a. What do you mean by chromatic aberration in lenses?

Three children of masses and their position on the merry go round

M1 = 22kg

M2 = 28kg

M3 = 33kg

They are all initially riding at the edge of the merry go round

Then, R1 = R2 = R3 = R = 1.7m

Mass of Merry go round is

M =105kg

Radius of Merry go round.

R = 1.7m

Angular velocity of Merry go round

ωi = 22 rpm

If M2 = 28 is moves to center of the merry go round then R2 = 0, what is the new angular velocity ωf

Using conservation of angular momentum

Initial angular momentum when all the children are at the edge of the merry go round is equal to the final angular momentum when the second child moves to the center of the merry go round  Then,

L(initial) = L(final)

Ii•ωi = If•ωf

So we need to find the initial and final moment of inertia

NOTE: merry go round is treated as a solid disk then I= ½MR²

I(initial)=½MR²+M1•R²+M2•R²+M3•R²

I(initial) = ½MR² + R²(M1 + M2 + M3)

I(initial) = ½ × 105 × 1.7² + 1.7²(22 + 28 + 33)

I(initial) = 151.725 + 1.7²(83)

I(initial) = 391.595 kgm²

Final moment of inertial when R2 =0

I(final)=½MR²+M1•R²+M2•R2²+M3•R²

Since R2 = 0

I(final) = ½MR²+ M1•R² + M3•R²

I(final) = ½MR² + (M1 + M3)• R²

I(final)=½ × 105 × 1.7² + ( 22 +33)•1.7²

I(final) = 151.725 + 158.95

I(final) = 310.675 kgm²

Now, applying the conservation of angular momentum

L(initial) = L(final)

Ii•ωi = If•ωf

391.595 × 22 = 310.675 × ωf

Then,

ωf = 391.595 × 22 / 310.675

ωf = 27.73 rpm

Answer: So, the final angular momentum is 27.73 revolution per minute

Answer:

5 m/s I hope it will help you

Explanation:

mark me as a brainlist answer

Answer: 5.12x10∧-4N

Explanation:

Force = I B L

L = 6.4m

Let Current (I) I₁ = I₂= 14A

Distance of the wire = 42cm = 0.42m

BUT

B = μ₀I / 2πr

=(2X10∧-7 X 12) / 0.42

       B =5.714×10∧-6T

Force = I B L

Force = 14x [5.714×10-6]×6.4

Force = 5.12x10∧-4N

Answer:

Following are the solution to the given question:

Explanation:

Applying the Snell Law:

when is the liquid's refractive index, is the air's refractive index.  

the liquid's refractive index,   the air's index of refraction.

It is the limiting case when , Inside the interphase of two mediums, light is scattered. Thus,

[tex]n_l \sin \theta_l = n_a \sin 90^\circ = n_a[/tex]

[tex]\theta_l = \arcsin \dfrac{n_a}{n_l} =\arcsin \dfrac{1}{1.38} = 46.4^\circ[/tex]

From  the incident angles [tex]\theta_l[/tex] is greater than 46.4°, that is the light reflected back into the liquid.

[tex]3.20×10^7\:\text{m/s}[/tex]

Explanation:

Let

[tex]L = 28.2\:\text{m}[/tex]

[tex]L' = 28.2\:\text{m} - 0.161\:\text{m} = 28.039\:\text{m}[/tex]

The Lorentz length contraction formula is given by

[tex]L' = L\sqrt {1 - \left(\dfrac{v^2}{c^2}\right)}[/tex]

where L is the length measured by the moving observer and L' is the length measured by the stationary Earth-based observer. We can rewrite the above equation as

[tex]\sqrt {1 - \left(\dfrac{v^2}{c^2}\right)} = \dfrac{L'}{L}[/tex]

Taking the square of the equation, we get

[tex]1 - \left(\dfrac{v^2}{c^2}\right) = \left(\dfrac{L'}{L}\right)^2[/tex]

or

[tex]1 - \left(\dfrac{L'}{L}\right)^2 = \left(\dfrac{v}{c}\right)^2[/tex]

Solving for v, we get

[tex]v = c\sqrt{1 - \left(\dfrac{L'}{L}\right)^2}[/tex]

[tex]\:\:\:\:=(3×10^8\:\text{m/s})\sqrt{1 - \left(\dfrac{28.039\:\text{m}}{28.2\:\text{m}}\right)^2}[/tex]

[tex]\:\:\:\:=3.20×10^7\:\text{m/s} = 0.107c[/tex]

(a) The maximum height reached by the ball from the ground level is 75.87m

(b) The time taken for the ball to return to the elevator floor is 2.21 s

The given parameters include:

To calculate:

(a) the maximum height of the projectile

total initial velocity of the projectile = 10 m/s + 20 m/s  = 30 m/s (since the elevator is ascending at a constant speed)

at maximum height the final velocity of the projectile (ball), v = 0

Apply the following kinematic equation to determine the maximum height of the projectile.

