Select from the following for the next two questions:
A virtual, inverted and smaller than the object
B real, inverted and smaller than the object
C virtual, upright and smaller than the object
D real, upright and larger than the object
E virtual, upright and larger than the object
F real, inverted and larger than the object
G virtual, inverted and larger than the object
H real, upright and smaller than the object
An object is placed 46.9 cm away from a converging lens. The lens has a focal length of 10.0 cm. Select the statement from the list above which best describes the image an objesthse place 46.9 cm away from a spherical convex mirror. The radius of curvature of the mirror is 20.0 cm. Select the statement from the An object is placed 46.9 cm away from a spherical convex mirror. The radius of curvature of the mirror is 20.0 cm. Select the statement from the list above which best describes the image.

Answer:

The magnetic force would be:

[tex]F\approx 2.625\,\,10^{-5}\,\,N[/tex]

Explanation:

Recall that the magnetic force on a charged particle (of charge q) moving with velocity (v) in a magnetic field B, is given by the vector product:

F = q v x B

(where the bold represents vectors)

the vector product involves the sine of the angle ([tex]\theta[/tex]) between the vectors, so we can write the relationship between the magnitudes of these quantities as:

[tex]F=q\,v\,B\,sin(\theta)[/tex]

Therefore replacing the known quantities for the first case:

[tex]F=q\,v\,B\,sin(\theta)\\2.98\,\,10^{-4} \,\,N=q\,(25.7\,\,m/s)\,B\,sin(90^o)\\2.98\,\,10^{-4} \,\,N=q\,(25.7\,\,m/s)\,B\\q\,\,B=\frac{2.98\,\,10^{-4} }{25.7} \,\frac{N\,\,s}{m}[/tex]

Now, for the second case, we can find the force by using this expression for the product of the particle's charge times the magnetic field, and the new velocity and angle:

[tex]F=q\,v\,B\,sin(\theta)\\F=q\,(4.64\,\,m/s)\,B\,sin(29.2^o)\\F=q\,B(4.64\,\,m/s)\,\,sin(29.2^o)\\F=\frac{2.98\,\,10^{-4} }{25.7} \,(4.64\,\,m/s)\,\,sin(29.2^o)\\F\approx 2.625\,\,10^{-5}\,\,N[/tex]

Answer:

The intensity at distance 2d from source is  [tex]I_1 = \frac{1}{4} * I[/tex]  

Explanation:

From the question we are told that

     The distance of the wave from point source is  d  

     The  intensity is  [tex]I[/tex]  

     The distance we are considering is  2d

Generally the intensity of a wave is mathematically represented as

            [tex]I = \frac{ P }{\pi d^2 }[/tex]    

Here P is power of point source      

Now when  d =  2d

          [tex]I_1 = \frac{ P }{\pi (2d)^2 }[/tex]        

           [tex]I_1 = \frac{ 1 }{4 } * \frac{ P }{\pi d^2 }[/tex]

    =>   [tex]I_1 = \frac{1}{4} * I[/tex]  

The intensity at a distance 2d from the source is equal to [tex]I'=\frac{I}{4}[/tex]

Given the following data:

To determine the intensity at a distance 2d from the source:

Mathematically, the intensity of a wave is given by the formula:

[tex]I=\frac{P}{\pi d^2}[/tex]

Where:

Since the distance is doubled (2d), we have:

[tex]I'=\frac{P}{\pi (2d)^2}\\\\I'=\frac{P}{4\pi (d)^2}\\\\I'=\frac{1}{4} \times \frac{P}{\pi (d)^2}\\\\I'=\frac{1}{4} \times I\\\\I'=\frac{I}{4}[/tex]

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Answer:

It will be half that if the first wave

Explanation:

Because the wave speed remains the same, the result of doubling the frequency is that the wavelength is half as large as it

Answer:

1.6×10²⁰

Explanation:

An ampere is a Coulomb per second.

1 A = 1 C / s

The amount of charge after 5 seconds is:

5.0 A × 5 s = 25 C

The number of electrons is:

25 C × (1 electron / 1.6×10⁻¹⁹ C) = 1.6×10²⁰ electrons

Answer:

A. 30.38°

B 5.04N

Explanation:

Using

F= ILBsin theta

2 .55N= 8.4Ax 0.5mx 1.2T x sintheta

Theta = 30.38°

B. If theta is 90°

Then

F= 8.4Ax 0.5mx 1.2x sin 90°

F= 5.04N

Answer:

The intensity of sound at rock concert is  10¹⁰ greater than that of a whisper.

