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Eddi Din [679]
3 years ago
15

The "escape velocity" from Earth (the speed required to escape Earth's gravity) is 2.5 × 104 miles per hour. What is this speed

in m/s? (1 mile = 1609 m)
Physics
1 answer:
ANEK [815]3 years ago
7 0

Answer:

2.5 × 10⁴ mi/h = 1.1 × 10⁴ m/s

Explanation:

Hi there!

We have the following equivalencies:

1 mile = 1609 m

1 hour = 3600 s

Then to convert miles to meters, we can multiply the given quantity in miles by ( 1609 m/ 1 mile) and we will obtain the same quantity in meters. In the same way, if we want to convert hours into seconds, we can multiply the given quantity in hours by (3600 s/ 1 hour) and we will obtain seconds.

Let´s convert miles per hour into m/s:

2.5*10^{4} \frac{mi}{h}  (\frac{1609m}{1mi})(\frac{1h}{3600s}) = 1.1 × 10⁴ m/s  (notice how the units mi and h cancel)

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Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
  • second charge, q_2=3\times 10^{-3}\ C
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Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

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\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

and

x=-1-0.4365=-1.4365\ m

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