[tex]v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m[/tex]

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection

h = h₁ + h₂ + h₃

h = 28 m + 2 m  +  45.87 m

h = 75.87 m

(b) The time taken for the ball to return to the elevator floor

Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m

Apply the following kinematic equation to determine the time to return to the elevator floor.

[tex]h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s[/tex]

To learn more about projectile calculations please visit: https://brainly.com/question/14083704

Explanation:

The heart rate of the astronaut is 78.5 beats per minute, which means that the time between heart beats is 0.0127 min. This will be the time t measured by the moving observer. The time t' measured by the stationary Earth-based observer is given by

[tex]t' = \dfrac{t}{\sqrt{1 - \left(\dfrac{v^2}{c^2}\right)}}[/tex]

a) If the astronaut is moving at 0.480c, the time t' is

[tex]t' = \dfrac{0.0127\:\text{min}}{\sqrt{1 - \left(\dfrac{0.2304c^2}{c^2}\right)}}[/tex]

[tex]\:\:\:\:=0.0145\:\text{min}[/tex]

This means that time between his heart beats as measured by Earth-based observer is 0.0145 min, which is equivalent to 69.1 beats per minute.

b) At v = 0.940c, the time t' is

[tex]t' = \dfrac{0.0127\:\text{min}}{\sqrt{1 - \left(\dfrac{0.8836c^2}{c^2}\right)}}[/tex]

[tex]\:\:\:\:=0.0372\:\text{min}[/tex]

So at this speed, the astronaut's heart rate is 1/(0.0372 min) or 26.9 beats per minute.

Answer:

UY Scuti is slightly larger than VY Canis Majoris

Explanation:

These stars are millions of miles away and cannot be seen by the naked eye.

Beetlejuice is another large star that can be seen by the eye.

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

We are being given that:

A pictorial view showing the diagrammatic representation of the information given in the question is being attached in the image below.

From the attached image below, the potential difference across the conducting element of the length (dx) moving with the velocity (v) appears to be perpendicular to the magnetic field (B).

The magnitude of the potential difference induced between the center of the blade in relation to either of its ends can be determined by using the derived formula from Faraday's law of induction which can be expressed as:

[tex]\mathsf{E = B\times l\times v}[/tex]

where;

From the diagram, Let consider the length of the conducting element (dx) at a distance of length (x) from the center O.

Then, the velocity (v) = ωx

The potential difference across it can now be expressed as:

[tex]\mathsf{dE = B*(dx)*(\omega x)}[/tex]

For us to determine the potential difference, we need to carry out the integral form from center point O to L/2.

∴

[tex]\mathsf{E = \int ^{L/2}_{0}* B (\omega x ) *(dx)}[/tex]

[tex]\mathsf{E = B (\omega ) \times \Big[ \dfrac{x^2}{2}\Big]^{L/2}_{0}}[/tex]

[tex]\mathsf{E = B (\omega ) * \Big[ \dfrac{L^2}{8}\Big]}[/tex]

Recall that,

[tex]\mathsf{E = (4\times 10^{-3}) * (17) \Big[ \dfrac{(1.5)^2}{8}\Big]}[/tex]

[tex]\mathsf{E = 19.13 mV}[/tex]

Thus, we can now conclude that the magnitude of the potential difference as a result of the rotation caused by the metal blade from the center to either of its ends is 19.13 mV.

Learn more about Faraday's law of induction here:

https://brainly.com/question/13369951?referrer=searchResults

Answer:

Simple harmonic motion is a special type of periodic motion or oscillation motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacemen

Answer:

B = 1.03 10⁻⁸ T

Explanation:

For an electromagnetic wave, the electric and magnetic fields must oscillate in phase so that they remain between them at all times, otherwise the wave will extinguish

This relational is expressed by the relation

           E /B = c

           B = E / c

let's calculate

            B = 3.10 / 3 10⁸

            B = 1.03 10⁻⁸ T

Answer:

Current =,charge / time

Charge = e = 1.6E-19 coulombs

t = T time for 1 revolution  (period)

v = S / T = distance traveled in 1 revolution / time for 1 revolution

T = S / v = 2 pi * 4.76E-10 / 2.43E5 = 1.23E-14

I = Q / T = 1.6E-19 / 1.23E-14 = 1.30E-5

Answer:

c) kinetic energy

Explanation:

Answer: C)  kinetic energy

Explanation:

The question is incomplete. The complete question is :

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .

Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:

1. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.

Solution :

Given :

[tex]A = 10 \ cm^2[/tex]

   [tex]$=0.0010 \ m^2$[/tex]

d = 10 mm

  = 0.010 m

Then, Capacitance,

[tex]$C=\frac{k \epsilon_0 A}{d}$[/tex]

[tex]$C=\frac{8.85 \times 10^{12} \times 3 \times 0.0010}{0.010}$[/tex]

[tex]$C=2.655 \times 10^{12} \ F$[/tex]

[tex]$U_1 = \frac{1}{2}CV^2$[/tex]

[tex]$U_1 = \frac{1}{2} \times 2.655 \times 10^{-12} \times (15V)^2$[/tex]

[tex]$U_1=2.987 \times 10^{-10}\ J$[/tex]

Now,

[tex]$C_k=\frac{1}{2} \frac{k \epsilon_0}{d} \times \frac{A}{2}$[/tex]

And

[tex]$C_{air}=\frac{1}{2} \frac{\epsilon_0}{d} \times \frac{A}{2}$[/tex]

In parallel combination,

[tex]$C_{eq}= C_k + C_{air}$[/tex]

[tex]$C_{eq} = \frac{1}{2} \frac{\epsilon_0 A}{d}(1+k)$[/tex]

[tex]$C_{eq} = \frac{1}{2} \times \frac{8.85 \times 10^{-12} \times 0.0010}{0.01} \times (1+3)$[/tex]

[tex]$C_{eq} = 1.77 \times 10^{-12}\ F$[/tex]

Then energy,

[tex]$U_2 =\frac{1}{2} C_{eq} V^2$[/tex]

[tex]$U_2=\frac{1}{2} \times 1.77 \times 10^{-12} \times (15V)^2$[/tex]

[tex]$U_2=1.99 \times 10^{-10} \ J$[/tex]

b). Now the charge on the [tex]\text{capacitor}[/tex] is :

[tex]$Q=C_{eq} V$[/tex]

[tex]$Q = 1.77 \times 10^{-12} \times 15 V$[/tex]

[tex]$Q = 26.55 \times 10^{-12} \ C$[/tex]

Now when the capacitor gets disconnected from battery and the [tex]\text{dielectric}[/tex] is slowly [tex]\text{removed the rest}[/tex] of the way out of the [tex]\text{capacitor}[/tex] is :

[tex]$C_3=\frac{A \epsilon_0}{d}$[/tex]

[tex]$C_3 = \frac{0.0010 \times 8.85 \times 10^{-12}}{0.01}$[/tex]

[tex]$C_3=0.885 \times 10^{-12} \ F$[/tex]

[tex]$C_3 = 0.885 \times 10^{-12} \ F$[/tex]

Without the dielectric,

[tex]$U_3=\frac{1}{2} \frac{Q^2}{C}$[/tex]

[tex]$U_3=\frac{1}{2} \times \frac{(25.55 \times 10^{-12})^2}{0.885 \times 10^{-12}}$[/tex]

[tex]$U_3=3.98 \times 10^{-10} \ J$[/tex]

Answer:

X = 10.8387 cm²/s

Explanation:

In this exercise, you're required to convert a value from one unit to another.

Converting 42 ft²/hr to cm²/s;

Conversion:

1 ft² = 929.03 cm²

42 ft² = X cm²

Cross-multiplying, we have;

X = 42 * 929.03

X = 39019.26 cm²

Next, we would divide by time in seconds.

1 hour = 3600 seconds

X = 39019.26/3600

X = 10.8387 cm²/s

Answer:

960 nm

Explanation:

Given that:

wavelength = 640 nm

For the second (2nd) dark spot;  the order of interference m = 1

Thus, the path length difference is expressed by the formula:

[tex]d sin \theta = (m + \dfrac{1}{2}) \lambda[/tex]

[tex]d sin \theta = (1 + \dfrac{1}{2}) 640[/tex]

[tex]d sin \theta = ( \dfrac{3}{2}) 640[/tex]

dsinθ = 960 nm

Answer:

u=20 m/s, T=4s

Explanation:

Given final velocity v= 0 m/s and displacement h= 20 m; acceleration due to gravity = 10 m/ s 2

From equation of motion 

v2=u2+2gs−u2=−2(10).20u=20m/s

and time t can be determined by the formula

t=gv−u=−10−20=2s

total time = 2× time of ascend=2×2=4s

it is helpful for you

Answer:

5 kilometers per hour

Explanation:

Speed = distance / time

Distance: 50km

Time: 10 hours

Speed = 50/10 = 5kph

Answer:

5kmph

Explanation:

if the bus traveled 50 km in 10 hours, we have to divide 50 by 10 to see how fast it traveled per hour.

