Double the resistance of the resistor by changing it from 10 Ω to 20 Ω. What happens to the current flowing through the circuit?

Due to the net electric flux that is moving outward through the closed box's surface, the box has a negative charge.

Given:-

The electric field is constant over all the faces.

From Gauss law:

[tex]\int \vec E \cdot d\vec A = \dfrac{Q}{e_o}[/tex]

[tex]\vec E =[/tex] An electric field throughout the cube.

[tex]d \vec A =[/tex] Elemental area of the cube surface.

[tex]Q =[/tex] Electric charge around the cube.

[tex]e_o =[/tex]  Electric constant.

[tex]\dfrac{Q}{e_o} = (-15-20-15+20+15+10)\\ \\ = -5A[/tex]

According to Gauss's law, the box contains a negative charge.

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The box's charge type (being positive, negative, or neutral) is determined by the net electric flux's direction. Outward flux signifies positive charge, whilst inward flux indicates a negative charge. Boxes with no net electric flux have no charge.

The determination of whether a closed box contains a positive charge, negative charge, or no charge is dependent on the direction of the net electric flux. If there is net electric flux passing outward through the surface of the box, this indicates the box contains a positive charge. Conversely, if the net electric flux is passing inward through the surface, the box contains a negative charge. Lastly, if there is no net electric flux (flux in cancelled by flux out), it suggests there is no charge in the box.

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Answer:

R

Explanation:

Given that,

Two stage rocket traveling at

V = 1210 m/s with respect to earth

First stage

When fuel Is run out , explosive bolts releases and push rocket backward at speed is

V1 = 40m/s relative to second stage

Therefore

V1 = 40 - V2

The first stage is 3 times as massive as the second stage

I.e Mass of first stage is 3 times the second stage

Let Mass of second stage be

M2 = M

Then, M1 = 3M

Velocity of second stage V2?

Applying conservation of linear momentum

Momentum before explosion = momentum after explosion

Momentum p=mv

Then,

(M1+M2)V = —M1•V1 + M2•V2

(3M+M)•1210 = —3M•(40-V2) +M•V2

4M × 1210 = —120M + 3M•V2 + MV2

4840M = —120M + 4M•V2

4840M + 120M = 4M•V2

4960M = 4M•V2

Then, V2 = 4960M / 4M

V2 = 1240 m/s

Answer:

R₁ = 50.77 Ω

Explanation:

Since, we know that:

Electric Power = P = VI

but from Ohm's Law:

V = IR

(or) I = V/R

Therefore,

P = V²/R

(OR) R = V²/P

where,

V = Battery Voltage

R = Resistance of combination

FOR SERIES COMBINATION:

R = Rs = (57 V)²/48 W

Rs = 67.69 Ω

but, we know that:

Rs = R₁ + R₂

R₁ + R₂ = 67.69 Ω

R₁ = 67.69 Ω - R₂  __________ eqn (1)

FOR PARALLEL COMBINATION:

R = Rp = (57 V)²/256 W

Rp = 12.69 Ω

but, we know that:

Rp = (R₁R₂)/(R₁ + R₂) = 12.69 Ω

using eqn (1) and value of R₁ + R₂, we get

Rp = 12.69  = R₂(67.69 - R₂)/67.69

859.08 = 67.69 R₂ - R₂²

R₂² - 67.69 R₂ + 859.08 = 0

Solving this quadratic equation we get the answers:

Either, R₂ = 50.76 Ω

Either, R₂ = 16.92 Ω

Since, it is stated in the question that R₁ > R₂. Therefore, we choose the second value. So,

R₂ = 16.92 Ω

using this value in eqn (1), we get:

R₁ = 67.69 Ω - 16.92 Ω

R₁ = 50.77 Ω

Final answer:

To calculate R1, first, find the total resistance for series (R1 + R2) and parallel (1/R1 + 1/R2) connections using the power formula P = V^2 / R, with provided power values for each case. Then solve the simultaneous equations.

Explanation:

The problem can be solved by using the formulas for resistances in series and parallel, as well as the formula for the power supplied by the battery.

In series, the resistances add up: Rtotal = R1 + R2. The power provided by the battery to the series circuit is given by P = V2 / Rtotal, where P is 48.0 W and V is 57.0 V. Rearranging to solve for Rtotal, we get Rtotal = V2 / P.

