Which law does the following statement express? "In all cases of electromagnetic induction, the induced voltages have adirection such that the currents they produce oppose the effect that produces them."

Answers

Answer 1

Answer:

Final answer:

Faraday's Law of electromagnetic induction states that induced voltages produce currents that oppose the change in the magnetic field.

Explanation:

The law that the statement expresses is Faraday's Law of electromagnetic induction.

According to Faraday's Law, whenever there is a change in the magnetic field through a conductor, it induces an electromotive force (EMF) or voltage across the conductor. This induced voltage creates a current that flows in a direction that opposes the change in magnetic field.

This phenomenon is described by Lenz's Law, which states that the induced current always flows in such a way as to produce a magnetic field that opposes the change in the external magnetic field.

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A friend tells you that a lunar eclipse will take place the following week, and invites you to join him to observe the eclipse through a high-powered telescope he owns. You are curious what the eclipse might look like from different perspectives in space. If the moon has a diameter of 2,159.14 miles, what is the maximum distance that it could be observed by the naked eye with enough detail that you could distinguish it from other celestial bodies (assuming that you have 20/20 vision)

Answers

Answer:

y = 80.2 mille

Explanation:

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

θ = 1.22 550 10⁻⁹ / 0.002

θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

tan θ = y / L

y = L tan θ

y = 2,389 10⁵ tan 3,355 10⁻⁴

y = 8.02 10¹ mi

y = 80.2 mille

This is the smallest size of an object seen directly by the eye

Final answer:

An individual with 20/20 vision can observe the moon from a maximum distance of around 6200 km or 3850 miles. Beyond this distance, it might be difficult to distinguish the moon from other celestial objects without using a telescope. The use of a telescope can expand this range significantly.

Explanation:

The detailed observation of a lunar eclipsed, when viewed without any form of optical aid like a telescope, is contingent on many factors, one of which is the human eye's angular resolution—the eye's ability to differentiate between two separate points of light. For an average human eye with 20/20 vision, the angular resolution is approximately 0.02 degrees.

To calculate the maximum distance at which the moon could be observed clearly with the eye, the formula for small angle approximation can be used, which in this context is: Distance = Size / Angle = (2159.14 miles) / (0.02 degrees in radians). This calculates to a distance of approximately 6200 km or 3850 miles.

Beyond this distance, distinguishing the moon from other celestial bodies might be challenging using just the eye. Utilizing a high-powered telescope would significantly extend this range by magnifying the image, allowing clearer detail over much greater distances.

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A sheet of aluminum alloy is cold-rolled 33 percent to a thickness of 0.096 in. If the sheet is then cold-rolled to a final thickness of 0.061 in., what is the total percent cold work done

Answers

Answer:

The total percent cold work done is 36.46%

Explanation:

Let initial metal thickness = T

Final metal thickness = t

The percent cold work done = WC

Then

%Wc = (T - t)/T × 100

% Wc = ( 0.096 - 0.061 )/0.096 ×100

Total %WC = 36.46%

Answer:

The total percent of cold work is 57.34%

Explanation:

Let x the initial thickness of the sheet. After 33% of cold working, the thickness is 0.096 in. Then:

x - 0.33x = 0.096

x = 0.143 in

the final thickness is equal to 0.061 in. The percent of cold work done is:

$percent-of-cold-work=((initial-thickness)-(final-thickness))/(initial-thickness)*100$

$percent-of-cold-work=(0.143-0.061)/(0.143) *100=57.34$ %

Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits

Answers

Answer:

The maximum number of bright spot is $n_(max) =5001$

Explanation:

From the question we are told that

The slit distance is $d = 1 \ mm = 0.001 \ m$

The wavelength is $\lambda = 400 \ nm = 400*10^(-9 ) \ m$

Generally the condition for interference is

$n * \lambda = d * sin \theta$

Where n is the number of fringe(bright spots) for the number of bright spots to be maximum $\theta = 90$

=> $sin( 90 )= 1$

$n = (d )/(\lambda )$

substituting values

$n = ( 1 *10^(-3) )/( 400 *10^(-9) )$

$n = 2500$

given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as

$n_(max) = 2 * n + 1$

The 1 here represented the central bright spot

$n_(max) = 2 * 2500 + 1$

$n_(max) =5001$

In an 8.00km race, one runner runs at a steady 11.8 km/hr and another runs at 15.0 km/hr. How far from the finish line is the slower runner when the faster runner finishes the race?

Answers

Answer:

The slower runner is 1.71 km from the finish line when the fastest runner finishes the race.

Explanation:

Given;

the speed of the slower runner, u₁ = 11.8 km/hr

the speed of the fastest runner, u₂ = 15 km/hr

distance, d = 8 km

The time when the fastest runner finishes the race is given by;

$Time = (Distance )/(speed)\n\nTime = (8)/(15) \n\nTime = 0.533 \ hr$

The distance covered by the slower runner at this time is given by;

d₁ = u₁ x 0.533 hr

d₁ = 11.8 km/hr x 0.533 hr

d₁ = 6.29 km

Additional distance (x) the slower runner need to finish is given by;

6.29 km + x = 8km

x = 8 k m - 6.29 km

x = 1.71 km

Therefore, the slower runner is 1.71 km from the finish line when the fastest runner finishes the race.

Early in the morning, when the temperature is 5.5 °C, gasoline is pumped into a car’s 53-L steel gas tank until it is filled to the top. Later in the day the temperature rises to 27 °C. Since the volume of gasoline increases more for a given temperature increase than the volume of the steel tank, gasoline will spill out of the tank. How much gasoline spills out in this case?

Answers

Answer:

Volume of gasoline spills out is 0.943 L.

Explanation:

Volumetric expansion of both gasoline and steel tank is :

$\beta_(gas)=9.5 *10^(-4)/K\n\beta_(steel \ gas)=3.6 * 10^(-5)/K.$ { source Internet}

We know expansion due to temperature change is :

$\Delta V=\beta*\Delta T* V$

For gasoline:

$\Delta V_g=0.98 \ L.\n$

Similarly for Steel tank:

$\Delta V_(steel \ gas)=0.037\ L$ .

Now, volume of gasoline spills out is equal to difference between expansion in volume.

$\Delta V_(gas)-\Delta V_(Steel \ gas)=0.98-0.037\ L=0.943\ L.$

Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculate the force of gravity that the space shuttle experiences

Answers

Final answer:

The force of gravity that the space shuttle experiences is 9.8 x 10^5 Newtons.

Explanation:

To calculate the force of gravity that the space shuttle experiences, we can use the equation F = mg, where F represents the force of gravity, m is the mass of the object, and g is the acceleration due to gravity (approximately 9.8 m/s² on Earth). In this case, the mass of the space shuttle is given as 1.0 x 10^5 kg. However, we need to convert the altitude of the shuttle into meters, so 200.0 km becomes 200,000 meters.

Now we can calculate the force of gravity:

F = (1.0 x 10^5 kg)(9.8 m/s²)

F = 9.8 x 10^5 N

Therefore, the space shuttle experiences a force of gravity of 9.8 x 10^5 Newtons.

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