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A thin Nichrome wire connected to an ammeter surrounds a region of time-varying magnetic flux, and the ammeter reads 13 amperes. If instead of a single wire we use a coil of thin Nichrome wire containing 23 turns, what does the ammeter read?

Answers

Answer 1

Answer:

Answer:

The current would be same in both situation.

Explanation:

Given that,

Current I = 13 A

Number of turns = 23

We need to calculate the induced emf

Using formula of induced emf is

$\epsilon=NA(dB)/(dt)$

For N = 1

$\epsilon=A(dB)/(dt)$

We need to calculate the current

Using formula of current

$i=(\epsilon)/(R)$

Put the value of emf

$i=(A(dB)/(dt))/(R)$

Now, if the number of turn is 22 , then induced emf would be

$\epsilon'=NA(dB)/(dt)$

Then the current would be

$i'=(\epsilon')/(NR)$

$i'=(NA(dB)/(dt))/(NR)$

$i'=(A(dB)/(dt))/(R)$

$i'=i$

Hence, The current would be same in both situation.

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Light from a sodium lamp (λ = 589 nm) illuminates two narrow slits. The fringe spacing on a screen 150 cm behind the slits is 4.0 mm. What is the spacing (in mm) between the two slits?

Answers

The spacing between the two slits is 0.221mm.

The spacing between the two slits is given as,

$D=(\lambda L)/(y)$

Where $\lambda$ is wavelength, y is fringe spacing and L is length of screen.

Given that, $\lambda=589nm,L=150cm,y=4mm$

Substitute in above equation.

$D=(589*10^(-9)*150*10^(-2) )/(4*10^(-3) )\n \nD=2.21*10^(-4) m\n\nD=0.221mm$

Hence, the spacing between the two slits is 0.221mm.

Learn more about the sodium lamp here:

brainly.com/question/24867424

An electric generator contains a coil of 140 turns of wire, each forming a rectangular loop 71.2 cm by 22.6 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 4.32 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1120 rev/min about an axis perpendicular to the magnetic field?

Answers

Answer:

11405Volt

Explanation:

To solve this problem it is necessary to use the concept related to induced voltage or electromotive force measured in volts. Through this force it is possible to maintain a potential difference between two points in an open circuit or to produce an electric current in a closed circuit.

The equation that allows the calculation of this voltage is given by,

$\epsilon = BAN \omega$

Where

B = Magnetic field

A= Area

N = Number of loops

$\omega$ = Angular velocity

Our values previously given are:

$N = 140$

$A = 71.2*10^(-2)m*22.6*10^(-2)m=0.1609m^2$

$B = 4.32 T$

$\omega = 1120 rev / min$

We need convert the angular velocity to international system, then

$\omega = 1120 rev/min$

$\omega = 1120rev/min*(2\pi)/(1rev)*(1min)/(60sec)$

$\omega = 117.2rad/s$

Applying the equation for emf, we replace the values and we will obtain the value.

$\epsilon = BAN \omega$

$\epsilon = (4.32)(0.1609)(140)*117.2$

$\epsilon = 11405Volt$

The equations for single-slit and multiple-slit interference both contain the variable θ. For the multiple-slit case, the angle is: a. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences constructive interference.
b. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences destructive interference.
c. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences constructive interference.
d. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences destructive interference.

Answers

Answer:

the answers the correct one is c

Explanation:

The diffraction pattern for a slit is

a sin θ = m λ

Where a is the width of the slit, λ the wavelength, m the order of destructive interference and θ the angle where the interference occurs.

The expression for multi-slit diffraction (diffraction grating) is

d sin θ = m λ

Where d is the distance between slits, λ the wavelength m the order of the diffraction maximums and θ the angle for these maximums.

When we compare the expressions of the answers the correct one is c

A 75 kg man starts climbing a ladder that leans against a wall. If the weight of the ladder is negligible, determine how far up the ladder the man can climb before the ladders starts to slip. The coefficient of friction on both surfaces is μS=0.25

Answers

The man can climb $\bold { X (max) = 0.25* L* tan \alpha }$ , before the ladders starts to slip.