50/10 = 5

therefore, the bus was traveling 5 km per hour

hope this helps :)

Answer:

as it 8s based upon to fundamental units distance and Time

Answer:

true

Explanation:

forces require an agent

you should always be able to identify what (the agent) is producing the force

Answer:

option D thinking so

Explanation:

okay na your whish

Answer:

Abby is standing (4.5^2 + 2.3^2)^1/2   from the far speaker

D2 = 5.05 m from the far speaker

The difference in distances from the speakers is

5.05 - 4.5 = .55 m     (Let y be wavelength, lambda)

n y = 4.5

(n + 1) y = 5.05 for the speakers to be in phase at smallest wavelength

y = .55 m          subtracting equations

f = v / y = 340 / .55 = 618 / sec     should be the smallest frequency

Answer:

a) ε = 14.7 μv

b) ε = 21 μv

Explanation:

Given the data in the question;

Diameter of solenoid; d = 3 cm

radius will be half of diameter,  so, r = 3 cm / 2 = 1.5 cm = 1.5 × 10⁻² m

Number of turns; N = 40 turns per cm = 4000 per turns per meter

Current; [tex]I[/tex] = 0.235 A

change in time Δt = 0.40 sec

Now,

We determine the magnetic field inside the solenoid;

B = μ₀ × N × [tex]I[/tex]  

we substitute

B = ( 4π × 10⁻⁷ Tm/A ) × 4000 × 0.235  

B = 1.1881 × 10⁻³ T

Now, Initial flux through the coil is;

∅₁ = NBA = NBπr²  

and the final flux  

∅₂ = 0        

so, the εmf induced ε = -Δ∅/Δt = -( ∅₂ - ∅₁ ) / Δt

= -( 0 - NBπr² ) / Δt  

= NBπr² / Δt    

a)

for N = 7

ε = [ 7 × ( 1.1881 × 10⁻³ ) × π( 1.5 × 10⁻² )² ] / 0.40

ε = 14.7 × 10⁻⁶ v

ε = 14.7 μv      

b)

for N = 10

ε = [ 10 × ( 1.1881 × 10⁻³ ) × π( 1.5 × 10⁻² )² ] / 0.40

ε = 21 × 10⁻⁶ v

ε = 21 μv  

A

Explanation:

1qdeeeeeeeeeeehhhhhhhhhwilffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff.

Answer:

Liquids evaporate faster as they heat up and more particles have enough energy to break away. The particles need energy to overcome the attractions between them. ... At this point the liquid is boiling and turning to gas. The particles in the gas are the same as they were in the liquid they just have more energy.

This question is incomplete, the complete question is;

A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?

Answer:

the current flows through the outer loop is 1.3 × 10⁻⁵ A

Explanation:

Given the data in the question;

Length [tex]l[/tex] = 11.34 cm = 0.1134 m

radius a = 1.85 cm = 0.0185 m

turns N = 1627

Net resistance [tex]R_{sol[/tex] = 144.9 Ω

radius b = 3.77 cm = 0.0377 m

[tex]R_o[/tex] = 1651.6 Ω

ε = 34.95 V

t = 2.58 μs = 2.58 × 10⁻⁶ s

Now, Inductance; L = μ₀N²πa² / [tex]l[/tex]

so

L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134

L = 0.003576665 / 0.1134

L = 0.03154

Now,

ε = d∅/dt = [tex]\frac{d}{dt}[/tex]( BA ) =  [tex]\frac{d}{dt}[/tex][ (μ₀In)πa² ]

so

ε = μ₀n [tex]\frac{dI}{dt}[/tex]( πa² )

ε = [ μ₀Nπa² / [tex]l[/tex] ] [tex]\frac{dI}{dt}[/tex]

ε = [ μ₀Nπa² / [tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]

I = ε/[tex]R_o[/tex] = [ μ₀Nπa² / [tex]R_o[/tex][tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]

so we substitute in our values;

I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]

I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )

I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]

I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]

I = 1.3 × 10⁻⁵ A

Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A

Answer:

1.28 *10^-5 A

Explanation:

Same work as above answer. Needs to be more precise

Answer:

The results of the experiment indicated a shift consistent with zero, and certainly less than a twentieth of the shift expected if the Earth's velocity in orbit around the sun was the same as its velocity through the ether.

Explanation: