Ask question

Ask question

All categories

vladimir1956 [14]

3 years ago

11

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-forme

d into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is 4.5 x 10 -3 Wb. What is the flux that passes through the circular loop?

1 answer:

Sindrei [870]3 years ago

5 0

I need to know this answer

You might be interested in

1.Ben is pushing a block up a ramp that is 3 meters high and 6 meters long. The block has a force of 49N. It is taking Ben 37 Ne

inna [77]

Idrk sorry... i’m doing this bc it’s asking me to. wish i could help you guys . i really do !

5 0

3 years ago

For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.

Kamila [148]

Answer:

A) E = 4.96 x 10³ eV

B) E = 4.19 x 10⁴ eV

C) E = 3.73 x 10⁹ eV

Explanation:

A)

For photon energy is given as:

$E = hv$

$E = \frac{hc}{\lambda}$

where,

E = energy of photon = ?

h = 6.625 x 10⁻³⁴ J.s

λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Therefore,

$E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}$

$E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})$

<u>E = 4.96 x 10³ eV</u>

<u></u>

B)

The energy of a particle at rest is given as:

$E = m_{0}c^2$

where,

E = Energy of electron = ?

m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (9.1 x 10^{-31} kg)(3 x 10^8 m/s)^2\\$

$E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\$

<u>E = 4.19 x 10⁴ eV</u>

<u></u>

C)

The energy of a particle at rest is given as:

$E = m_{0}c^2$

where,

E = Energy of alpha particle = ?

m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (6.64 x 10^{-27} kg)(3 x 10^8 m/s)^2\\$

$E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\$

<u>E = 3.73 x 10⁹ eV</u>

8 0

2 years ago

Four point charges, each of magnitude 2.38 µC, are placed at the corners of a square 75.2 cm on a side. If three of the charges

poizon [28]

Answer:

The Electric Force on Negative Charge is 2.968 N

Explanation:

charge on each corner, q = 2.38 micro coulomb

Side of square, a = 75.2 cm

Coulombic constant, K = 8.98755 x 10^9 Nm²/C²

sides of the square are A,B,C and D

and all sides of a square are equal so

AB = BC = CD = DA = 75.2 cm = 0.752 m

Diagonal, AC = BD = 1.414 x 0.752 = 1.06 m

Electric field at D due to charge at A

EA= Kq÷AB^2

= 8.98755×10^9 × 9.87×10^-6 ÷ 0.752^2

EA= 156863.82 N/C

Similarly Electric field at D due to charge C

EC=Kq÷CD^2

= 8.98755×10^9 ×9.87×10^-6 ÷ 0.752^2

EC= 156863.82 N/C

Electric field at D due to charge at BB

EB=Kq÷BD^2

EB=8.98755×10^9 × 9.87×10^-6 ÷ 1.06^2

EB=78949.01 N/C

Resolve the compoents

Ex = EA + EB cos 45

Ex = 156863.82 + 78949.01 x 0.707

Ex = 212689.2 N/C

Ey = EC + EB Sin 45

Ey = 156863.82 + 78949.01 x 0.707

Ey = 212689.2 N/C

The resultant electric field is

E = 1.414 x 212689.2 = 300787.95 N/C

the electric force on the negative charge is

F = q x E

F = 9.87 x 10^-6 x 300787.95

F = 2.968 N

7 0

1 year ago

Match the following.

Ksivusya [100]

Answer:

La verdad no entendi bien pero igual esta muy bien, Escoge lo que sientas en el corazón

Explanation:

Haz lo que sientas eso es lo mejor

3 0

2 years ago

the two ropes are used to vertically lower a 255 kg piano from exactly 4 m form a seocnd sotry window to the ground how much wor

Ilya [14]

Complete Question

The Question diagram is attached below

Answer:

a) $W_{Fg}= 12500 Nm$

b) $W_{T_1}= - 6339.3Nm$

c) $W_{T_2}= - 3662.8Nm$

Explanation:

From the question we are told that:

Mass $m=255kg$

Distance $d=4m$

Generally the equation for Work done is mathematically given by

$W=F*d$

For $F_g$

$W_{Fg}=2500 x 5.3$

$W_{Fg}= 12500 Nm$

For $T_1$

$W_{T_1}= - {1830 sin(60) x 4}$

$W_{T_1}= - 6339.3Nm$

For $T_2$

$W_{T_2}= - {1295 sin(45) x 4}$

$W_{T_2}= - 3662.8Nm$

6 0

2 years ago