Characteristics help scientists ________ objects.

Answer:

Explanation:

energy emitted by  source per second  = .5 J

Eg = 1.43 eV .

Energy converted into radiation = .5 x .12 = .06 J

energy of one photon = 1.43 eV

= 1.43 x 1.6 x 10⁻¹⁹ J

= 2.288 x 10⁻¹⁹ J .

no of photons generated = .06 / 2.288 x 10⁻¹⁹

= 2.6223 x 10¹⁷

wavelength of photon λ = 1275 / 1.43 nm

= 891.6 nm .

momentum of photon = h / λ  ;  h is plank's constant

= 6.6 x 10⁻³⁴ / 891.6 x 10⁻⁹

= .0074 x 10⁻²⁵ J.s

Total momentum of all the photons generated

= .0074 x 10⁻²⁵  x 2.6223 x 10¹⁷

= .0194 x 10⁻⁸ Js

b ) spectral width in terms of wavelength = 30 nm

frequency width = ?

n = c / λ  , n is frequency , c is velocity of light and λ is wavelength

differentiating both sides

dn = c x dλ / λ²

given dλ = 30 nm

λ = 891.6 nm

dn = 3 x 10⁸ x 30 x 10⁻⁹ / ( 891.6  x 10⁻⁹ )²

= 11.3 x 10¹² Hz .

c )

10 nW = 10  x 10⁻⁹ W

= 10⁻⁸ W .

energy of 50 dB

50 dB = 5 B

I / I₀ = 10⁵   ;   decibel scale is logarithmic , I is energy of sound having dB = 50 and  I₀ = 10⁻¹² W /s

I = I₀ x 10⁵

= 10⁻¹² x 10⁵

= 10⁻⁷ W

= 10 x 10⁻⁸ W

power required

= 10⁻⁸ + 10 x 10⁻⁸ W

= 11  x 10⁻⁸ W.

Final answer:

The question involves calculating the wavelength and diffraction patterns of electrons in a single slit experiment. The student would use the de Broglie wavelength formula and the principles of single slit diffraction to find the dimensions of the central maximum. It demonstrates the wave-particle duality exhibited by electrons.

Explanation:

The student is seeking to understand how to find the wavelength and diffraction patterns of electrons. The context involves an electron beam passing through a single slit, creating a diffraction pattern on a distant screen. From the given potential difference, one could calculate the de Broglie wavelength of the electrons, as their velocity can be determined under the assumption that they are nonrelativistic. The central maximum's width on the diffraction pattern can be deduced using the slit width and the wavelength.

The phenomenon of diffraction and interference highlighted in the question is a demonstration of the wave-like properties of electrons, referred to as wave-particle duality. The experimental setup often includes narrow slits whose sizes are comparable to the wavelength of electrons, resulting in observable wave effects such as constructive and destructive interference.

The angular positions of the minima and maxima in the diffraction pattern are crucial for determining the dimensions of the pattern. The de Broglie wavelength plays a significant role in these calculations, linking the microscopic quantum world to observable macroscopic patterns.

Determining the radius of the satellite TV dish to achieve a specific electric field requires knowing the intensity and power of the broadcast signal as well as the dish's area. The radius can be calculated from the formula for the area encompassed by the dish, but without additional information about the broadcast power or spread area, a specific radius cannot be provided.

To determine how large the radius R of the satellite TV receiver dish must be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver, we need to relate the intensity of the electromagnetic wave to the electric field amplitude and the area of the dish. The intensity (I) is related to the electric field strength (E) by the relationship I =  rac{1}{2} extZ_0[tex]E^2[/tex], where Z0 is the impedance of free space (approximately 377 ohms).

The power received by the dish (Pr) is the product of the intensity and the area of the dish, Ad: Pr = I  imes Ad. Given that the receiver has an area of 5 cm2 and the required electric field amplitude is 0.1 mV/m, we would solve for the radius R of the dish using the formula for area of a circle, A = \\(pi)[tex]R^2[/tex].

However, to solve this problem, we would need additional information such as the power broadcast by the satellite and over what area this power is spread. With our current information, we cannot provide an exact answer, but typically residential satellite dishes have diameters a little less than half a meter to effectively receive TV signals.

Answer:

8cm

Explanation:

Here, two disc are identical and rolling on the horizontal surface

Also,while disc is in rolling motion its kinetic energy is sum of rotational kinetic energy and transnational kinetic energy.