For parallel resistances, we have 1/Rtotal = 1/R1 + 1/R2. The power in the parallel case is P = V2 / Rtotal. With V and P given as 57.0 V and 256 W, respectively, we can solve for Rtotal.

With both Rtotal values calculated, we can set up two equations with two unknowns (R1 and R2), knowing that R1 > R2. Solving these simultaneous equations gives us the value of R1.

Answer:

(a)  277.05 Ω

(b) 15.39 Ω

(c) 3.76 W

Explanation:

(a)

Applying,

P = V²/R.......................... Equation 1

Where P = Power dissipated by the string. V = Voltage source, R = equivalent resistance of the light string

Make R the subject of the equation

R = V²/P................... Equation 2

Given: V = 130, P = 61  W

Substitute into equation 2

R = 130²/61

R = 277.05 Ω

(b) The resistance of a single light is given as

R' = R/18 (since the light are connected in series and the are identical)

Where R' = Resistance of the single light.

R' = 277.05/18

R' = 15.39 Ω

(c)

Heat dissipated in a single light is given as

P' = I²R'..................... Equation 3

Where P' = heat dissipated in a single light, I = current flowing through each light.

We can calculate for I using

P = VI

make I the subject of the equation

I = P/V

I = 61/130

I = 0.469 A.

Also given: R' = 15.39 Ω

Substitute into equation 3

P' = 0.496²(15.39)

P' = 3.76 W

(a)The equivalent resistance of the light string is 277.05 Ω

(b)The power is dissipated in a single ligh15.39 Ω

(c)The resistance of the light string now3.76 W

(a) P is = V²/R.......................... Equation 1

Where P is = Power dissipated by the string.

Then V = Voltage source,

After that R is = the equivalent resistance of the light string

Now Make R the subject of the equation

R is = V²/P................... Equation 2

Then Given: V = 130,

P = 61 W

After that Substitute into equation 2

Then R = 130²/61

Therefore, R = 277.05 Ω

(b) When The resistance of a single light is given as

R' is = R/18 (since the light are connected in series and are identical)

Now Where R' is = Resistance of the single light.

R' is = 277.05/18

Therefore, R' is = 15.39 Ω

(c) When Heat dissipated in a single light is given as

P' is = I²R'..................... Equation 3

Where P' is = heat dissipated in a single light,

Then I = current flowing through each light.

Now We can calculate for I using

P is = VI

Now we make I the subject of the equation

After that I = P/V

Then I = 61/130

I is = 0.469 A.

Also given: R' is = 15.39 Ω

Then Substitute into equation 3

P' is = 0.496²(15.39)

Therefore, P' is = 3.76 W

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The electric potential at point A in the electric field= 0.099 x 10 ⁻¹v

Explanation:

Given data,

charge = 5.5 x 10¹² C

k =9.00 x 10⁹

The electric potential V of a point charge can found by,

V= kQ / r

Assuming, r=5.00×10⁻² m

V= 5.5 x 10⁻¹²C x  9.00 x 10⁹ / 5.00×10⁻² m

V=  49.5 x 10⁻³/ 5.00×10⁻²

Electric potential V=  0.099 x 10⁻¹v

The electric potential at a point in the electric field created by a point charge is calculated by the formula V = kQ/r. The provided constants are [tex]Q = 5.5 \times 10^{-12}\, \text{C}[/tex] and [tex]k = 9.00 \times 10^9[/tex] The distance r from the point charge is necessary to calculate the electric potential.

The electric potential at point A in the electric field created by a point charge can be found using the formula V = kQ/r, where V is the electric potential, k is Coulomb's constant, Q is the value of the point charge, and r is the distance from the point charge.

Given, the charge [tex]Q = 5.5 \times 10^{-12}\, \text{C}[/tex], and Coulomb's constant [tex]k = 9.00 \times 10^9[/tex]

The distance r is not given in the question. Therefore, we can't calculate the electric potential at point A. However, if distance r from the point charge were provided (in meters), we could substitute k, Q, and r into the formula and solve for V.

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Answer:

Catastrophic events of weather related outcomes.

Explanation:

The mental map of a person from a certain place may change due to long periods of time outside, since this gives the mind time to forget certain important details, they may also change according to people's experience and perception, places, regions and environments since these places are also changing and the perception we had is no longer the same. A mental map is a first person perspective of an area that an individual possesses. This type of subconscious map shows a person how a place looks and how to interact with it.

