PHYSICS COLLEGE

) A circular coil of diameter 20. cm, with 16. turns is in a 0.13 Tesla field. (a) Find the total flux through the coil when the field is perpendicular to the coil plane. (b) If the coil is rotated in 10. ms so its plane is parallel to the field, find the average induced emf.

Answers

Answer 1

Answer:

Answer:

(a) 0.0041 weber

(b) 0.41 volt

Explanation:

diameter of coil, d = 20 cm

radius of coil, r = half of diameter = 10 cm = 0.1 m

magnetic field strength, B = 0.13 tesla

(a)

The angle between the normal of the coil and the magnetic field is 0°.

Magnetic flux, Ф = B x A x Cos 0°

Ф = 0.13 x 3.14 x 0.1 x 0.1 x 1

Ф = 0.0041 Weber

(b)

angle between the magnetic field and the normal of the coil is 90°.

time, t = 10 ms = 0.01 s

final flux = B x A x cos 90° = 0

induced emf = rate of change of magnetic flux

e = (0.0041 - 0) / 0.01

e = 0.41 Volt

Answer 2

Answer:

Answer:

a) $\phi=0.4084\ T.m^2$

b) $emf=653.44\ V$

Explanation:

Given:

diameter of the coil, $d=20\ cm=0.2\ m$

no. of turns in the coil, $N=16$

magnetic field strength to which the coil is subjected, $B=0.13\ T$

time taken by the coil to rotate from normal the field to parallel, $t=10* 10^(-3)\ s$

The flux through the coil can be given as:

$\phi=BA$

where:

$A=$ area enclosed by the section of the coil

$\phi=0.13* \pi* (0.2^2)/(4)$

$\phi=0.4084\ T.m^2$

When the coil is rotated there is change in flux which lead to an induced emf in the coil according to the Faraday's law:

$emf=N(d\phi)/(t)$

where:

$d\phi=$ change in the flux

here the flux changes from maximum value to zero when the coil becomes parallel to the field lines because then there is no field line intercepting the coil area.

$emf=16* (0.4084)/(0.01)$

$emf=653.44\ V$

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The animal that is hunted and consumed is considered the

Answers

Predator? like they hunt their prey

Answer:

prey

Explanation:

If you want to play a tune on wine glasses, you’ll need to adjust the oscillation frequencies by adding water to the glasses. This changes the mass that oscillates (more water means more mass) but not the restoring force, which is determined by the stiffness of the glass itself. If you need to raise the frequency of a par- ticular glass, should you add water or remove water?

Answers

Answer:Reducing mass i.e. water

Explanation:

Frequency For given mass in glass is given by

$f=(1)/(2\pi )\sqrt{(k)/(m)}$

where k =stiffness of the glass

m=mass of water in glass

from the above expression we can see that if mass is inversely Proportional to frequency

thus reducing mass we can increase frequency

A 1.5m wire carries a 7 A current when a potential difference of 87 V is applied. What is the resistance of the wire?

Answers

Answer:

$R\approx12.43 \,\, \Omega$

Explanation:

We can use Ohm's Law to find the resistance R of a wire that carries a current I under a given potential difference:

$V=I\,\,R\nR = (V)/(I) \nR=(87)/(7) \nR\approx12.43 \,\, \Omega$

Answer:

Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

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Name at 2 areas of physics that make video games possible

Answers

Answer:

projectiles

electromagnetic

Answer:

Explanation:

física cuántica y Quantum Moves

G Water enters a house through a pipe 2.40 cm in diameter, at an absolute pressure of 4.10 atm. The pipe leading to the second-floor bathroom, 5.20 m above, is 1.20 cm in diameter. The flow speed at the inlet pipe is 4.75 m/s a) What is the algebraic expression for flow speed in the bathroom?
b) Calculate the flow speed in the bathroom.
c) What is algebraic expression for the pressure in the bathroom?
d) Calculate the water pressure in the bathroom. Report your answer in the (atm) unit.

Answers

Answer:

A) A₁ V₁ = A₂V₂

B) V₂ = 19 m /s

C) P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂² + (h₂ - h₁ )ρg

D) P₂ = 1.88 atm

Explanation:

A) From the piaget's theory of conservation of volume, we can calculate the rate of flow of water from;

A₁ V₁ = A₂V₂

Where;

A₁ and A₂ are area of cross section V₁ and V₂ are velocity of flow at two places along pipe.