Explanation:

The intensity of sound is given by;

[tex]I(dB) = 10Log(\frac{I}{I_o} )[/tex]

where;

I is the intensity of the sound

I₀ is the threshold of sound intensity = 1 x 10⁻¹² W/m²

The intensity of sound at a rock concert

[tex]120 = 10Log(\frac{I}{1*10^{-12}} )\\\\12 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{12}\\\\I = 1*10^{-12} *10^{12}\\\\I = 1*10^0\\\\I =1 \ W/m^2[/tex]

The intensity of sound of a whisper

[tex]20 = 10Log(\frac{I}{1*10^{-12}} )\\\\2 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{2}\\\\I = 1*10^{-12} *10^{2}\\\\I = 1*10^{-10}\\\\I =10^{-10} \ W/m^2[/tex]

Thus, the intensity of sound at rock concert is  10¹⁰ greater than that of a whisper.

Answer:

factor that bug maximum KE change is 0.52284

Explanation:

given data

vertical distance = 6.5 cm

ripples decrease to =  4.7 cm

solution

We apply here formula for the KE of particle that executes the simple harmonic motion that is express as

KE = (0.5) × m × A² × ω²     .................1

and kinetic energy is  directly proportional to square of the amplitude.

so

[tex]\frac{KE2}{KE1} = \frac{A2^2}{A1^2}[/tex]      .............2

[tex]\frac{KE2}{KE1} = \frac{4.7^2}{6.5^2}[/tex]

[tex]\frac{KE2}{KE1}[/tex] = 0.52284

so factor that bug maximum KE change is 0.52284

The factor does the bug's maximum KE change should be considered as the 0.52284.

Since

vertical distance = 6.5 cm

ripples decrease to =  4.7 cm

So here we apply the given formula

KE = (0.5) × m × A² × ω²     .................1

here,

kinetic energy is directly proportional to square of the amplitude.

So,

= 4.7^2/ 6.5^2

= 0.52284

hence, The factor does the bug's maximum KE change should be considered as the 0.52284.

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Explanation:

Given that,

Separation between two long parallel conductors, r = 11 cm = 0.11 m

Current in first wire, [tex]I_1=3\ A[/tex]

Current in second wire, [tex]I_2=8\ A[/tex]

(a) The magnetic field created by I₁ at the location of I₂ is given by :

[tex]B_{12}=\dfrac{\mu_o I_1}{2\pi r}\\\\B_{12}=\dfrac{4\pi \times 10^{-7}\times 3I_1}{2\pi \times 0.11}\\\\B_{12}=5.45\times 10^{-6}\ T[/tex]

(b) Magnetic force per unit length exerted by [tex]I_1[/tex] on [tex]I_2[/tex] is given by :

[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 3\times 8}{2\pi \times 0.11}\\\\\dfrac{F}{l}=4.36\times 10^{-5}\ N/m[/tex]

(c) The magnetic field created by I₂ at the location of I₁ is given by :

[tex]B_{21}=\dfrac{\mu_o I_2} {2\pi r}\\\\B_{12}=\dfrac{4\pi \times 10^{-7}\times 8}{2\pi \times 0.11}\\\\B_{12}=1.45\times 10^{-5}\ T[/tex]

Hence, this is the required solution.

Answer:

 I = 7.96 W / m²

Explanation:

The light bulb emits a power of P = 100W, this power is distributed over the surface of a sphere, thus the emission is in all directions.

Intensity is defined by power per unit area

            I = P / A

The area of ​​a sphere is

         A = 4π r²

we substitute

         I = P / (4π r²)

in this case it tells us that the distance is r = 1 m

let's calculate

        I = 100 / (4π 1²)

        I = 7.96 W / m²

Answer:

The density is  [tex]\rho_z = 2544 \ kg /m^3[/tex]

Explanation:

From the question we are told that

    The mass of the rock is  [tex]m_r = 1.80 \ kg[/tex]

     The  tension on the string is  [tex]T = 10.8 \ N[/tex]

Generally the weight of the rock is  

        [tex]W = m * g[/tex]

