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elena-s [515]
3 years ago
5

Can you come up with a mathematical relationship, based on your data that shows the relationship between distance from the charg

es and electric field strength?
Physics
1 answer:
zalisa [80]3 years ago
4 0

Answer:

Explanation:

This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.

If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).

But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).

A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".

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Answer:

Part a)

M = 2.31 \times 10^{30} kg

Part b)

M = 1.15 M_s

Explanation:

Part a)

As we know that the period of revolution of the planet is given by

T = 2\pi\sqrt{\frac{r^3}{GM}}

now we know that

r = 10.5 \times 10^6 km = 1.05 \times 10^{10} m

T = 6.3 days

T = 5.44 \times 10^5 s

now we have

5.44 \times 10^5 = 2\pi\sqrt{\frac{(1.05\times 10^{10})^3}{(6.67 \times 10^{-11})M}}

M = 2.31 \times 10^{30} kg

Part b)

As we know that mass of sun is given as

M_s = 2.0 \times 10^{30} kg

now we have

\frac{M}{M_s} = \frac{2.31 \times 10^{30}}{2 \times 10^{30}}

M = 1.15 M_s

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