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Leya [2.2K]
3 years ago
12

What would the electrostatic force be for two 0.005C charges 50m apart?

Physics
1 answer:
irakobra [83]3 years ago
3 0

Answer:

90 N

Explanation:

The electrostatic force between two charges is given by:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

In this problem we have

q1 = q2 = 0.005 C

r = 50 m

So the electrostatic force is

F=(9\cdot 10^9 N m^2 C^{-2})\frac{(0.005 C)^2}{(50 m)^2}=90 N

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Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6
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Answer:

The answer to the question is as follows

The  acceleration due to gravity for low for orbit is  9.231 m/s²

Explanation:

The gravitational force is given as

F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }

Where F_{G} = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹\frac{Nm^{2} }{kg^{2} }

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by  F_{G} and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and  F_{G} = \frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} } = 9.231 N

Therefore the acceleration due to gravity =  F_{G} /mass  

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is  9.231 m/s²

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ENGLISH:

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3 years ago
5. If a spring with a 40 N/m spring constant is used to push a 10 kg ball to a speed of 2 m/s,
kolbaska11 [484]

By compressing the spring a distance <em>x</em> (in m), you are storing 1/2 <em>k</em> <em>x</em> ² (in J) of potential energy, which is converted completely into kinetic energy 1/2 <em>m v</em> ², where

• <em>k</em> = 40 N/m = spring constant

• <em>m</em> = 10 kg = mass of the ball

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So we have

1/2 <em>k</em> <em>x</em> ² = 1/2 <em>m</em> <em>v</em> ²

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