Therefore,

KE = [tex]\frac{1}{2}[/tex]mv²+ [tex]\frac{1}{2}[/tex][tex]Iw[/tex]²

For pure rolling of disc we have: [tex]v=Rw[/tex]

[tex]I=\frac{1}{2} mR[/tex]²

By substituting in KE eq, we get

KE = [tex]\frac{1}{2}[/tex]mv²+ [tex]\frac{1}{2}[/tex]([tex]\frac{1}{2} mR[/tex]²)([tex]\frac{v^{2} }{R^{2} }[/tex])

KE= [tex]\frac{1}{2} mv^{2} + \frac{1}{4} mv^{2}[/tex]

The total kinetic energy will convert into gravitational potential energy when disc roll over the inclined surface.

mgH=[tex]\frac{1}{2} mv^{2} + \frac{1}{4} mv^{2}[/tex] =>[tex]\frac{3}{4} mv^{2}[/tex]

[tex]mv^{2}[/tex]= 4/3mgH

If another disc rolls up on frictionless inclined plane then it will lose all its translational kinetic energy but rotational kinetic energy will remain as it is as there is no torque on the disc

Therefore, mgh=  1/2mv²

1/2mv²= 4/3mgH

mgh=2/3mgH

h=2/3H

Height = 12cm is given

h= 8cm

Thus,  Disk B reaches a height of 8cm above the floor.

Answer:

Magnitude of magnetic field is 1.5 x 10^(-8) T in the positive y-direction

Explanation:

From maxwell's equations;

B = E/v

Where;

B is maximum magnitude of magnetic field

E is maximum electric field

v is speed of light which has a constant value of 3 x 10^(8) m/s

We are given, E = 4.5 V/m

Thus; B = 4.5/(3 x 10^(8))

B = 1.5 x 10^(-8) T

Now, for Electric field, vector E to be in the positive x-direction, the product of vector E and vector B will have to be in the positive z-direction when vector B is in the positive y-direction

Thus,

Magnitude of magnetic field is 1.5 x 10^(-8) T in the positive y-direction

Magnitude of magnetic field in the space at given instant in time is [tex]\bold { 1.5 x 10^{-8}\ T}[/tex]  in the positive y-direction.

From Maxwell's equations,

[tex]\bold {B = \dfrac Ev}[/tex]

Where;

B - maximum magnitude of magnetic field = ?

E- maximum electric field  = 4.5 V/m

v- speed of light =  3 x 10^(8) m/s

Put the values in the formula,

[tex]\bold {B = \dfrac {4.5}{3 x 10^8}}\\\\\bold {B = 1.5 x 10^{-8}\ T}[/tex]

When Electric field, is in the positive x-direction, vector B is in the positive y-direction and  the product of vector E and vector B will have to be in the positive z-direction.

Therefore, magnitude of magnetic field in the space at given instant in time is [tex]\bold { 1.5 x 10^{-8}\ T}[/tex]  in the positive y-direction.

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The largest mass that block C can have so that blocks A and B still slide together, given the values, is approximately 10.2 kg.

The subject of this question is a physics problem involving static friction, mass, and gravity. The largest mass that block C can have so that blocks A and B still slide together can be calculated using the principle of static friction and the relevant equation:

fs_max = µs (mA + mB)g, with fs_max representing the maximum force of static friction, µs=0.75 being the coefficient of static friction, and g=9.8 m/s² being the acceleration due to gravity. Here mA=8 kg and mB=5 kg are the masses of block A and B respectively.

To keep A and B together, the tension (T) in the string must be less than or equal to fs_max. As T is also equal to the weight (mCg) of block C, from where we can find mC ≤ fs_max / g. Substituting the given values and calculations, we find that the maximum mass of block C should be around 10.2 kg for blocks A and B to slide together.

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Final answer:

The maximum mass that block C can have to prevent sliding between block A and B is calculated using the static friction coefficient [tex](\mu_s)[/tex] and the weight of block B [tex](m_B)[/tex], which results in a mass of 3.75 kg for block C.

Explanation:

To figure out the largest mass that block C can have without block A and B sliding relative to each other, we must use the static friction force and Newton's second law. The static friction force [tex](f_s)[/tex] is what keeps block B from sliding on block A.