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Answer:

The radius of curvature of the ion's orbit is 0.59 meters

Explanation:

Given that,

Mass of the 24 Mg ion, [tex]m=3.983\times 10^{-26}\ kg[/tex]

Potential difference, V = 3 kV

Magnetic field, B = 526 G

Charge on single ionized ion, [tex]q=1.6\times 10^{-19}\ C[/tex]

The radius of the the path traveled  by the charge is circular. Its radius is given by :

[tex]r=\dfrac{mv}{Bq}[/tex]

v is speed of particle.

v can be calculated using conservation of energy as :

[tex]\dfrac{1}{2}mv^2=qV\\\\v=\sqrt{\dfrac{2qV}{m}} \\\\v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^3}{3.983 \times 10^{-26}}} \\\\v=1.26\times 10^5\ m/s[/tex]

Radius,

[tex]r=\dfrac{3.983 \times 10^{-26}\times 1.26\times 10^5}{0.0526\times 1.6\times 10^{-19}}\\\\r=0.59\ m[/tex]

So, the radius of curvature of the ion's orbit is 0.59 meters.

To find the time it takes for the current to drop to 12% of its value and the energy dissipated by the circuit, we can use the equations for the decay of current in an RL circuit and the energy in an inductor respectively.

To determine the time it takes for the current in the circuit to drop to 12% of its value at t = 0, we need to use the equation for the decay of current in an RL circuit. The equation is given by I(t) = I(0) * exp(-t/τ), where I(t) is the current at time t, I(0) is the initial current, t is the time, and τ is the time constant. In this case, we can use I(t) = I(0) * exp(-t/τ) = 0.12 * I(0). To find the time, we rearrange the equation as t = -τ * ln(0.12).

The energy dissipated by the circuit can be calculated using the equation for the energy in an inductor, which is given by E = 1/2 * L * I(0)^2, where E is the energy, L is the inductance, and I(0) is the initial current. In this case, we can substitute the values given to find the energy dissipated.

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The time needed for the current in the circuit to drop to 12% of its value at t = 0 is approximately 20.63 ms. The energy dissipated in the circuit during that time is approximately 7.22 mJ.

Given, initial current I0 can be obtained from Ohm's law, I0 = V / R = 108 / 1100 = 0.09818 A. The circuit value diminishes exponentially over time and can be expressed as I = I0 * e-Rt/2L. Therefore, to find out when the current drops to 12% of its initial value we set I / I0 = 0.12 = e-Rt/2L, and solve for t. After calculation, we find t ≈ 20.63 ms.

As for part (b), the energy dissipated in an RL circuit over time is given by W = 1/2 * L * I2, where I is the current at time t, which is given by the relation: I = I0 * e-Rt/2L. Performing the integration over the time period t = 0 to 20.63 ms, we find that the energy dissipated is approximately 7.22 mJ.

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Answer:

[tex]6.75\mu C/m^2[/tex]

Explanation:

We are given that

Diameter,d=[tex]1\mu m=1\time 10^{-6} m[/tex]

[tex]1\mu m=10^{-6} m[/tex]

Radius,r=[tex]\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m[/tex]

Density,[tex]\rho=900kg/m^3[/tex]

Total number of electrons,n=39

Charge on electron =[tex]1.6\times 10^{-19} C[/tex]

Total charge=[tex]q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C[/tex]

Distance,s=2mm=[tex]2\times 10^{-3} m[/tex]

Mass =[tex]density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg[/tex]

Initial velocity,u=0

Final speed,v=4.5 m/s

[tex]v^2-u^2=2as[/tex]

[tex](4.5)^2-0=2a(2\times 10^{-3})[/tex]

[tex]20.25=4a\times 10^{-3}[/tex]

[tex]a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2[/tex]

Force,F=ma

[tex]qE=ma[/tex]

[tex]q(\frac{\sigma}{2\epsilon_0})=ma[/tex]

[tex]\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}[/tex]

[tex]\epsilon_0=8.85\times 10^{-12}[/tex]

[tex]\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2[/tex]

Wind is driven due to the differences in air pressure around the Earth.

To find the answer, we need to know about the wind flow.

Thus, we can conclude that wind is driven due to the differences in air pressure around the Earth.

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Answer:

a)  the geometric optics is adequate (Ray)

b) wave optics should be used for the second case

Explanation:

In general, the approximation of the geometric optics is adequate when the dimensions of the system are much greater than the wavelengths and the wave optics should be used for cases in which the size of the system is from the length of the wave.

Let's apply this to our case

a) in this case the size of the system is d = 26 cm=0.26 m and the wavelength is alm = 530 10⁺⁹ m

   in this case

                    d >>> λ

therefore the geometric optics is adequate (Ray)

b) in this case the system has a size of d = 114 10⁻⁹ m with a wavelength of λ= 722 nm

for this case d of the order of lam

therefore wave optics should be used for the second case

Answer: 0.666 V

Explanation:

Given

Radius of the loop, r = 14.8 cm = 0.148 m

Magnetic field present, B = 0.814 T

Rate of shrinking, dr/dt = 88.4 cm/s = 0.884 m/s

emf = dΦ/dt , where Φ = BA

emf = d(BA)/dt, where A = πr²

emf = d(Bπr²)/dt

if B is constant, then

emf = Bπ d(r²)/dt, on differentiating, we have,

emf = Bπ * 2r dr/dt

emf = 2πrB dr/dt, now if we substitute the values, we have

emf = 2 * 3.142 * 0.148 * 0.814 * 0.884

emf = 6.284 * 0.106

emf = 0.666 V

To solve the problem it is necessary to apply Ohm's law. From this it is established that the voltage is the equivalent to the product between the current and the resistance, therefore we have to,

[tex]V = IR[/tex]

Here,

V = Voltage

I = Current

R = Resistance

Rearranging to find the resistance,

[tex]R = \frac{V}{ I}[/tex]

Replacing,

[tex]R = \frac{100V}{20A}[/tex]

[tex]R = 5\Omega[/tex]

Therefore the resistance should be [tex]5\Omega[/tex]

The resistance should be [tex]\bold { 5\Omega}[/tex] in order to limit the emf to 100 V.

Ohm's law:

It states that the voltage is the equival to the product of the current and the resistance.

[tex]\bold{V =I\times R}[/tex]

Where,

V = Voltage = 100 Volts

I = Current  = 20 Ampere

R = Resistance

Put the values and solve it for R

[tex]\bold {R =\dfrac{100}{20}}\\\\\bold {R = 5\Omega}[/tex]

Therefore, the resistance should be [tex]\bold { 5\Omega}[/tex] in order to limit the emf to 100 V.

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Answer:

larger, because her angular speed is larger.

Explanation:

The rotational kinetic energy is proportional to the square of the angular velocity while it is linearly proportional to the moment of inertia. So the increase of angular speed will have a larger effect of the kinetic energy than the decrease of the moment of inertia.

Answer:

Rotational kinetic energy will Increase

Explanation:

Rotational kinetic energy KE is

KE = 1/2 x I x w^2

Where I is moment of inertia,

w is angular velocity.

It can be seen that increasing angular velocity increases rotational kinetic energy.

Answer:

Explanation:

In a scenario where a particle of charge overrightarrow{B} enters a magnetic field with a velocity overrightarrow{V}, it experiences a force overrightarrow{F} given by: overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B}).

implies F=BqVSin\theta.

Where theta is the angle between the velocity vector of the particle and the magnetic field vector.

When a particle enters the magnetic field at an angle 90, it moves in a circular path as it experiences a centripetal force, given by F=\frac{mV^2}{R}.

Where R is the radius of the circle, V is its velocity and m is its mass

Thus, magnetic force becomes F=BqVSin90^o=\frac{mV^2{R}\implies R=\frac{mV}{Bq}.

The equation changes as below, when velocity is doubled, let us assume that the radius is given by R_1.

R_1=\frac{2mV}{Bq}=2R.

Therefore, it is obvious that the velocity of a charged particle in a circular arc is directly proportional to the radius of the arc. The radius of the circular arc doubles when the velocity of the charged particle in the circular orbit doubles only if the mass, charge and magnetic field of the particle remains constant

Hence when velocity is doubled radius of the circle also gets doubled.

When the speed of the particle doubles, the radius of the arc also doubles.

