A wire carries a current of 0.66 A. This wire makes an angle of 58° with respect to a magnetic field of magnitude 5.50 10-5 T. The wire experiences a magnetic force of magnitude 7.10 10-5 N. What is the length of the wire?

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:

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:

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:

option D thinking so

Explanation:

okay na your whish

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:

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

[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]

Answer:

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

Explanation:

Given;

mass A, Ma = 2 kg

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

Mass M, Mm = 4 kg

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

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

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

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

The initial kinetic of the two masses;

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

The final kinetic energy of the two masses;

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

The change in kinetic energy is calculated as;

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

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

Answer:

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) 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

A

Explanation:

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Answer:

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

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 years old 7 months and 21 days

Explanation:

Answer:

lina marcela medina de jurado

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:

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:

Answer:

true

Explanation:

forces require an agent

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

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:

The total length of wire is 0.24 m.

Explanation:

Number of turns, N = 270

magnetic field, B = 0.48 T

frequency, f = 60 Hz

rms value of emf = 120 V

maximum value of emf, Vo = 1.414 x 120 = 169.68 V

let the area of square is A and the side is L.

The maximum emf is given by

Vo = N B A w

169.68 = 270 x 0.48 x A x 2 x 3.14 x 60

A = 3.5 x 10^-3 m^2

So,

L = 0.0589 m

Total length of wire, P = 4 L = 4 x 0.0589 = 0.24 m

Answer: (d)

Explanation:

Given

15 W is equivalent to 60 W light that is, it save 45 W

So, for 4 hours it is, [tex]4\times 45=180\ W.hr[/tex]

For 30 days, it becomes

[tex]\Rightarrow 180\times 30=5400\ W.hr\\\Rightarrow 5.4\ kWh[/tex]

Thus, [tex]5.4\ kWh[/tex] is saved in 30 days

option (d) is correct.

Explanation:

Energy utilization focuses on technologies that can lead to new and potentially more efficient ways of using electricity in residential, commercial and industrial settings—as well as in the transportation sector

Answer:

K = 2 10⁻⁸ J

Explanation:

Let's solve this exercise in parts, we start by finding the charge on each plate of the capacitor

          C = Q / ΔV

           C = ε₀ A / d

          ε₀ A / d = Q / ΔV

          Q = ε₀ A ΔV / d        (1)

indicate the potential difference ΔV₁ = 12 V, the distance between the plates d₁ = 3 mm = 0.003 m,  

as the power supply is disconnected and the capacitor is ideal the charge remains constant

in the second part we separate the plates at d₂ = 5 mm = 0.005 m, using equation 1

          ΔV₂ = [tex]\frac{Q d_2}{ \epsilon_o A}[/tex]

we substitute the equation for Q

         ΔV₂ = [tex]\frac{d_2}{\epsilon_o A} \ \frac{\epsilon_o A \Delta V }{d_1}[/tex]

         ΔV₂ = [tex]\frac{d_2}{d_1} \ \Delta V_1[/tex]

in the third part we use the concepts of energy

starting point. Test charge near positive plate

          Em₀ = U = q ΔV₂

final point. Test charge near negative plate

          Em_f = K

energy is conserved

          Em₀ = Em_f

          q ΔV₂ = K

          K = q ΔV₁ [tex]\frac{d_2}{d_1}[/tex]

we calculate

          K = 1 10⁻⁹  12  0.005/0.003

          K = 2 10⁻⁸ J

Answer:

"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web

Explanation:

tbh up above ✅

Answer:

cubic meter

Explanation:

Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion

Answer:

geosynchronous orbit (GEO) is a prograde, low inclination orbit about Earth having a period of 23 hours 56 minutes 4 seconds. A spacecraft in geosynchronous orbit appears to remain above Earth at a constant longitude, although it may seem to wander north and south.

Answer:

a)   R₁ = 14.1 Ω,   b)  R₂ =  19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

           V = i (R₁ + R₂)

with R₁ connected

           V = (i + 0.5) R₁

with R₂ connected

           V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

           V = i ( [tex]\frac{V}{ i+0.5} + \frac{V}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{1}{i+0.5} + \frac{1}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }[/tex] ) =  [tex]\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}[/tex]

           i² + 0.75 i + 0.125 = 2i² + 0.75 i

           i² - 0.125 = 0

           i = √0.125

           i = 0.35355 A

with the second equation we look for R1

          R₁ = [tex]\frac{V}{i+0.5}[/tex]

          R₁ = 12 /( 0.35355 +0.5)

          R₁ = 14.1 Ω

with the third equation we look for R2

          R₂ = [tex]\frac{V}{i+0.25}[/tex]

          R₂ =[tex]\frac{12}{0.35355+0.25}[/tex]

          R₂ =  19.9 Ω

Answer:

tables, charts and graphs

Explanation:

Answer:

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Explanation:B

Answer:

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