=>     [tex]W = 1.80 * 9.8[/tex]

=>   [tex]W = 17.64 \ N[/tex]

Now the upward force(buoyant force) acting on the rock  is mathematically evaluated as  

        [tex]F_f = W - T[/tex]

substituting values

       [tex]F_f = 17.64 - 10.8[/tex]

      [tex]F_f = 6.84 \ N[/tex]

This buoyant force is mathematically represented as

      [tex]F_f = \rho * g * V[/tex]

Here  [tex]\rho[/tex] is the density of water and it value is [tex]\rho = 1000\ kg/m^3[/tex]

 So

         [tex]V = \frac{F_f}{ \rho * g }[/tex]

        [tex]V = \frac{6.84}{ 1000 * 9.8 }[/tex]

        [tex]V = 0.000698 \ m^3[/tex]

Now for this rock to flow the upward force (buoyant force) must be equal to the length

      [tex]F_f = W[/tex]

      [tex]\rho_z * g * V = W[/tex]

Here z is smallest density of a liquid in which the rock will float

=>     [tex]\rho_z = \frac{W}{ g * V}[/tex]

=>   [tex]\rho_z = \frac{17.64}{ 0.000698 * 9.8}[/tex]

=>   [tex]\rho_z = 2544 \ kg /m^3[/tex]

Answer:

Using

A. .E = σ/εo = (q/A)/εo = = q/Aεo so if q = 2q, then

Ef/Ei = 2

B. If L is 2L then Ef = q/4Aεo and

Ef/Ei = 1/4

C. The electric field strength is not effected by d and as long as σ is unchanged, Ef/Ei = 1

Answer:

b. the particle must be moving parallel to the magnetic field.

Explanation:

The magnetic force on a moving charged particle is given by;

F = qvBsinθ

where;

q is the charge of the particle

v is the velocity of the particle

B is the magnetic field

θ is the angle between the magnetic field and velocity of the moving particle.

When is the charge is stationary the magnetic force on the charge is zero.

Also when the charge is moving parallel to the magnetic field, the magnetic force is zero.

Therefore, when a moving charged particle experiences no magnetic force, we can definitely conclude that the particle must be moving parallel to the magnetic field.

b. the particle must be moving parallel to the magnetic field.

Given that,

A charged particle enters a magnetic field with an angle theta .

If [tex]\theta=90^{\circ}[/tex]

We know that,

If the angle is 90° then the charged particle enters perpendicular to the B.

B is magnetic field.

The charged particle will be follow of the circular path.

If the angle is greater than 0 and less than 90° then the charged particle will be show the helical path.

Hence, This is required answer.

Answer:

λ_A = 700 nm ,   m_B = m_a 2

Explanation:

The expression that describes the diffraction phenomenon is

         a sin θ = m λ

where a is the width of the slit, lam the wavelength and m an integer that writes the order of diffraction

a) They tell us that now lal_ A m = 1

         a sin θ = λ_A

coincidentally_be m = 2

          a sin  θ = m λ_b

as the two match we can match

         λ _A = 2 λ _B

         λ_A = 2 350 nm

         λ_A = 700 nm

b)

For lam_B

       a sin  λ_A  = m_B  λ_B

For lam_A

        a sin θ_A = m_ λ_ A

to match they must have the same angle, so we can equal

           m_B  λ_B = m_A  λ_A

           m_B = m_A  λ_A / λ_B

           m_b = m_a 700/350

           m_B = m_a 2

Answer:

a)14Hz

b)26.6m/s

Explanation:

a)we were given

the first harmonics frequencies as 280 Hz

The second harmonic frequency as 294 Hz.

The fundamental frequency is equal to the gap which means the distance that exist between the harmonics, then

the fundamental frequency=(294 - 280 = 10 Hz)

= 14Hz

b) We know the frequency and the wavelength of the sound wave (

We were told that the wavelength must be twice the length of the tube then, velocity can be calculated as

And fundamental frequency= 14Hz, and distance of 1.90 m then

v = f*2L = (14Hz)*2*(1.90 m) = 26.6m/s

Therefore, the speed of sound in the gas in the tubes is 26.6m/s

Answer:

cognitive behavioral therapy

Explanation:

Answer:

D) diffraction

Explanation:

Corona is an optical phenomenon produced by the diffraction of sunlight or moonlight, as light moves through water droplets in the atmosphere.

This phenomenon produces one or more diffuse concentric rings of light around the Sun or Moon, usually seen as colored circles.

Therefore, the explanation for these phenomena of colored concentric circles, sometimes seen with the Sun or the Moon involves diffraction.

Answer:

1.24Mev

Explanation:

Using

E= hc/lambda

= (6.62x10^-19) x(3x10^8m/s)/(1x10^-12) x 1.602x10^-9

= 1.24Mev

Answer:

It bends away from the normal

Explanation:

From Snell's law of Refraction, when a ray passes from a medium of lower Refractive index to a medium with higher Refractive index, the Refractive ray will bend towards the normal. However, when the ray passes from a medium of higher Refractive index to a medium of lower Refractive index, the Refractive ray will bend away from the normal.

Now, from the question we are told that Light passes from a material with index of refraction 1.3 into one with index of refraction 1.2.

This means from a higher Refractive index to a lower one and from Snell's law as earlier said, the refracted ray will bend away from the normal

The refracted ray is seen to bend away from the normal.

Let us recall that an optically denser medium will have a higher refractive index. This means that the medium with a refractive index of  1.3  is the denser medium and the medium with a refractive index of  1.2 is the less dense medium.

From the statement in the question, we can boldly say that light is travelling from a denser to less dense medium given the values of the refractive index given. When light is travelling from a denser to a less dense medium, the refracted ray bends away from the normal.

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Answer:

potential energy is a stored energy or energy of position (gravitational).

Kinetic energy is a energy of motion.

Explanation:

in the formula K is for the kinetic and the P stand for the potential.

Answer:

1. e m fmax = 0.00598 Volt

2. Imax = 0.000854 Amp

Explanation:

1. Find the maximum induced emf.

e m fmax =

Given that e m fmax = N*A*B*w

N = 1

A = 2 cm^2 = 0.0002 m^2

f = 143 rotation per minute = 143/min

f = (143/min) * (1 min/60 sec) = 2.38/sec

w = 2Πf = 2 * Π * 2.38 = 14.95 rad/sec

B = 2T

e m fmax = N*A*B*w

e m fmax = 1 * 0.0002 * 2 * 14.95

e m fmax = 0.00598 Volt.

2. Find the maximum current through the bulb.

Imax = e m fmax / R

Where R is the total resistance in the circuit is 7 Ω.

Imax = 0.00598/7 = 0.000854 Amp.

Imax = 0.000854 Amp

1) The maximum induced EMF in the loop of wire is; EMF_max = 9.52 × 10^(-4) V

2) The maximum current through the bulb is;

I_max = 1.36 × 10^(-4) A

We are given;

Number of turns; N = 1

Magnitude of magnetic field; B = 2 T

Area; A = 2 cm² = 0.0002 m²

Angular frequency; ω = 143 /min = 2.38 /s

Resistance; R = 7 Ω.

1) Formula for maximum induced EMF is;

EMF_max = NAωB

Plugging in the relevant values gives;

EMF_max = 1 × 0.0002 × 2.38 × 2

EMF_max = 9.52 × 10^(-4) V

2) Formula for maximum current through the bulb is given as;

I_max = EMF_max/R

Plugging in the relevant values;

I_max = (9.52 × 10^(-4))/7

I_max = 1.36 × 10^(-4) A

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Answer:

Ultraviolet light from the sun.

Explanation:

This is an example of radiation.

Answer:

X-Ray

Explanation:

x-Ray is an example of radiation.

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:

halve the slit separation

Explanation:

As we know that

In YDS experiment, the equation of fringe width is as follows

[tex]\beta = \frac{\lambda D}{d}[/tex]

where,

D denotes the separation in the middle of screen and slits

d denotes the distance in the middle of two slits

And to increase the Δx we have to decrease the d i.e, the distance between the two slits

Hence, the first option is correct

Answer:

Explanation:

Answer:

a) for the torque to be maximum, sin should be maximum

i.e (sinФ)maximum = 1

b) therefore the Maximum torque is

Tmax = 0.1838 × 1 = 0.1838  N.m

c) Given the torque is 71.0% of its maximum value; Ф  = 45.24⁰ ≈ 45⁰

Explanation:

