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Alenkinab [10]
3 years ago
6

Question 3 You see the coordinates 5 E longitude, 10 N latitude. You do not need to look at a map in order to deduce that this l

ocation is a. near both the equator and the prime meridian Selected: b. near the International Date Line as well as the North PoleThis answer is incorrect. c. near the International Date Line and the prime meridian d. near the equator but quite far from the prime meridian e. near both the equator and the International Date Line
Physics
2 answers:
olchik [2.2K]3 years ago
8 0

Answer:

a. near both the equator and the prime meridian

Explanation:

Lets first understand what is equator and prime meridian.

We have drawn an imaginary grid on the Earth for navigational and other purposes. This grid has two coordinates : Latitude and Longitude. Equator is the largest latitude circle that divides the Earth in two equal hemispheres. This is taken as 0° for latitude. Latitudes increase as we move towards North/South pole which lies at 90°N and 90° S respectively. Longitudes are the lines running from north pole to south pole. The longitude that crosses from Greenwich is taken as 0°. This longitude is also known as prime meridian. The value of longitudes vary from 0° to 180° East and 180° West.

Now, the given coordinates are: 5° E and 10° N. It is quite evident that this location will be close to equator and prime meridian.

san4es73 [151]3 years ago
5 0

Answer:

a. Near both the equator and the prime meridian.

Explanation:

The  equator is at 0 degrees latitude and the prime meridian is 0 degrees longitude.

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Consider frames X and Y:

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In the shadow of a tree with a dense, leafy canopy, one sees a number of light spots. Surprisingly, they all appear to be circul
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The characteristics of the diffraction phenomenon allow to find the result for the shape of the points of light that you pass the tree is:

  • The shape of the dots is circular because it is in the range of far-field diffraction.

Diffraction is the phenomenon where the undulatory part of the light becomes evident, it is the interference of the waves that make up each ray of light, for this phenomenon to occur it must be fulfilled that the wavelength is of the order of the space where pass the light.

In the leafy tree it has many leaves, but there are spaces between them, some of these spaces are small and it fulfills the diffraction condition, therefore we see bright spots and not a continuous shadow.

Diffraction can be classified depending on the distance to the observer:

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In this case, the distance from the leaves to the observer is large, therefore we are in the case of far-field diffraction and since the edge of the leaves that forms the diffraction is closed, the observable shape is a circle.

In conclusion using the characteristics of the diffraction phenomenon we can find the result for the shape of the points of light that pass the tree is:

  • The shape of the dots is circular because it is in the range of far-field diffraction.

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3 years ago
A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa
Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

Q=P\tan\phi...(I)

We need to calculate the \phi

Using formula of \phi

\phi=\cos^{-1}(Power\ factor)

Put the value into the formula

\phi=\cos^{-1}(0.6)

\phi=53.13^{\circ}

Put the value of Φ in equation (I)

Q=600\tan(53.13)

Q=799.99\ kVAR

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

Q'=600\tan(0.95)

Q'=9.94\ KVAR

We need to calculate the difference between Q and Q'

Q''=Q-Q'

Put the value into the formula

Q''=799.99-9.94

Q''=790.05\ KVAR

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0
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