Physics, 05.09.2020 20:01 djdjdjdbdbjx

Calculate earth's average speed relative to the sun.

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Answer from: perezz97

Average speed of the earth relative to the sun = 30045.17 m/s

Explanation:

To calculate the average speed of the earth relative to the sun, we would have to divide the distance traveled from earth to sun by the time of travel. Thus;

Average speed of the earth = Distance travelled from earth to sun/total time of travel.

Now, formula for distance traveled by earth = 2πr

Where r is the average distance from the Earth to the sun and is given by 1.508 × 10^(11) m

Thus;

distance traveled by earth = 2π × 1.508 × 10^(11) = 947504344322.6816 m

Total time is time taken for the earth to complete a revolution around the sun. Time = 365 days

Converting this to seconds gives;

Since 24 hours make a day, we have;

365 × 24 × 60 × 60 = 3.1536 × 10^(7) s

Thus;

Average speed of the earth = 947504344322.6816/(3.1536 × 10^(7)) = 30045.17 m/s

Answer from: Quest

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia

Answer from: Quest

The answer is 4f======since electrostatic force is given by $f_{elec} = \dfrac{kq_1 q_2}{r^2}$ we let f be the electrostatic force when r = 1 m $f = \dfrac{kq_1 q_2}{1^2} = kq_1q_2$ if we let $f_{0.5}$ denote the electrostatic force when the distance is decreased so that r = 0.5m, then $\begin{aligned} f_{0.5} & = \dfrac{kq_1 q_2}{0.5^2} \\ & = \frac{1}{0.5^2} \cdot kq_1q_2 \\ & = \frac{1}{\left( \frac{1}{2}\right)^2} \cdot kq_1q_2 \\ & = 4 \cdot \boxed{ kq_1q_2 } \end{aligned}$ we established earlier that f = kq₁q₂. therefore $f_{0.5} = 4 f$ the force will now be 4f.

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Calculate earth's average speed relative to the sun....