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GREYUIT [131]
3 years ago
14

Describe a technology used in space exploration.

Physics
1 answer:
Alex777 [14]3 years ago
3 0

Answer:

High speed optical communication technology

To be able to communicate from the space to the earth and from earth to space is one of the most essential features required during space exploration.

Explanation:

Space exploration involves going into the space, beyond the earth's atmosphere. Landing on other planets and studying their details, going into deeper space beyond the planets to discover new cosmic events or structures is all a part of space exploration.

The key to analyse the studies and observations is being able to communicate the data collected, photos taken etc to the launch centers or space centers on earth. The space centers on earth should also be able to communicate with the persons or the satellites in space.

This is made possible using the optical communication technology which involves the use of optical fibers, lasers etc, since high speeds are more efficient during communication

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Two wires with equal lengths are made of pure copper. The diameter of wire A is three times the diameter of wire B. When 8 kg ma
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Answer: c. YA < YB

Explanation:

The formula for Young’s modulus is = Tensile stress / Tensile strain

Tensile stress = Force x Length

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Tensile strain = Area x change in length

Area = pi x D2 / 4 ; Pi = 3.14

Change in length = L2 – L1 (New length – Initial length)

Given parameters:

Length of wire A = Length of wire B, (let’s use 2meters for the calculation)

For wire A, Diameter = 3 x Wire B diameter

Assuming Diameter of wire B = 1meter

Therefore, diameter of wire A = 1 x 3 = 3meters

It is said that wire B stretches more than wire A when the man of 8kg is placed on both

For wire B, let’s assume new length is = 4m

For wire A let’s assume new length is = 3m.

(i) Tensile strain of wire A =  

Area of wire A = 3.14 x (32)/4 = 7.065m2

Change in length = 3m - 2m = 1m.

Therefore, tensile strain = 7.065m2 x 1m = 7.065m3

Young’s modulus for wire A (YA) = 160kgm2/s divided by 7.065m3  

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(ii) Tensile strain of wire B =

Area of wire B = 3.14 x (12)/4 = 0.785m2

Change in length = 4m – 2m = 2m

Therefore, tensile strain = 0.785m2 x 2m = 1.57m3

Young’s modulus for wire B (YB) = 160kgm2/s divided by 1.57m3

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From the calculations above, we see that YA is less than YB (YA < YB). This is true given that wire A has a greater diameter than wire B which in turn impacts the Area of the wire since the diameter is directly proportional to area and the area is inversely proportional to the young’s modulus.

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