Given that; Diameter is 8.40 cm,

Radius (R) = D/2 = 8.40/2 = 4.20 cm = 0.042 m

Number of turns (N) = 17

Current in the loop (I) = 3.20 A

Magnetic field (B) = 0.610 T

Let the angle between the loop's area vector A and the magnetic field B be

Now. the area of the loop is;

A = πR²

A = 3.14 ( 0.042 )²

A =  0.005539 m²

Torque on the loop (t) = NIABsinФ

t = 17 × 3.20 ×0.005539 × 0.610 × sinФ

t = 0.1838sinФ N.m

for the torque to be maximum, sin should be maximum

i.e (sinФ)maximum = 1

therefore the Maximum torque is

Tmax = 0.1838 × 1 = 0.1838  N.m

Given the torque is 71.0% of its maximum value

t = 0.71 × tmax

t = 0.71 × 0.1838

t = 0.1305

Now

0.1305 N.m =  0.1838 sinФ N.m

sinФ = 0.1305 / 0.1838

sinФ = 0.71001

Ф = sin⁻¹ 0.71001

Ф  = 45.24⁰ ≈ 45⁰

Answer:

[tex]B=2.74\times 10^{-10}\ T[/tex]

Explanation:

It is given that,

A planar electromagnetic wave is propagating in the +x direction.The electric field at a certain point is, E = 0.082 V/m

We need to find the magnetic vector of the wave at the point P at that instant.

The relation between electric field and magnetic field is given by :

[tex]c=\dfrac{E}{B}[/tex]

c is speed of light

B is magnetic field

[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.082}{3\times 10^8}\\\\B=2.74\times 10^{-10}\ T[/tex]

So, the magnetic vector at point P at that instant is [tex]2.74\times 10^{-10}\ T[/tex].

The magnetic vector of the wave at the point P at that instant is [tex]2.73 \times 10^{-10}T[/tex]

The formula relating electric field and the magnetic field is given as;

[tex]c=\frac{E}{B}[/tex]

From the formula shown:

[tex]B=\frac{E}{c}\\B=\frac{0.082}{3.0\times 10^8}\\B=2.73 \times 10 ^{-10}T[/tex]

Hence the magnetic vector of the wave at the point P at that instant is [tex]2.73 \times 10^{-10}T[/tex]

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Answer:

During acceleration, you are moving across a distance over a time, but also increasing how fast we are doing it. Therefore, it means by how many meters per second the velocity changes every second

Explanation:

Answer:

hello the needed diagram is missing attached below is the diagram and the detailed solution

The tension in BC = [tex]\frac{\sqrt{2} }{4} mg[/tex]

Explanation:

ATTACHED BELOW IS THE DETAILED SOLUTION T THE GIVEN PROBLEM

Ma = mg - T/ [tex]\sqrt{2}[/tex]  equation 1

Ma = 3T / [tex]\sqrt{2}[/tex]   equation 2

equate both equations to determine the tension on BC

Answer:

The wavelength is [tex]\lambda_2 = 534 *10^{-9} \ m[/tex]

Explanation:

From the question we are told that

   The wavelength of the first light is  [tex]\lambda _ 1 = 587 \ nm[/tex]

    The order of the first light that is being considered is  [tex]m_1 = 10[/tex]

     The order of the second light that is being considered is  [tex]m_2 = 11[/tex]

Generally the distance between the fringes for the first light is mathematically represented as

      [tex]y_1 = \frac{ m_1 * \lambda_1 * D}{d}[/tex]

 Here  D is the distance from the screen

 and    d  is the distance of separation of the slit.

      For the second light the distance between the fringes is  mathematically represented as

         [tex]y_2 = \frac{ m_2 * \lambda_2 * D}{d}[/tex]

Now given that both of the light are passed through the same double slit

       [tex]\frac{y_1}{y_2} = \frac{\frac{m_1 * \lambda_1 * D}{d} }{\frac{m_2 * \lambda_2 * D}{d} } = 1[/tex]

=>    [tex]\frac{ m_1 * \lambda _1 }{ m_2 * \lambda_2} = 1[/tex]

=>     [tex]\lambda_2 = \frac{m_1 * \lambda_1}{m_2}[/tex]

=>    [tex]\lambda_2 = \frac{10 * 587 *10^{-9}}{11}[/tex]

=>   [tex]\lambda_2 = 534 *10^{-9} \ m[/tex]