This force is given by [tex]f_s = \mu_s \times N[/tex], where μ_s is the coefficient of static friction and N is the normal force. Since block B is at rest on block A, N is equal to the weight of block B, [tex]N = m_B \times g[/tex].

For block A and block B to accelerate together, the tension in the string (T) caused by the weight of block C must not exceed the static friction force. Therefore, the maximum force that can be applied by block C before block B starts sliding is the static friction force: T ≤ f_s.

So, the maximum weight (and hence mass) that block C can have is when [tex]T = f_s[/tex]. Since [tex]T = m_C \times g[/tex] for block C, we have m_C × g = μ_s × m_B × g. Canceling out g and solving for m_C gives us the formula [tex]m_C = \mu_s \times m_B[/tex].

Substituting the given values, we get [tex]m_C = 0.750 \times 5.00 kg = 3.75 kg[/tex]. Hence, the largest mass that block C can have without causing sliding between block A and B is 3.75 kg.

Answer:

Explanation:

Given a particle of mass

M = 1.7 × 10^-3 kg

Given a potential as a function of x

U(x) = -17 J Cos[x/0.35 m]

U(x) = -17 Cos(x/0.35)

Angular frequency at x = 0

Let find the force at x = 0

F = dU/dx

F = -17 × -Sin(x/0.35) / 0.35

F = 48.57 Sin(x/0.35)

At x = 0

Sin(0) =0

Then,

F = 0 N

So, from hooke's law

F = -kx

Then,

0 = -kx

This shows that k = 0

Then, angular frequency can be calculated using

ω = √(k/m)

So, since k = 0 at x = 0

Then,

ω = √0/m

ω = √0

ω = 0 rad/s

So, the angular frequency is 0 rad/s

At 0.108 m distance from the center of the sphere does the electric field have the same magnitude as it has at P.

Given that,

Radius of the sphere R= 0.30 m

Distance from the center of the sphere r= 0.50 m

Electric field = 15000 N/C

r > R and for this value of r.

Let density of charge = [tex]\rho[/tex]

therefore, we have,

[tex]k*\rho*(4/3*(\pi*0.3^3))/0.5^2 = k*\rho*(4/3*(\pi*r^3))/r^2[/tex]

where r is the distance from centre.

Now, we have,

the field inside a sphere is given by [tex]\(kqx/r^3\)[/tex]

the field outside the sphere is given by [tex]\(kq/x^2\)[/tex]

so equating the two equations,

[tex]kq*x/(0.3^3)=kq/(0.5^2)[/tex]

or,  x=0.108m

So, After solving we get distance= 0.108 m.

Hence, At 0.108 m distance from the center of the sphere does the electric field have the same magnitude as it has at P.

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The electric field of a nonconducting sphere is directly proportional to the distance from its center when inside the sphere and inversely proportional to the square of the distance when outside. To have the same magnitude of 15,000 N/C as at point P (0.50m), the point must be outside the sphere at a distance that depends on the sphere's total charge.

The question asked is based on the concept of electric fields in Physics. Firstly, we need to understand that the electric field inside a uniformly charged nonconducting sphere is directly proportional to the distance from the center of the sphere (It follows the equation E = k*r, where E is the electric field, k is a constant, and r is the distance from the center). At the point P (0.50m), the electric field E is given as 15,000 N/C.

So, for the electric field to have the same magnitude at another point, this point must be outside the sphere. This is because the electric field will decrease once we move out of the sphere (Outside the sphere, the electric field falls off as 1/r^2, so to achieve the same magnitude of 15,000 N/C we have to move farther away from the sphere). The exact distance depends on the total charge of the sphere, which is not given in the question.

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The complete question is;

A circular coil consists of N = 410 closely winded turns of wire and has a radius R = 0.75 m. A counterclockwise current I = 2.4 A is in the coil. The coil is set in a magnetic field of magnitude B = 1.1 T.

a. Express the magnetic dipole moment μ in terms of the number of the turns N, the current I, and radius

R.

b. Which direction does μ go?

Answer:

A) μ = 1738.87 A.m²

B) The direction of the magnetic moment will be in upward direction.