A - point at the top of the ladder

B - point at the bottom of the ladder

C - point where the man is positioned in the ladder

L- the length of the ladder

α - angle between ladder and ground

x - distance between C and B

The forces act on the ladder,

Horizontal reaction force (T) of the wall against the ladder

Vertical (upward) reaction force (R) of ground against the ladder.

Frictionalhorizontal ( to the left ) force (F)

Vertical( downwards) of the man,

mg = 75 Kg x 9.8 m/s² = 735 N

in static conditions,

∑Fx = T - F = 0 Since, T = F

∑Fy = mg - R = 0 Since, 735 - R = 0, R = 735

∑ Torques(b) = 0

In point B the torque produced by forces R and F is Zero

Then:

∑Torques(b) = 0

And the arm lever for each force,

mg = 735

Since, ∑Torques(b) = 0

$\bold {735* x* cos\alpha = F* L* sin\alpha }$ Since,T = F

$\bold {F = \frac {735* x* cos\apha }{L* sin\alpha }}$ $\bold { \frac {cos \alpha } { sin\alpha }= cot\alpha =\frac 1{tan\alpha}}$

$\bold {F = \frac {735* x* cos\apha }{L }}$

$\bold {F = 735* x* tan\alpha }}$

F < μR the ladder will starts slipping over the ground

μ(s) = 0.25

$\bold { X (max) = 0.25* L* tan \alpha }$

Therefore, the man can climb $\bold { X (max) = 0.25* L* tan \alpha }$ , before the ladders starts to slip. \

To know more about Torque,

brainly.com/question/6855614

Answer:

x (max) = 0,25*L*tanα

Explanation:

Letá call

A: point at the top of the ladder

B: the point at the bottom of the ladder

C: point where the man is up the ladder

L the length of the ladder

α angle between ladder and ground

"x" distance between C and B

Description

The following forces act on the ladder

Point A: horizontal (to the right) reaction (T) of the wall against the ladder

Point B : Vertical (upwards) reaction (R) of ground against the ladder

frictional horizontal ( to the left ) force (F)

Point C : Weight (vertical downwards)) of the man mg

mg = 75 Kg * 9,8 m/s² mg = 735 [N]

Then in static conditions:

∑Fx = T - F = 0 ⇒ T = F

∑Fy = mg - R = 0 ⇒ 735 - R = 0 ⇒ R = 735

∑Torques(b) = 0

Note: In point B the torque produced by forces R and F are equal to 0

Then:

∑Torques(b) = 0

And the arm lever for each force is:

mg = 735

d₁ for mg and d₂ for T

cos α = d₁/x then d₁ = x*cosα

sin α = d₂ / L then d₂ = L*sinα

Then:

∑Torques(b) = 0 ⇒ 735*x*cosα - T*L*sinα = 0

735*x*cosα = T*L*sinα

T = F then 735*x*cosα = F*L*sinα

F = (735)*x*cosα/L*sinα cos α / sinα = cotgα = 1/tanα

F = (735)*x*cotanα/L or F = (735)*x/L*tanα

When F < μ* R the ladder will stars slippering over the ground

μ(s) = 0,25 0,25*R = 735*x/L*tanα

x = 0,25*R*tanα*L/735

But R = mg = 735 then

0,25*L*tanα = x

Then x (max) = 0,25*L*tanα

On a 30degrees day, there is an explosion. The sound is heard 3.4s after seeing the flash. How far away was the explosion

Answers

Answer:

Distance = 1.2 km (Approx)

Explanation:

Given:

Temperature (T) = 30°C

Total Time taken (t) = 3.4 Sec

We know that sound increases in the air sound increase nearly 0.60 m/s for every sound with temperature.

Speed of sound = 331 m/s

So,

V = (331 + 0.60T) m/s

V = 331+(0.60×300C) m/s

V = 349 m/s

Distance = speed × time

Distance = 349 × 3.4

Distance = 1,186.6 m

Distance = 1.2 km (Approx)

When the sum of all the forces acting on a block on an inclined plane is zero, the blockA) must be at rest
B) must be accelerating
C) may be slowing down
D) may be moving at constant speed

Answers

Answer:

hmmm thats too hard for me.

Explanation:

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