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A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field, such that the normal to the coil is parallel to the magnetic field. The coil is then rotated through an angle of 90˚, so that the normal becomes perpendicular to the magnetic field. The coil has an area of 1.5 × 10-3 m2, 50 turns, and a resistance of 180 Ω. During the time while it is rotating, a charge of 9.3 × 10-5 C flows in the coil. What is the magnitude of the magnetic field?

Answers

Answer 1

Answer:

Answer:

3.4 x 10^-4 T

Explanation:

A = 1.5 x 10^-3 m^2

N = 50

R = 180 ohm

q = 9.3 x 106-5 c

Let B be the magnetic field.

Initially the normal of coil is parallel to the magnetic field so the magnetic flux is maximum and then it is rotated by 90 degree, it means the normal of the coil makes an angle 90 degree with the magnetic field so the flux is zero .

Let e be the induced emf and i be the induced current

e = rate of change of magnetic flux

e = dФ / dt

i / R = B x A / t

i x t / ( A x R) = B

B = q / ( A x R)

B = (9.3 x 10^-5) / (1.5 x 10^-3 x 180) = 3.4 x 10^-4 T

Answer 2

Answer:

Final answer:

The magnitude of the magnetic field can be calculated using Faraday's Law of electromagnetic induction, by setting up and solving an equation involving the number of turns in the coil, the area of the coil, and the time it takes for the coil to rotate.

Explanation:

To calculate the magnitude of the magnetic field, we can use Faraday's Law of electromagnetic induction, which can be expressed as E = d(N∙Φ )/dt, where E represents the induced EMF, N is the number of turns, and Φ is the magnetic flux (flux equals the product of the magnetic field B, the area A through which it passes and the cosine of the angle between B and A).

Given the information in the problem, we know that E = Q/R ∙ t. Since the coil is rotated through 90 degrees, it goes from being parallel to being perpendicular to the field, resulting in a change in magnetic flux of BNA. We can set up the equation E = d(NBA)/dt = Q/R ∙ t = [(50 turns) ∙ (1.5 × 10-3 m²) ∙ B)/(t)]

We can solve this equation to determine the magnitude of the magnetic field B. Remember, always double-check your calculations to ensure their accuracy.

Learn more about Magnetic Field here:

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As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.73 g/m and a 1.30 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.30 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 200 Hz . Next, with the 1.30 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 313 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work.What value of g will you report back to headquarters?

Answers

Answer:

The value of g is $g =76.2 m/s^2$

Explanation:

From the question we are told that

The mass of the weight is $m = 1.30 kg$

The spring constant $k = 1.73 g/m = 1.73 *10^(-3) \ kg/m$

The second harmonic frequency is $f = 100 \ Hz$

The number of oscillation is $N = 200$

The time taken is $t = 315 \ s$

Generally the frequency is mathematically represented as

$f = (v)/(\lambda)$

At second harmonic frequency the length of the string vibrating is equal to the wavelength of the wave generated

$l = \lambda$

Noe from the question the vibrating string is just half of the length of the main string so

Let assume the length of the main string is $L$

So $l = (L)/(2)$

The velocity of the vibrating string is mathematically represented as

$v = \sqrt{(T)/(\mu) }$

Where T is the tension on the string which can be mathematically represented as

$T = mg$

$v = \sqrt{(mg)/(k) }$

Then

$f = (v)/((L)/(2) )$

=> $v = (fL )/(2)$

=> $\sqrt{(mg)/(k) } = (fL)/(2)$

=> $g = (f^2 L^2 \mu)/(4m)$

substituting values

$g = ((100) * (1.73 *10^(-3) ))/((4 * 1.30)) L^2$

$g = 3.326 m^(-1) s^(-2) L^2$

Generally the period of oscillation is mathematically represented as

$T_p = 2 \pi \sqrt{(L)/(g) }$

=> $L = (T^2 g)/(4 \pi ^2)$

The period can be mathematically evaluated as

$T_p = (t)/(N)$

substituting values

$T_p = (315)/(200)$

$T_p = 1.575 \ s$

Therefore

$L = (1.575^2 * g )/(4 \pi ^2)$

$L = 0.0628 ^2 g$

$g = 3.326 m^(-1) s^(-2) L^2$

substituting for L

$g = 3.326 ((0.0628) g)^2$

=> $g = (1)/((3.326)* (0.0628)^2)$

$g =76.2 m/s^2$

A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its kinetic energy at circled A? 1.2635 Correct: Your answer is correct. J (b) What is its speed at circled B? 4.54 Correct: Your answer is correct. m/s (c) What is the net work done on the particle by external forces as it moves from circled A to circled B?

Answers

Answer:

a). $E_(kA)=1.2635 J$

b). $V_(B)=4.535(m)/(s)$

c). Δ $E_(t)=8.4635 J$

Explanation:

ΔE=kinetic energy

a).

$E_(kA)=(1)/(2)*m*v_(A) ^(2) \n v_(A)=1.9 (m)/(s)\n m=0.70kg\nE_(kA)=(1)/(2)*0.70kg*(1.9 (m)/(s))^(2) \nE_(kA)=1.2635 J$

b).

$E_(kB)=(1)/(2)*m*v_(B) ^(2)$

$V_(B)^(2)=(E_(kB)*2)/(m) \nV_(B)=\sqrt{(E_(kB)*2)/(m)} \nV_(B)=\sqrt{(7.2J*2)/(0.70kg)} \nV_(B)=4.53 (m)/(s)$

c).

net work= EkA+EkB

$E_(t)=E_(kA)+ E_(kB)\nE_(t)=1.2635J+7.2J\nE_(t)=8.4635J$

A straight trail with a uniform inclination of 16° leads from a lodge at an elevation of 600 feet to a mountain lake at an elevation of 7,000 feet. What is the length of thetrail (to the nearest foot)?
O A. 6,658 ft
OB. 25,396 ft
OC. 7,282 ft
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Answers

I believed the answer is d

Shameeka is studying for an exam she took the notable about calcium and chlorine which are known to for my comic born's which shameekas error?

Answers

The individual calcium atom has a positive and not negative, 2 charge

Answer:

The individual calcium atom has a positive, not negative, 2 charge.

Explanation:

Did the quiz also had it on the unit test on edgunity.

Hope this helps guys!

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

Which of the following statements correctly compares the relationship between the earth, its atmosphere and radiation?1. The earth is cooled and its atmosphere is heated by solar radiation.

2. The earth is heated and its atmosphere is cooled by terrestrial radiation.

3. The earth is cooled and its atmosphere is heated by terrestrial radiation.

4. The earth is heated and its atmosphere is cooled by solar radiation.

Don't answer unless you know for sure. Thank you so much!

Answers

Answer: The option 4 is correct answer.

Explanation:

Terrestrial radiation is a long wave electromagnetic radiation. It originates from the earth and its atmosphere.

The sun emits a huge amount of energy. It travels across the space. The atmosphere is not directly heated by the solar radiation. It is heated by the terrestrial radiation that the planet itself emits.

When the land is heated then it emits radiation which heats up the atmosphere.

The earth is cooled and its atmosphere is heated by terrestrial radiation.

Therefore, the relationship between the earth, its atmosphere and radiation is correctly compared by statement 4.

The earth is cooled and its atmosphere is heated by terrestrial radiation.

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