Explanation:

We are given;

The number of circular coils;

N = 410

The radius of the coil;R = 0.75m

The current in the coils; I = 2.4 A

The strength of magnetic field;

B =1.1T

The formula for magnetic dipole moment is given as;

μ = NIA

Where;

N is number of turns

I is current

A is area

Now, area; A = πr²

So, A = π(0.75)²

Thus,plugging in relevant values, the magnetic dipole moment is;

μ = 410 * 2.4 * π(0.75)²

μ = 1738.87 A.m²

B) According to Fleming's right hand rule, the direction of the magnetic moment comes out to be in upward direction.

The circular coil with specific parameters immersed in the magnetic field is analyzed according to the principles of electromagnetism. The right-hand rule is used to determine the direction of the magnetic field, and formulas are used to calculate the magnetic force and torque on the coil given the counterclockwise current and magnetic field strength.

In the described case, we are dealing with a circular coil of N=410 turns, with a radius R = 0.75m, carrying a counterclockwise current I = 2.4A. This coil is set in a magnetic field B = 1.1T, that points to the right. When a current flows in a wire, it creates a magnetic field around it. The direction of magnetic field can be determined by the right-hand rule, where your thumb points in the direction of the current and your fingers curl in the direction of the magnetic field.

In a circular loop, there is a simple formula for calculating the magnetic field strength at the center of the loop. If we consider the magnetic field created in this circular loop wire, its strength and directionality would vary. The magnetic force on this current-carrying conductors is given by F = I x B, where I is the current and B is the magnetic field.

The net torque on a current-carrying loop of any shape in a magnetic field is given by t = μ × B, where μ is the magnetic dipole moment and B is the magnetic field strength. The orientation and magnitude of magnetic field would cause varying effects in the coil.

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The energy an object has as a result of motion is known as kinetic energy.A force must be applied to an object in order to accelerate it.We must put in effort in order to apply a force.After the work is finished, energy is transferred to the item, which then moves at a new, constant speed.

Solve the problem ?

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Final answer:

To determine the maximum distance the stone can slide on a rough surface, divide the kinetic energy by the friction force. The maximum distance is 48 meters.

Explanation:

To determine the maximum distance the stone can slide on a rough surface, we need to calculate the work done by the friction force.

Work done = Force x Distance

Given that the friction force is 0.50 N and the stone has 24 J of kinetic energy, we can rearrange the equation to solve for distance:

Distance = Kinetic energy / Friction force

Plugging in the values, we get:

Distance = 24 J / 0.50 N = 48 meters

Therefore, the maximum distance the stone can slide on the rough surface is 48 meters.

Answer:

Explanation:

The original frequency of sound being emitted f₀ = 1800

Its velocity towards the observer v ( let )

Apparent frequency f = 2130

velocity of sound = V

 [tex]f=f_0\times\frac{V}{(V - v)}[/tex]

Placing the given values

[tex]2130=1800\times\frac{343}{(343 - v)}[/tex]

1.1833 = [tex]\frac{343}{343 - v}[/tex]

1.1833 v = 62.87

v = 53.13 m /s .

b ) In the second case

formula for apparent frequency

[tex]f=f_0\times\frac{V+v}{(V - v)}[/tex]

Substituting the values

[tex]f=1800\times\frac{343+53}{(343 - 53)}[/tex]

= 2458 Hz .

Answer:

(a) 8Ω (b)  Ratio = Parra/P8 ohm = 1

Explanation:

Solution

Recall that,

An high-fidelity amplifier has one output for a speaker of resistance of = 8 Ω

Now,

(a) How can  two 8-Ω speakers be  arranged, when one =  4-Ω speaker, and one =12-Ω speaker

The Upper arm is : 8 ohm, 8 ohm

The Lower arm is : 12 ohm, 4 ohm

The Requirement is  = (16 x 16)/(16 + 16) = 8 ohm

(b) compare  your arrangement  power output of with the power output of a single 8-Ω speaker

The Ratio = Parra/P8 ohm = 1

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:

[tex]\Delta s = 0.978\,m[/tex]

Explanation:

The pinball-spring system is modelled after the Principle of Energy Conservation:

[tex]U_{g,1} + U_{k,1} + K_{1} = U_{g,2} + U_{k,2} + K_{2}[/tex]

[tex]-(0.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right) \cdot \Delta h + \frac{1}{2}\cdot \left(120\,\frac{N}{m}\right)\cdot (-0.2\,m)^{2} = 0[/tex]

The height reached by the pinball is:

[tex]\Delta h = 0.489\,m[/tex]

The distance travelled by the pinball is:

[tex]\Delta s =\frac{0.489\,m}{\sin 30^{\circ}}[/tex]

[tex]\Delta s = 0.978\,m[/tex]

Answer:

The answer is "[tex]1.75 \cdot (2.60 \hat{i} + 3.5 \hat{j}) \times 10^{11} \ \ \frac{m}{s^2} \\[/tex] ".

Explanation:

Formula of acceleration =  

[tex]\frac{F_e}{m_e} =\frac{-e(\underset{E}{\rightarrow} + \underset{V}{\rightarrow} \times \underset{B}{\rightarrow})}{m_e}[/tex]

values:

[tex]\underset{E}{\rightarrow} = (2.60 \hat{i} + 5.90 \hat{j}) \frac{V}{m} \\\\\underset{B}{\rightarrow} = 0.400 k \ T \\\\\underset{V}{\rightarrow} = 6.0 \hat {i} \ \ \frac{m}{s} \\\\\ apply \ value \ in \ above \ formula: \\\\ \frac{F_e}{m_e} =\frac{e}{m_e} \cdot (2.60 \hat{i} + 5.90 \hat{j}+6.0 \hat {i} \times 0.4\hat {k} ) \\\\ \frac{F_e}{m_e} =\frac{e}{m_e} \cdot (2.60 \hat{i} + 5.90 \hat{j}- 2.4 \hat {j}) \\\\\therefore \frac{e}{m_e} = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-21}} \\\\[/tex]

[tex]\frac{F_e}{m_e} = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-21}} \cdot (2.60 \hat{i} + 5.90 \hat{j}- 2.4 \hat {j}) \\\\\frac{F_e}{m_e} = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-21}} \cdot (2.60 \hat{i} + 3.5 \hat{j}) \\\\\frac{F_e}{m_e} = 1.75 \cdot (2.60 \hat{i} + 3.5 \hat{j}) \times 10^{11} \ \ \frac{m}{s^2} \\\\[/tex]

The acceleration of the electron can be found by using the net force formula and Newton's second law. The given values of velocity and magnetic field can be used to calculate the force experienced by the electron due to the magnetic field. The force can then be divided by the mass of the electron to find the acceleration.

The acceleration of the electron can be found using the formula for the net force on a charged particle moving in a magnetic field.

The force experienced by the electron due to the magnetic field is given by the equation F = qvB, where F is the force, q is the charge of the particle, v is the velocity, and B is the magnetic field. Plugging in the given values, we have F = (1.60 x 10-19 C)(6.00 x 107 m/s)(0.500 T).

To determine the acceleration, we can use Newton's second law, F = ma. Rearranging the equation, we have a = F/m, where a is the acceleration and m is the mass of the electron. Plugging in the values, we get a = [(1.60 x 10-19 C)(6.00 x 107 m/s)(0.500 T)] / (9.11 x 10-31 kg).

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Answer:

Vacuoles......final answer

Final answer:

To maximize the power output of a circuit with a 5V battery and three resistors, the resistors should be placed in parallel because this arrangement supplies more current than a series combination, thereby maximizing power.

Explanation:

To maximize the power output for a circuit with a given voltage source, the resistors should be arranged to draw the highest current possible. For resistors connected in parallel, the total resistance of the circuit decreases, as opposed to when they are in series, where it increases. Since power (P) is calculated using the formula P = V^2 / R, where V is the voltage and R is the resistance, minimizing R will maximize P.

Because a parallel configuration guarantees that each resistor gets the full voltage of the battery, the correct answer is: The student should place all resistors in parallel because the battery supplies more current to a parallel combination than to a series combination. This is because in a parallel circuit, the voltage across each resistor is equal to the voltage of the source, which maximizes the current through each resistor due to Ohm's law (I = V/R), and therefore maximizes the total current in the circuit.

Answer option (a) is thus correct: The student should place all resistors in parallel because the battery supplies more current to a parallel combination than to a series combination.

Answer:

37.45 km/hr

Explanation:

To solve this, we use the law of conservation of momentum in two directions (x, and y).

in x direction

1100 * v * cosθ = 1100 * 20 * Cos30 + 1300 * 23 * cos46

1100 * vcosθ = 22000 * 0.866 + 29900 * 0.695

1100 * vcosθ = 19052 + 20780.5

1100 * vcosθ = 39832.5

vcosθ = 39832.5 / 1100

vcosθ = 36.21

In the y direction

1100 * v * Sinθ = 1100 * 20 sin30 - 1300 * 23 sin46

1100 * vsinθ = 22000 * 0.5 - 29900 * 0.719

1100 * vsinθ = 11000 - 21498.1

1100 * vsinθ = -10498.1

vsinθ = -10498.1 / 1100

vsinθ = -9.54

Since we are looking for v, then

v² = vcos²θ + vsin²θ

v² = 36.21² + (-9.54²)

v² = 1311.16 + 91.01

v² = 1402.17

v = √1402.17

v = 37.45 km/hr

Thus, the initial speed of the lighter car is 37.45 km/hr

Answer:

1. If turning off one device turns off all the devices then it means home is wired in series.

2. Headlights of a car are connected in parallel.

3. Fuse is installed in series with a device.

4. The device will get damaged to excessive current

5. Refer to the explanation

Explanation:

1. How do you know that homes are not wired in series?

We use parallel wiring in home wiring so that each device gets the same voltage, if we use a series connection then each device will get different voltage depending upon its resistance. Moreover, in case you want to use more than 1 device then you would have to turn on both of them to complete the circuit, these are the reasons we dont use series wiring at homes.

2. Are the two headlights of a car in series or parallel?

The headlights of a car are wired in parallel so that even if one headlight gets damaged and stops working, the other headlight keeps on working. If it was wired in series then both would have stopped working when any of them gets damaged.

3. Are fuses put in parallel or in series with devices they are meant to protect?

Fuses are always connected in series, when a high fault current flows through the fuse, it gets melted and breaks the path so that the fault current doesn't flow through the device.

4. Why is it unsafe to replace a fuse that keeps burning out with one which has a higher amp rating?

The ampere rating of the fuse is selected with respect to the device it is connected to. Lets say you have a device which cannot withstand a current of 10A. So you have connected a fuse of 10A rating in series with the device. The fuse got burnt out several times now if you decided to replace the fuse with a higher ampere rating lets say 12A then what would happen? it means that now a current of 10A can flow through the device which will damage the device for sure.

5. A bulb and pair of batteries might have the batteries connected to each other in series or parallel. What are the advantages of each arrangement?

Advantages of series arrangement:

Advantages of parallel arrangement:

Although copper wire #2 is twice as long and twice the radius of copper wire #1, the resistance in both wires is the same due to the fact that the increase in length is offset by the quadrupling of the cross-sectional area.

The resistance of a wire is inversely proportional to its cross-sectional area and directly proportional to its length. The formula for resistance is R = ρL/A, where R is the resistance, ρ is the resistivity, L is the length, and A is the cross-sectional area. In this case, the length of wire #2 is twice that of wire #1 (2L), but the radius (and therefore the cross-sectional area) is also twice as large. The cross-sectional area of a wire is πr², so doubling the radius actually quadruples the area. So although the length of wire #2 is doubled, the cross-sectional area is quadrupled, meaning that overall, the resistance of wire #1 is the same as wire #2.

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The resistance of wire #1 is twice as high as that of wire #2. So, the correct statement is: The resistance of wire #1 is twice as high as that of wire #2. Hence the correct option is 3.

To understand the resistance of the two copper wires, we need to use the formula for resistance:

Resistance [tex](R) = \rho \times (L / A)[/tex]

where ρ (rho) is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.

For wire #1, the length is L and the radius is b. The cross-sectional area (A1) of wire #1 is given by:

[tex]A_1 = \pi \times b^2[/tex]

Thus, the resistance of wire #1 (R1) is:

[tex]R_1 = \rho \times (L / (\pi \times b^2))[/tex]

For wire #2, the length is 2L and the radius is 2b. The cross-sectional area (A2) of wire #2 is:

[tex]A_2 = \pi \times (2b)^2 = 4 \times \pi \times b^2[/tex]

Thus, the resistance of wire #2 (R2) is:

[tex]R_2 = \rho \times (2L / (4 \times \pi \times b^2)) = (\rho \times L) / (2 \times \pi \times b^2) = R_1 / 2[/tex]

[tex]w + h = 28[/tex]

[tex]l \times w \times h = v[/tex]

[tex]v = w \times h \times 24[/tex]

[tex]w = 28 - h[/tex]

[tex]h(28 - h) \times 24 = v[/tex]

[tex]24( - {h}^{2} + 28h) = v[/tex]