The magnetic force on the particle is calculated as follows;

[tex]F = qvB[/tex]

The centripetal force on the particle is calculated as follows;

[tex]F_c = \frac{mv^2}{R}[/tex]

The speed of the particle is calculated as follows;

[tex]\frac{mv^2}{R} = qvB\\\\mv = qBR\\\\v = \frac{qBR}{m} \\\\\frac{v_1}{R_1} = \frac{v_2}{R_2}[/tex]

when the speed of the particle doubles;

[tex]\frac{v_1}{R_1} = \frac{2v_1}{R_2} \\\\R_2v_1 = 2R_1v_1\\\\R_2 = 2R_1[/tex]

Thus, we can conclude that when the speed of the particle doubles, the radius of the arc also doubles.

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Answer:

[tex]\frac{g_{2}}{g_{1}} = \frac{1}{4}[/tex]

Explanation:

The period of the simple pendulum is:

[tex]T = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex]

Where:

[tex]l[/tex] - Cord length, in m.

[tex]g[/tex] - Gravity constant, in [tex]\frac{m}{s^{2}}[/tex].

Given that the same pendulum is test on each planet, the following relation is formed:

[tex]T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}[/tex]

The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:

[tex]\frac{g_{2}}{g_{1}} = \left(\frac{T_{1}}{T_{2}} \right)^{2}[/tex]

[tex]\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}[/tex]

[tex]\frac{g_{2}}{g_{1}} = \frac{1}{4}[/tex]

To solve this problem we will apply the concepts related to angular resolution based on wavelength and the slit width at this case of the cat's pupils. This relationship is given as,

[tex]\theta = \frac{\lambda}{a}[/tex]

Here,

[tex]\lambda[/tex] =  Wavelength

a = Slit width

[tex]\theta = \frac{519*10^{-9}m}{0.55*10^{-3}m}[/tex]

[tex]\theta = 9.43*10^{-4} Rad[/tex]

Therefore the angular resolution is [tex]9.43*10^{-4}rad[/tex]

Final answer:

The angular resolution for a cat observing birds can be estimated using the formula θ = 1.22 λ / D, with a pupil width of 0.550 mm and a wavelength of 519 nm, yields an angular resolution of approximately 1.152 × 10⁻⁶ radians.

Explanation:

The question refers to angular resolution in optics, specifically related to the diffraction limit of a cat's eyes observing birds. In physics, the angular resolution for a circular aperture, like the pupil of a cat's eye, can be estimated using the formula θ = 1.22 λ / D, where θ is the angular resolution in radians, λ is the wavelength of the light, and D is the diameter of the aperture (the pupil in this case). Given that the cat's pupil has a slit width of a = 0.550 mm (which we will use in place of the diameter for this rough estimate) and the light has an average wavelength of λ = 519 nm, the angular resolution θ can be calculated as follows:

θ = 1.22 × 519 x 10⁻⁹ m / 0.550 x 10^-3 m

θ = 1.22 × 519 / 550 × 10⁻⁶

θ = 1.22 × 0.944 × 10⁻⁶

θ = 1.152 × 10⁻⁶ radians

Given the specified conditions, this calculation estimates the angular resolution for the cat's eyes. Note, however, that this is a simplification and doesn't take into account the complexities of a vertical slit pupil versus a circular aperture.

Answer: 2.44*10^3 km

Explanation:

NOTE: We would be solving this question in "cm" instead of the usual "m"

1 mi² = 2.59 km²

2.59 km² = 2.59*10^10 cm²

Given, area of South Dakota

A = 7.588*10^4 mi²

A = 7.588*10^4 * 2.59*10^10

A = 1.97*10^15 cm² is the area of South Dakota

S = πd²/4

S = 3.142 * 0.635² / 4

S = 3.142 * 0.4/4

S = 3.142 * 0.1

S = 0.3142 cm² is the area of 1 marshmallow

1.97*10^15 / 0.3142 = 6.27*10^15, thus is the number of marshmallows in one single layer to cover the state area

6.02*10^23 / 6.27*10^15 = 9.6*10^7, this is the number of layers

9.6*10^7 * 2.54 = 2.44*10^8, this is the height in cm

Height in km is 2.44*10^3 km

Answer:

The magnitude of the magnetic force on the particle is 67.86 N.          

Explanation:

Given that,

Charge, [tex]q=34.6\ \mu C=34.6\times 10^{-6}\ C[/tex]

Speed of particle, v = 68.4 m/s in +x direction

Magnetic field, [tex]B=(0.466 j+0.876 k)\ T[/tex]

We need to find the magnitude of the magnetic force on the particle. The magnetic force is given by :

[tex]F=q(v\times B)\\\\F=34.6\times 10^{-6}(68.4i+0+0\times (0+0.466j+0.876 k))\\\\F=\begin{pmatrix}0&-59.9184&31.8744\end{pmatrix}\\\\F=(-59.9184j+31.8744k)\ N[/tex]

The magnitude of magnetic force on the particle is :

[tex]|F|=\sqrt{(-59.9184)^2+(31.8744)^2} \\\\|F|=67.86\ N[/tex]

So, the magnitude of the magnetic force on the particle is 67.86 N.          

Answer:

1. downward

2. to the left

3. downward and to the left

Explanation:

This is gotten by using vector law of triangle Addition which states that If 2 vectors acting simultaneously on a body are represented both in magnitude and direction by 2 sides of a triangle taken in an order then the resultant(both magnitude and direction) of these vectors is given by 3rd side of that triangle taken in opposite order.

Light is incident in air (n = 1) and refracted into a medium with an index of refraction of 2. Light incident parallel to the central axis is focused at a point 24cm from the surface

When a light  ray passes through from one medium to another medium ,due to the difference in medium index, the light deflect from its original path .

This deflection  from its path is called refraction and The phenomenon of refraction can be seen in sound and water waves.

Cornea reflect the light before light reaches to the retina. Change in the speed of light when it travels from one medium to other medium then due to this there is change in the path of light, this phenomenon is known as refraction of light.

Equation for refraction

n2/v- n1/u = n2-n1/R

= u = infinity ( parallel)

2/v= (2-1)/R

V = 2R

V= 24cm

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Answer:

Look up attached file

Explanation:

Final answer:

To hold the projectile in position, the force required is 20 Newtons.

Explanation:

To calculate the force required to hold the projectile in position, we need to consider the acceleration of the projectile in the pipe.

From the given information, we can calculate the acceleration using the formula:

Acceleration = Change in velocity / Time taken

Since the mean velocity is given as 6 m/s, we can use this as the final velocity and assume the initial velocity is 0 m/s.

Substituting the values in the formula, we have:

Acceleration = (6 m/s - 0 m/s) / 0.3 m = 20 m/s^2

Now we can calculate the force using Newton's second law:

Force = Mass * Acceleration

Assuming the mass of the projectile is 1 kg, we have:

Force = 1 kg * 20 m/s^2 = 20 N

The maximum possible wavelength for constructive interference from two shifted loudspeakers at given interference maxima positions is calculated to be 0.5738 m.

The question is about the phenomenon of constructive interference of sound waves emitted from two loudspeakers that are shifted apart. The problem requires identifying the maximum possible wavelength of the sound waves produced by the speakers.

Constructive interference occurs when two waves meet and their crests (high points) and troughs (low points) align. This occurs effectively when the path difference is either zero or a multiple of the wavelength. When the path difference is specifically an integral multiple of the wavelength, the interference is constructive and has a maximum value.

Given that the loudspeakers are moved to a distance of 3.0 m sideways and 7.0 m forward, the distance separating the two is calculated using Pythagoras's theorem as √((7)^2+(3)^2) = √58 m. The maxima of the constructive interference are at positions 14, 12, and 34 the distance between the speakers, thus the maximum possible wavelength is 2 * √58 / 34 = 0.5738 m.

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Answer:

L=N/I*magnetic flux

=N/I*BA=N*mu(0)NIA/2pi*R=mu(0)N^2/2pi*R)*Pi*a^2

=mu(0)N^2a^2/2R

Explanation:

The self-inductance L of the toroid is mu(0)N^2a^2/2R.

Since

The toroidal inductor has a circular cross-section of radius a. The toroid has N turns and radius R.

The toroid is narrow ( a≪R )

So,

L=N/I*magnetic flux

=N/I*BA=N*mu(0)NIA/2pi*R=mu(0)N^2/2pi*R)*Pi*a^2

=mu(0)N^2a^2/2R

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The energy of a photon is the product of the planks constant and frequency. The photon energy of this light is 151 keV.

It is the product of the planks constant and frequency.