B) Using the formula given in A above, we obtain;

π x 1.2² x 4.75 = π x 0.6² x V₂

V₂ x 0.36 = 6.84

V₂ = 6.84/0.36

V₂ = 19 m /s

c ) To find pressure we shall apply Bernoulli's theorem in fluid dynamics;

P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂² + (h₂ - h₁ )ρg

Where;

P₁ and P₂ are pressure at ground and second floor respectively

v₁ and v₂ are velocity at ground and second floor respectively

h₁ and h₂ are height at ground and second floor respectively ρ is density of water.

Thus, plugging in the relevant values to obtain;

4.1 x 10⁵ + (1/2 x 1000 x 4.75²) = P₂ + (1/2 x 1000 x 19²) + (5.2 x 1000 x 9.8)

(4.1 x 10⁵) + (0.11 x 10⁵) = P₂ + (1.8 X 10⁵) + (0.51 X 10

P₂ = 1.9 X 10⁵ N/m² = 1.88 atm

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 4.90 m away from the slits.a. What is the distance Δ ymax-max between the first maxima (on the same side of the central maximum) of the two patterns?
b. What is the distance Δymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?

Answers

Answer:

a)Δy = 81.7mm

b)Δy = 32.7cm

Explanation:

To calculate the distance between any point of the interference pattern, simply use the trigonometric ratio of the tangent:

$Tan \theta = (y)/(D)$

where D is the separation between the slits and the screen where the interference pattern is observed.

a) In this case:

Δy = |y1max (λ1) − y1max (λ2)|

Δy = $|(D\lambda _1)/(d) - (D\lambda _2)/(d) |$

Δy = $D |(d/20)/(d) - (d/15)/(d) |$

Δy = $D |(1)/(20) - (1)/(15) |$

Δy = $4.90 |(1)/(20)- (1)/(15) |$

Δy = 81.7mm

The separation between these maxima is 81.7 mm

Δy = |y₂max (λ1) − y₂max (λ2)|

Δy = $D|(2(d/20))/(d) - (5(d/15))/(2d) |$

Δy = $4.90|(1)/(10) - (1)/(6) |$

Δy = 32.7cm

The separation between the maximum interference of the 2nd order (2nd maximum) of the pattern produced by the laser 1 and the minimum of the 2nd order (3rd minimum) of the pattern produced by the laser 2 is 32.7 cm.

Final answer:

We can solve the problem using the concepts of waveinterference and the formulas for maxima and minima positions (i.e., y = L*m*λ/d and y = L*(m+1/2)*λ/d respectively). The difference between the first maxima of the two patterns is 4.9/60 m and the difference between the second maximum of laser 1 and the third minimum of laser 2 is also 4.9/60 m.

Explanation:

The problem described deals with wave interference and can be addressed using the formulas for path difference and phasedifference.

To answer part a, we need to find the difference between the positions of the first maxima for the two lasers. The position of any maxima in an interference pattern can be found using the formula: y = L * m * λ / d, where L is the distance from the slits to the screen, m is the order of the maxima, λ is the wavelength, and d is the slit separation.

So for the first laser (λ1=d/20) the position of the first maxima would be y1 = 4.9m * 1 * (d/20) / d =4.9/20 m.

And for the second laser (λ2 = d/15) the position of the first maxima would be y2= 4.9m * 1 * (d/15) / d =4.9/15 m.

Then, the distance Δ ymax-max between the first maxima of the two patterns is y2-y1= 4.9/15 m - 4.9/20 m = 4.9/60 m.

Answering part b involves finding the positions of the second maximum of laser 1 and the third minimum of laser 2. The position of any minimum in an interference pattern can be calculated using the formula: y = L * (m+1/2) * λ / d. For the second maximum of laser 1, we have y1max2 = 4.9 m * 2 * (d/20) / d = 4.9/10 m. For the third minimum of laser 2, we have y2min3 = 4.9m * (3.5) * (d/15)/d = 4.9*7/30 m. The difference Δymax-min is y2min3-y1max2= 4.9*7/30 m - 4.9/10 m = 4.9/60 m.

Learn more about Wave Interference here:

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