Answer:

The angle of incidence is  [tex]\theta_i =42.6^o[/tex]

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

   The distance between the mirror and the wall is  [tex]a = 33.7 cm[/tex]

    The height of the above the mirror is  [tex]b = 36.7 cm[/tex]

Generally the angle which the reflected ray make with the mirror is mathematically evaluated as

             [tex]\alpha =tan ^{-1} (\frac{b}{a})[/tex]

substituting values

             [tex]\alpha = tan ^{-1}( \frac{36.7}{33.7})[/tex]

            [tex]\alpha =47.4^o[/tex]

From the diagram we can deduce that  the angle of incidence is

             [tex]\theta_i = 90 - \alpha[/tex]

So          [tex]\theta_i = 90 - 47.4[/tex]

              [tex]\theta_i =42.6^o[/tex]

Answer:

the signs of heat and work are; -Q and -W

Explanation:

The first law of thermodynamics is given by; ΔU = Q − W

where;

ΔU is the change in internal energy of a system,

Q is the net heat transfer (the sum of all heat transfer into and out of the system)

W is the net work done (the sum of all work done on or by the system).

Now, The system in this case is the tire and since the air gets warmer, heat must have left the system. Therefore Q is negative (-Q).

Since work is done by the system, W remains negative.

Thus, the signs of heat and work are; -Q and - W

Answer:

an orange fringe at 0°, yellow fringes at ±50° and red fringes farther out.

Explanation:

In the visible spectrum- red to violet, red has the highest wavelength.

The maximum internsity  for diffraction grating is given by,

Sinθ[tex](_{max} )[/tex] = mλ/d

It is concluded that the angle of diffraction increases with increase in wavelength'λ' .  So, red fringe will be farthest from the center, orange light will be at the center and yellow fringe will be at 50°.

Therefore, The new pattern consists of : an orange fringe at 0°, yellow fringes at ±50° and red fringes farther out.

Answer:

the  magnitude of the dipole moment is 3.5*10^-11Cm

Explanation:

The dipole moment is given by the following formula:

[tex]\mu=qr[/tex]

r: distance between the centers of the charges = 500mm

q: charges of the bowling balls = 0.70nC

By replacing you obtain:

[tex]\mu=(0.70*10^{-9}C)(500*10^{-4}m)=3.5*10^{-11}Cm[/tex]

hence, the magnitude of the dipole moment is 3.5*10^-11Cm

Answer:

[tex]L(t) = 0.125 (cos (\frac{\pi}{14.765}(t+7)) + 0.125[/tex]

Explanation:

The expression for the  trigonometric function is :

L(t) = A (cos (B(t - C)))+ D   ----- equation (1)

where ;

[tex]A = \frac{max-min}{2}[/tex]

[tex]A = \frac{0.25-0}{2}[/tex]

A = 0.125

D = [tex]\frac{0+.025}{2}[/tex]

D = 0.125

Period of the lunar cycle = 29.53

Then;

[tex]\frac{2 \pi}{B} = 29.53[/tex]

[tex]29.53 \ \ B = 2 \pi[/tex]

[tex]B = \frac{2 \pi}{29.53}[/tex]

[tex]B = \frac{\pi}{29.53}[/tex]

Also; we known that December 25 is 7 days before January 1.

Then L(-7) = 0.025

Plugging all the values into trigonometric function ; we have:

[tex]0.125 ( cos ( \frac{\pi}{14.765}((-7)-C)))+0.125 = 0.25 \\ \\ \\ ( cos ( \frac{\pi}{14.765}((-7)-C))) = \frac{0.25-0.125}{0.125}[/tex]

[tex]( cos ( \frac{\pi}{14.765}((-7)-C))) = 1[/tex]

[tex]( \frac{\pi}{14.765}((-7)-C))= cos^{-1} (1)[/tex]

[tex]}((-7)-C))=0[/tex]

[tex]C= -7[/tex]

[tex]L(t) = 0.125 (cos (\frac{\pi}{14.765}(t-(-7))) + 0.125[/tex]

[tex]L(t) = 0.125 (cos (\frac{\pi}{14.765}(t+7)) + 0.125[/tex]

The formula for the trigonometric function that models the illumination [tex]\( L(t) \)[/tex] of the moon t days after January 1, 2016, is:

[tex]\[ L(t) = 0.125 + 0.125 \cos\left(\frac{2\pi}{29.53}(t - 777 - 25)\right) \][/tex]

The moon's illumination cycle can be modeled using a cosine function because it is periodic and symmetric about the maximum (full moon) and minimum (new moon) points. The maximum illumination is 0.250 lux, and the minimum is 0 lux. The period of the cycle is 29.53 days.