[tex]E = \dfrac {hc}\lambda[/tex]

Where,

[tex]h[/tex] -Plank's constant  = [tex]\bold {6.63\times 10^{-34}\rm \ J/s}[/tex]

[tex]c[/tex]- speed of light = [tex]\bold { 3\times 10^8 \rm \ m/s }[/tex]

[tex]\lambda[/tex][tex]\labda[/tex]- wavelength = 8.2 pm = [tex]\bold {8.2 \times 10^{-12 }\rm \ m}[/tex]

Put the values in the equation,

[tex]E = \dfrac { 6.63\times 10^{-34}\rm \ J/s\times 3\times 10^8 \rm \ m/s }{ 8.2 \times 10^{-12 }\rm \ m}\\E = 151 \rm \ keV[/tex]

Therefore, the photon energy of this light is 151 keV.

Learn more about photon energy:

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Answer: 13 cm

Explanation:

Given

Inductance of the solenoid, L = 3.01•10⁻⁶ H

Width of the wire, x = 1.13 m

Length of the tube, z = ?

Now, we know that

L = μ₀N²A/z, where

N = x/2πr, making r subject of formula,

r = x/2πN

Also,

A = πr², on substituting for A, we have

A = πx²/4π²N²

Now finally, we substitute in the initial equation and solve

L = μ₀N²A/z

L = μ₀N²πx²/4π²N²z

L = μ₀x²/4z, making z subject of formula, we have

z = μ₀x²/4L

z = 4π*10⁻⁷ * 1.13² /4 * 3.01*10⁻⁶

z = (4π*10⁻⁷ * 1.2769) / (4 * 3.01*10⁻⁶)

z = 1.6*10^-6 / 1.2*10^-5

z = 0.13 m

Therefore, the length of the tube should be 13 cm

The length of the tube should be 469 centimeters.

To find the required length of the tube for Tia's solenoid, we can use the formula for the inductance of a solenoid:

[tex]\[ L = \frac{{\mu_0 \cdot N^2 \cdot A}}{l} \][/tex]

Where:

L = inductance of the solenoid (3.01 μH)

[tex]\( \mu_0 \)[/tex] = permeability of free space (4π × 10^-7 T*m/A)

N = number of turns of wire

A = cross-sectional area of the solenoid

[tex]\( l \)[/tex] = length of the solenoid

We're given [tex]\( L \)[/tex] and [tex]\( N \cdot l \)[/tex] (the total length of wire wound around the tube). We need to find [tex]\( l \)[/tex], the length of the tube.

First, let's rearrange the formula to solve for [tex]\( l \)[/tex]:

[tex]\[ l = \frac{{\mu_0 \cdot N^2 \cdot A}}{{L}} \][/tex]

We know the cross-sectional area ( A ) can be represented as [tex]\( A = \pi r^2 \)[/tex], where ( r ) is the radius of the tube. Since the wire is uniformly wound, we can calculate ( A ) using the length of the wire and the number of turns.

[tex]\[ A = \frac{{\text{{Length of wire}}}}{{\text{{Number of turns}}}} \][/tex]

Given that the wire length is 1.13 m and there are [tex]\( N \)[/tex] turns, [tex]\( A = \frac{{1.13}}{{N}} \)[/tex].

Substituting this into the equation for [tex]\( l \)[/tex], we get:

[tex]\[ l = \frac{{\mu_0 \cdot N^2 \cdot \left(\frac{{1.13}}{{N}}\right)}}{{L}} \][/tex]

[tex]\[ l = \frac{{\mu_0 \cdot 1.13}}{{L}} \][/tex]

Now, plug in the known values:

[tex]\[ l = \frac{{4\pi \times 10^{-7} \cdot 1.13}}{{3.01 \times 10^{-6}}} \][/tex]

[tex]\[ l = \frac{{4\pi \times 1.13}}{{3.01}} \][/tex]

[tex]\[ l = \frac{{4 \times 3.1416 \times 1.13}}{{3.01}} \][/tex]

[tex]\[ l = \frac{{14.1376}}{{3.01}} \][/tex]

[tex]\[ l = 4.69 \, \text{m} \][/tex]

Finally, convert this length to centimeters:

[tex]\[ l = 469 \, \text{cm} \][/tex]

So, the length of the tube should be 469 centimeters.

Depending on its size, composition, and the eccentricity of its orbit, that scanty description could apply to a planet, an asteroid, a comet, a meteoroid, or another star.

Answer:

Explanation: check the attached document for step by step solution.