen = 0.661 m.

For small angles sinθ ≈ tanθ

So, mλ/d = L/D

making L subject of the formula, we have

L = mλD/d

L = 1 × 455 × 10⁻⁹ m × 0.661 m ÷  1/5640 × 10⁻² m per line

L = 1 × 455 × 10⁻⁹ m × 0.661 m  × 5640 × 10² line per m

L = 1696258.2 × 10⁻⁷ m

L = 0.16963 m

L ≅ 0.17 m

So, for centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is w = 2L.

So, w = 2 × 0.17 m

w = 0.34 m

w = 34 cm

So for the centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is 34 cm.

Learn more about diffraction grating here:

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Answer:

the change in the kinetic energy of the system is -42.47 J

Explanation:

Given;

mass A, Ma = 2 kg

initial velocity of mass A, Ua = 15 m/s

Mass M, Mm = 4 kg

initial velocity of mass M, Um = 7 m/s

Let the common velocity of the two masses after collision = V

Apply the principle of conservation of linear momentum, to determine the final velocity of the two masses;

[tex]M_aU_a + M_mU_m = V(M_a + M_m)\\\\(2\times 15 )+ (4\times 7) = V(2+4)\\\\58 = 6V\\\\V = \frac{58}{6} = 9.67 \ m/s[/tex]

The initial kinetic of the two masses;

[tex]K.E_i = \frac{1}{2} M_aU_a^2 \ + \ \frac{1}{2} M_mU_m^2\\\\K.E_i = (0.5 \times 2\times 15^2) \ + \ (0.5 \times 4\times 7^2)\\\\K.E_i = 323 \ J[/tex]

The final kinetic energy of the two masses;

[tex]K.E_f = \frac{1}{2} M_aV^2 \ + \ \frac{1}{2} M_mV^2\\\\K.E_f = \frac{1}{2} V^2(M_a + M_m)\\\\K.E_f = \frac{1}{2} \times 9.67^2(2+ 4)\\\\K.E_f = 280.53 \ J[/tex]

The change in kinetic energy is calculated as;

[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 280.53 \ J \ - \ 323 \ J\\\\\Delta K.E = -42.47 \ J[/tex]

Therefore, the change in the kinetic energy of the system is -42.47 J

Answer

Explanation

:giác mạc

Answer:

I current in a coil,,,,,,

Answer:D?

Explanation:

Sorry if i'm wrong....

Answer:

   t = 2.09 10⁻³ s

Explanation:

We must solve this problem in parts, first we look for the acceleration of the electron and then the time to travel the distance

let's start with Newton's second law

        ∑ F = m a

the force is electric

        F = q E

we substitute

        q E = m a

        a = [tex]\frac{q}{m} \ E[/tex]

        a = [tex]\frac{1.6 \ 10^{-19}}{ 9.1 \ 10^{-31} } \ 7.8 \ 10^{-9}[/tex]

        a = 1.37 10³ m / s²

now we can use kinematics

        x = v₀ t + ½ a t²

indicate that rest starts v₀ = 0

        x = 0 + ½ a t²

        t = [tex]\sqrt{\frac{2x}{a} }[/tex]

        t = [tex]\sqrt{\frac {2 \ 3 \ 10^{-3}}{ 1.37 \ 10^3} }[/tex]

        t = 2.09 10⁻³ s

Answer:

Explanation:

A bullet being shot out of a gun tends to leave tiny amounts of the bullet behind due to friction between the bullet and the gun barrel.

A roller skater pushing requires the conversion of food chemical energy to muscle contraction energy. This conversion increases the body temperature and sweat is excreted to counteract the heat increase. The evaporation of the sweat causes a slight decrease in body mass.

A jet plane taking off consumes some of the fuel carried onboard to provide thrust. The products of combustion become part of the exhaust stream leaving the airplane rearward providing forward thrust.

Answer:

a politician is a person active in party politic or person holding os seeking an elected seat in government.politicans purpose, support,and create laws that govern the land and by extension ,it's people.

Answer:

I = 15,645. kg*m/s or 15,645 N*s

Explanation:

I = m(^v)

I = 1050kg((26.2m/s-11.3m/s)

I = 15,645. kg*m/s

Answer:

(a) emf = 0.507 V

(b) emf = 0.0507 V

(c) emf = 0.00234 V

Explanation:

Given;

number of turns of the coil, N = 40 turns

diameter of the coil, d = 11 cm

radius of the coil, r = 5.5 cm = 0.055 m

magnitude of the magnetic field, B = 0.4 T

The magnitude of the induced emf is calculated as;

[tex]emf = - N\frac{d\phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux= BA \\\\A \ is the \ area \ of \ the \ coil = \pi r^2 = \pi (0.055)^2 = 0.0095 \ m^2\\\\emf = - N \frac{dB.A}{dt} = -NA\frac{dB}{dt} \\\\emf = -NA\frac{(B_2 - B_1)}{t} \\\\emf = NA \frac{(B_1 - B_2)}{t} \\\\the \ final \ magnetic \ field \ is \ reduced \ to \ zero;\ B_2 = 0\\\\emf = \frac{NAB_1}{t}[/tex]

(a) when the time, t = 0.3 s

[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{0.3} = 0.507 \ V[/tex]

(b) when the time, t = 3.0 s

[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{3} = 0.0507 \ V[/tex]

(c) when the time, t = 65 s

[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{65} = 0.00234 \ V[/tex]

Answer:

473.15

Explanation:

(A)

Explanation:

We can see that the resistors are connected in parallel so all of them have the same voltage of 100 V. We also know that

[tex]P = VI[/tex]

Since resistor Y dissipates 100 W of power, we can solve for the current as

[tex]I = \dfrac{P}{V} = \dfrac{100\:\text{W}}{100\:\text{V}} = 1.0\:\text{A}[/tex]

The current in resistor Y is

Answer:

The dimensional formula of Young's modulus is [ML^-1T^-2]

Answer:

G.oogle : The dimensional formula for Young’s modulus is:

A. [ML−1T−2]A. [ML−1T−2]

B. [M0LT−2]B. [M0LT−2]

C. [MLT−2]C. [MLT−2]

D. [ML2T−2]

Answer:

4

Explanation:

Answer:

Answer:

Part a)

23.1 Nm

..........

Answer:

[tex]v=0.9833\ c[/tex]

Explanation:

The density changes means that the length in the direction of the motion is changed.

Therefore,

[tex]$\text{Density} = \frac{m}{lwh}$[/tex]

Given :

Side,  b = h = 0.13 m

Mass, m = 3.3 kg

Density = 8100 [tex]kg/m^3[/tex]

So,

[tex]$8100=\frac{3.3}{l \times 0.13 \times 0.13}$[/tex]

[tex]$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$[/tex]

l = 0.024 m

Then for relativistic length contraction,

[tex]$l= l' \sqrt{1-\frac{v^2}{c^2}}$[/tex]

[tex]$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$[/tex]

[tex]$0.184= \sqrt{1-\frac{v^2}{c^2}}$[/tex]

[tex]$0.033= 1-\frac{v^2}{c^2}}$[/tex]

[tex]$\frac{v^2}{c^2}= 0.967$[/tex]

[tex]$\frac{v}{c}=0.9833$[/tex]

[tex]v=0.9833\ c[/tex]

Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).

Answer:

x = 0.9 m

Explanation:

For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive

          ∑ τ = 0

          60 1.5 - 78 1.5 + 30 x = 0

where x is measured from the left side of the fulcrum

           90 - 117 + 30 x = 0

           x = 27/30

           x = 0.9 m       

In summary the center of mass is on the side of the lightest weight x = 0.9 m

Answer:

The answer is "80 cm".

Explanation:

The distance of 21 cm between the speaker's effect of high strength but a spacing of 61 cm corresponds to a zero to zero intensity, that also is, the waves are all in phase with others [tex]\Delta \ x_1 = 21 \ cm[/tex] this is out of phase [tex]\Delta\ x_2 = 61\ cm[/tex]

[tex]\therefore\\\\\Delta\ x_2 -\Delta\ x_1 = \frac{\lambda}{2}\\\\\lambda= 2( \Delta\ x_2 -\Delta\ x_1)[/tex]

   [tex]= 2 ( 61\ cm - 21\ cm)\\\\ = 2(40\ cm)\\\\= 80\ cm[/tex]

Complete Question

The Question diagram is attached below

Answer:

a)  [tex]W_{Fg}= 12500 Nm[/tex]

b)  [tex]W_{T_1}= - 6339.3Nm[/tex]

c)  [tex]W_{T_2}= - 3662.8Nm[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=255kg[/tex]

Distance [tex]d=4m[/tex]

Generally the equation for Work done is mathematically given by

[tex]W=F*d[/tex]

For [tex]F_g[/tex]

[tex]W_{Fg}=2500 x 5.3[/tex]

[tex]W_{Fg}= 12500 Nm[/tex]

For [tex]T_1[/tex]

[tex]W_{T_1}= - {1830 sin(60) x 4}[/tex]

[tex]W_{T_1}= - 6339.3Nm[/tex]

For [tex]T_2[/tex]

[tex]W_{T_2}= - {1295 sin(45) x 4}[/tex]

[tex]W_{T_2}= - 3662.8Nm[/tex]

Answer:

The answer is "0.00466 N".

