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3 years ago

What is the repulsive force between two pith balls that are 10.0 cm apart and have equal charges of -40.0 NC?

Physics

1 answer:

galina1969 [7]3 years ago

5 0

Answer:

So the repulsive force between the pith ball will be $1.44\times 10^{-3}N$

Explanation:

We have given that the pith ball have the equal charge q = -40 nC = $-40\times 10^{-9}C$

Distance between the charges = 10 cm =0.1 m

According to coulombs law $F=\frac{KQ_1Q_2}{R*2}$

$F=\frac{9\times 10^9\times -40\times 10^{-9}\times -40\times 10^{-9}}{0.1^2}=1440000\times 10^{-9}=1.44\times 10^{-3}N$

So the repulsive force between the pith ball will be $1.44\times 10^{-3}N$

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(NEED HELP BADLY)

spin [16.1K]

Answer:

The answer to your question is below

Explanation:

Data 1

mass 1 = 250

mass 2 = 250 kg

gravity constant = 6.67 x 10⁻¹¹ Nm²/kg²

distance = 8 m

Formula

$F = G\frac{m1m2}{r^{2} }$

Substitution

$F = 6.67 x 10^{-11} \frac{250 x 250}{8^{2} }$

Result

F = 0.000000065 N

Data 2

mass 1 = 1000 kg

mass 2 = 1000 kg

distance = 5 m

Substitution

$F = 6.67 x 10^{-11} \frac{1000 x 1000 }{5^{2} }$

Result

F = 0.000002667 N

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4 years ago

An airplane travels 2400 km at a speed of 600 km/h, decreases its speed to 400 km/h for the next 1200 km and travels the last 2,

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Forget the numbers and think about how speed works for a second. If a plane flies at 600km/h, this means that it will fly exactly 600 km in 1 hour. So, how long would it take to fly 2400 km??

2400/600 = 4hours.

Do this for all three and add up the hours.

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Average speed is the total distance / total time.

So,

Add up all the distances and divide by 12.

(2400+1200+2500) / 12

Average speed: 508.33333333 km/h

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3 years ago

If Mercury, moved two orbital paths closer to the sun: what would be different? List serval differences, ideas.......

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If Mercury, or any planet, were somehow moved to an orbit closer
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-- its speed in orbit would be greater,

-- the distance around its orbit would be shorter,

-- its orbital period ("year") would be shorter,

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3 years ago

Two forces are acting on an object.

jarptica [38.1K]

Answer:

8N -6N = 2N

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3 years ago

Read 2 more answers

The planet Uranus has a radius of 25,360 km and a surface acceleration due to gravity of 9.0 m/s^2 at its poles. Its moon Mirand

AlexFokin [52]

Answer:

$8.67791\times 10^{25}\ kg$

$0.34589\ m/s^2$

$0.07903\ m/s^2$

Explanation:

M = Mass of Uranus

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Uranus = 25360 km

h = Altitude = 104000 km

$r_m$ = Radius of Miranda = 236 km

m = Mass of Miranda = $6.6\times 10^{19}\ kg$

Acceleration due to gravity is given by

$g=\dfrac{GM}{r^2}\\\Rightarrow M=\dfrac{gr^2}{G}\\\Rightarrow M=\dfrac{9\times 25360000^2}{6.67\times 10^{-11}}\\\Rightarrow M=8.67791\times 10^{25}\ kg$

The mass of Uranus is $8.67791\times 10^{25}\ kg$

Acceleration is given by

$a_m=\dfrac{GM}{(r+h)^2}\\\Rightarrow a_m=\dfrac{6.67\times 10^{-11}\times 8.67791\times 10^{25}}{(25360000+104000000)^2}\\\Rightarrow a_m=0.34589\ m/s^2$

Miranda's acceleration due to its orbital motion about Uranus is $0.34589\ m/s^2$

On Miranda

$g_m=\dfrac{Gm}{r_m^2}\\\Rightarrow g_m=\dfrac{6.67\times 10^{-11}\times 6.6\times 10^{19}}{236000^2}\\\Rightarrow g_m=0.07903\ m/s^2$

Acceleration due to Miranda's gravity at the surface of Miranda is $0.07903\ m/s^2$

No, both the objects will fall towards Uranus. Also, they are not stationary.

6 0

3 years ago