To find the midline of the function, we take the average of the maximum and minimum illuminations:

[tex]\[ \text{Midline} = \frac{\text{Max illumination} + \text{Min illumination}}{2} = \frac{0.250 + 0}{2} = 0.125 \][/tex]

The amplitude of the function is the distance from the midline to the maximum or minimum, which is half the difference between the maximum and minimum illuminations:

[tex]\[ \text{Amplitude} = \frac{\text{Max illumination} - \text{Min illumination}}{2} = \frac{0.250 - 0}{2} = 0.125 \][/tex]

The period of the function is the length of one complete cycle, which is given as 29.53 days. To convert this to radians, we use the formula:

[tex]\[ \text{Radian frequency} = \frac{2\pi}{\text{Period}} = \frac{2\pi}{29.53} \][/tex]

The phase shift occurs because the full moon does not happen on January 1, 2016. It happens on December 25, 2015, which is 777 days before January 1, 2016, plus an additional 25 days to account for the days after December 25 until January 1. Therefore, the phase shift is:

[tex]\[ \text{Phase shift} = 777 + 25 \][/tex]

Putting all this together, the formula for the illumination L(t) as a function of time t (in days) is:

[tex]\[ L(t) = \text{Midline} + \text{Amplitude} \cdot \cos\left(\text{Radian frequency} \cdot (t - \text{Phase shift})\right) \]\[ L(t) = 0.125 + 0.125 \cos\left(\frac{2\pi}{29.53}(t - 777 - 25)\right) \][/tex]

This function models the illumination of the moon t days after January 1, 2016, with the correct midline, amplitude, period, and phase shift.

Explanation:

The electrical force between charges is given by :

[tex]F_e=\dfrac{kq_eq_p}{r^2}[/tex]

[tex]q_e\ and\ q_p[/tex] are charge on electron and proton respectively.

[tex]F_e=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1^2}\\\\F_e=2.3\times 10^{-28}\ N[/tex]

The Gravitational force between masses is given by :

[tex]F_G=\dfrac{Gm_em_p}{r^2}[/tex]

[tex]m_e\ and\ m_p[/tex] are masses of electron and proton respectively.

[tex]F_G=\dfrac{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}{1^2}\\\\F_G=1.01\times 10^{-67}[/tex]

Ratio of electrical to the gravitational force is :

[tex]\dfrac{F_e}{F_G}=\dfrac{2.3\times 10^{-28}\ N}{1.01\times 10^{-67}\ N}\\\\\dfrac{F_e}{F_G}=2.27\times 10^{39}[/tex]

Hence, this is the required solution.

Final answer:

To find the magnitude of the electrical force between an electron and proton separated by 1 m, we use Coulomb's Law with given constants. The electrical force is calculated to be approximately 2.30 x 10⁻¹⁰ N.

Explanation:

To calculate the magnitude of the electrical force between an electron and proton separated by 1 meter, we use Coulomb's Law which is given by the formula:

F = k * |q¹ * q²| / r²

Where F is the force in Newtons (N), k is the Coulomb constant (8.98755 x 10⁹ N*m²/C²), q1 and q2 are the charges of the proton and electron respectively, and r is the separation distance in meters. Since both the proton and the electron have an elemental charge of 1.60² x 10⁻¹⁹ C, albeit with opposite signs, their charges can be multiplied to give their product in Coulomb's equation. The separation r is 1 meter.

So the magnitude of the electrical force F is calculated as:

F = (8.98755 x 109 N*m²/C²) * (1.60² x 10⁻¹⁹ C)² / 12m²

F ≈ 2.30 x 10¹⁰ N

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed  v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ   = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number =  2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

so

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2  m

and when x =  x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is  0.0549 m

Answer:

Explanation:

This is a problem based on interference pf light waves.

wavelength of light λ = 553 nm

slit separation d = .134 x 10⁻³ m

screen distance D = 4.5 m

for fourth order bright fringe, path- length  difference = 4 x λ

= 4 x 553

= 2212 nm .

= 2.212 μm