PHYSICS COLLEGE

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s , the induced emf in the second coil has a magnitude of 1.60×10−3 V . Part A What is the mutual inductance of the pair of coils? MM = nothing H Request Answer Part B If the second coil has 22 turns, what is the flux through each turn when the current in the first coil equals 1.25 A ? ΦΦ = nothing Wb

Answers

Answer 1

Answer:

To solve this problem we need to use the emf equation, that is,

$E=m(dI)/(dT)$

Where E is the induced emf

I the current in the first coil

M the mutual inductance

Solving for a)

$M=(E)/((dI)/(dT))\nM=(1.6*10^(-3))/(0.245)=6.53*10^(-3)H$

Solving for b) we need the FLux through each turn, that is

$\Phi=(MI)/(N)$

Where N is the number of turns in the second coil

$\Phi=(6.53*10^(-3)*1.25)/(22)=3.71*10^(-4)Wb$

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A convex lens is placed on a flat glass plate and illuminated from above with monochromatic red light. When viewed from above, concentric bands of red and dark are observed. What does one observe at the exact center of the lens where the lens and the glass plate are in direct contact?A) a darkspotB) a bright spot thatis some color other than redC) a bright redspotD) a rainbow of color

Answers

Answer:

After passing through the glass plate, the red light disperses and meets at point.

The convex lens has two refracting surfaces, and convex kens is called as converging lens. So, at the exact center of the lens, one observes a Dark spot.

Thus, the correct option is a) one observes a dark spot.

Answer:

The answer is: A) a darkspot

Explanation:

When the red light passes through the glass plate, it is scattered. the convex lens (convergent lens) has two refractive surfaces, therefore, in the center of the lens, a characteristic dark spot would be observed.

when a ball is dropped off a cliff in free fall, it has an acceleration of 9.8 m/s^2. what is its acceleration as it gets closer to the ground

Answers

acceleration due to gravity is contract for the purposes of this question, so the acceleration would remain at 9.8 m/s^2

If the ball, the cliff, and the ground are all on the Earth, and everything is bathed in an ocean of air, then the ball's acceleration will decrease as it falls, because of the friction of air resistance. If it has far enough to fall, it's possible that its acceleration may even become zero, and the ball settle on a constant speed (called "terminal velocity") before it hits the ground.

But until we get to College-level Physics and Engineering, we ALWAYS ignore that stuff, and assume NO AIR RESISTANCE. The ball is in FREE FALL, and the ONLY force acting on it is the force of gravity. We also assume that the distance of the fall is small enough so that the value of gravity is constant over the entire fall.

Under those assumptions, there's nothing present to change the acceleration of the falling ball. It's 9.81 m/s² when it rolls off the edge of the cliff, and it's 9.81 m/s² when it hits the ground.

A machine can make doing work easier by reducing the force exerted, changing the distance over which the force is exerted, or changing the direction of the force.True OR False

HELP ME!!!!!!¡!!!!

Answers

I believe the correct answer is true. A machine can make doing work easier by reducing the force exerted, changing the distance over which the force is exerted, or changing the direction of the force. Hope this answer the question.

You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.09 mm and place your screen 8.61 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.53 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength ???? expressed in nanometers?

Answers

Answer:

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Explanation:

The formula of the Young's Double Slit experiment is given as follows:

$\Delta x = (\lambda L)/(d)\n\n\lambda = (\Delta x d)/(L)$

where,

λ = wavelength = ?

L = distance between screen and slits = 8.61 m

d = slit spacing = 1.09 mm = 0.00109 m

Δx = distance between consecutive bright fringes = $(4.53\ cm)/(10)$ = 0.00453 m

Therefore,

$\lambda = ((0.00453\ m)(0.00109\ m))/(8.61\ m)$

λ = 5.734 x 10⁻⁷ m = 573.4 nm

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