Explanation:

[tex]F=(B \times i) L\\\\[/tex]

therefore the smaller side is parallel to magnetic field  

[tex]\therefore \\\\F= B i L\ \sin\ 'o'=0 \ N[/tex]

calculating the force on the layer side:

[tex]\to F=0.15 \times 1.7 \times 0.0183 \times \sin 90^{\circ}=0.00466\ N\\\\[/tex]

Therefore [tex]F_o[/tex] the net force on the  rectangular loop [tex]= 0.00466 \ N[/tex]

Answer:

The net electric field at the midpoint is 6.85 x 10^7 N/C.

Explanation:

q = − 8.3 μC

q' = + 7.8 μC

d =  9.2 cm

d/2 = 4.6 cm

The electric field due to the charge q at midpoint is

[tex]E = \frac{k q}{r^2}\\\\E = \frac{9\times 10^9\times 8.3\times 10^{-6}}{0.046^2}\\\\E = 3.53\times 10^7 N/C[/tex] leftwards

The electric field due to the charge q' at midpoint is

[tex]E' = \frac{k q}{r^2}\\\\E' = \frac{9\times 10^9\times 7.8\times 10^{-6}}{0.046^2}\\\\E' = 3.32\times 10^7 N/C[/tex]

The resultant electric field at mid point is

E'' = E + E' = (3.53 + 3.32) x 10^7 = 6.85 x 10^7 N/C

Answer:

a) 0.46875

b) 8 km

Explanation:

Smokestack height ( H ) = 50 m

speed of pollutant / wind speed = 3 m/s

Half width = 2 [tex]\sqrt{2Dt }[/tex] = 50 m  ---- ( 1 )

a) If L = 2 km

value of turbulent diffusion coefficient D

back to equation 1

50 = 2 √ 2 * D * ( 2000/3 )

2500 = 4 * 2 * D * ( 2000/3 )

D = 2500 / ( 8 * ( 2000/3 )  )

   = 0.46875

where : time to travel ( t ) = Distance / speed = 2000 / 3

b) when the smoke stack = 50 * 2 = 100 m

L = 800 m = 8 km

attached below is the detailed solution

Answer:

The charges are + 74.3 μC and - 74.3 μC

Explanation:

Let the charges be q and q'.

Since the charges initially attract each other with a force of 1.39 N, the force of attraction is given by

F = kqq'/r² where k = 9 × 10⁹ Nm²/C² and r = distance between the charges = 0.189 m

When the charges are brought together, they share their charge equally and have a net charge of (q + q')/2 each.

They now repel each other.

So, the magnitude of the force of repulsion is given by

F' = k[(q + q')/2][(q + q')/2]/r²

F' = k[(q + q')²/4r²

Since the magnitude of the force of attraction and repulsion are the same, we have that

F = F'

kqq'/r² = k[(q + q')²/4r²

qq' = (q + q')²/4

(q + q')² = 4qq'

q² + 2qq' + q'² = 4qq'

q² + 2qq' - 4qq' + q'² = 0

q² - 2qq' + q'² = 0

(q - q')² = 0

q - q' = 0

q = q'

Substituting q = q' into F, we have

F = kqq'/r²

F = kq²/r²

making q subject of the formula, we have

q² = Fr²/k

q = √(Fr²/k)

q = r√(F/k)

Substituting the values of the variables into the equation, we have

q = 0.189 m√(1.39 N/9 × 10⁹ Nm²/C²)

q = 0.189 m√(0.15444 × 10⁻⁹ Nm²/C²)

q = 0.189 m(0.3923 × 10⁻³ C/m)

q = 0.0743 × 10⁻³ C

q = 74.3 × 10⁻³ × 10⁻³ C

q = 74.3 × 10⁻⁶ C

q = 74.3 μC

Since q and q' initially attract, it implies that they initially had opposite charges.

So, q = 74.3 μC and q' = -74.3 μC

So, the charges are + 74.3 μC and - 74.3 μC

Your friend would have a better experience of this event, than you .

An eclipse is produced when a planetary body moves in front of another planetary body and is visible from a third planetary body. Considering the sun, moon, and earth's locations in relation to one another during the time of the eclipse,

there are various types of eclipses in our solar system. For instance, a lunar eclipse occurs when the earth passes between the moon and the sun.

For the solar eclipse to happen the light from the sun is obstructed by the moon observing from the earth.

The buddies left Earth because they could view the whole eclipse, but you were on the moon and only saw parts of the eclipse turn black.

To learn more about the eclipse from here, refer to the link;

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The magnitude of the forces acting at the top are;

[tex]\mathbf{F_{Top, \ x}}[/tex] = 132.95 N

[tex]\mathbf{F_{Top, \ y}}[/tex] = 0

The magnitude of the forces acting at the bottom are;

[tex]\mathbf{F_{Bottom, \ x}}[/tex] = [tex]\mathbf{ F_f}[/tex] = -132.95 N

[tex]\mathbf{F_{Bottom, \ y}}[/tex] = 784.8 N

The known parameters in the question are;

The mass of the person, m₁ = 70.0 kg

The length of the ladder, l = 6.00 m

The mass of the ladder, m₂ = 10.0 kg

The distance of the base of the ladder from the house, d = 2.00 m

The point on the roof the ladder rests = A frictionless plastic rain gutter

The location of the center of mass of the ladder, C.M. = 2 m from the bottom of the ladder

The location of the point the person is standing = 3 meters from the bottom

g = The acceleration due to gravity ≈ 9.81 m/s²

The required parameters are;

The magnitudes of the forces on the ladder at the top and bottom

The strategy to be used;

Find the angle of inclination of the ladder, θ

At equilibrium, the sum of the moments about a point is zero

The angle of inclination of the ladder, θ = arccos(2/6) ≈ 70.53 °C

Taking moment about the point of contact of the ladder with the ground, B gives;

[tex]\sum M_B[/tex] = 0

Therefore;

[tex]\sum M_{BCW}[/tex] = [tex]\sum M_{BCCW}[/tex]

Where;

[tex]\sum M_{BCW}[/tex] = The sum of clockwise moments about B

[tex]\sum M_{BCCW}[/tex] = The sum of counterclockwise moments about B

Therefore, we have;

[tex]\sum M_{BCW}[/tex] = 2  × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81

[tex]\sum M_{BCCW}[/tex] = [tex]F_R[/tex] × √(6² - 2²)

Therefore, we get;

2  × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81  = [tex]F_R[/tex] × √(6² - 2²)

[tex]F_R[/tex]  = (2  × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81)/(√(6² - 2²)) ≈ 132.95

The reaction force on the wall, [tex]F_R[/tex] ≈ 132.95 N

We note that the magnitude of the reaction force at the roof, [tex]F_R[/tex] = The magnitude of the frictional force of bottom of the ladder on the floor, [tex]F_f[/tex] but opposite in direction

Therefore;

[tex]F_R[/tex] = [tex]-F_f[/tex]

[tex]F_f[/tex] = - [tex]F_R[/tex] ≈ -132.95 N

Similarly, at equilibrium, we have;

∑Fₓ = [tex]\sum F_y[/tex] = 0

The vertical component of the forces acting on the ladder are, (taking forces acting upward as positive;

[tex]\sum F_y[/tex] = -70.0 × 9.81 - 10 × 9.81 + [tex]F_{By}[/tex]

∴ The upward force acting at the bottom, [tex]F_{By}[/tex] = 784.8 N

Therefore;

The magnitudes of the forces at the ladder top and bottom are;

At the top;

[tex]\mathbf{F_{Top, \ x}}[/tex] = [tex]F_R[/tex] ≈ 132.95 N←

[tex]\mathbf{F_{Top, \ y}}[/tex] = 0 (The surface upon which the ladder rest at the top is frictionless)

At the bottom;

[tex]\mathbf{F_{Bottom, \ x}}[/tex] = [tex]F_f[/tex] ≈ -132.95 N →

[tex]\mathbf{F_{Bottom, \ y}}[/tex] = [tex]F_{By}[/tex] = 784.8 N ↑

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Answer:

salt

Explanation:

salt is a component for many acid base reactions