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3 years ago

8

The electrostatic force between two-point charges is F. The charge on each point

1 answer:

Svetllana [295]3 years ago

5 0

<h3>Explanation:</h3><h3><em><u>AG</u></em><em><u> office</u></em><em><u> and</u></em><em><u> unlike</u></em><em><u> the</u></em><em><u> other</u></em><em><u> hand</u></em><em><u> the</u></em><em><u> poem</u></em><em><u> by</u></em><em><u> kho</u></em><em><u> </u></em><em><u>o</u></em><em><u>f</u></em><em><u> the</u></em><em><u> poem</u></em><em><u> by</u></em><em><u> kho</u></em><em><u> in</u></em><em><u> a</u></em><em><u> new</u></em><em><u> job</u></em><em><u> so</u></em><em><u> that</u></em><em><u> the</u></em><em><u> other</u></em><em><u> side</u></em><em><u> is</u></em><em><u> no</u></em><em><u> par</u></em><em><u> massage</u></em><em><u> therapy</u></em><em><u> in</u></em><em><u> a</u></em><em><u> few</u></em><em><u> things</u></em><em><u> that</u></em><em><u> are</u></em><em><u> you</u></em><em><u> doing</u></em><em><u> recently</u></em><em><u> so</u></em><em><u> much</u></em><em><u> to</u></em><em><u> say</u></em><em><u> the</u></em><em><u> other</u></em><em><u> day</u></em><em><u> that</u></em><em><u> are</u></em><em><u> in</u></em><em><u> kho</u></em><em><u> gaye</u></em><em><u> aur</u></em><em><u> Gk</u></em><em><u> and</u></em><em><u> hai</u></em><em><u> aur</u></em><em><u> papa</u></em><em><u> ko</u></em><em><u> bhi</u></em><em><u> nahi</u></em><em><u> kuch</u></em><em><u> bhi</u></em><em><u> hai</u></em><em><u> aur</u></em><em><u> Gk</u></em><em><u> ek</u></em><em><u> baar</u></em><em><u> fir</u></em><em><u> against</u></em><em><u> the</u></em><em><u> spirit</u></em><em><u> in</u></em><em><u> kho</u></em><em><u> gaye</u></em><em><u> hai</u></em><em><u> to</u></em><em><u> wo</u></em><em><u> nahi</u></em><em><u> kuch</u></em><em><u> nahi</u></em><em><u> kiya</u></em><em><u> hai</u></em><em><u> GSV</u></em><em><u> aur</u></em><em><u> bhajo</u></em><em><u> na</u></em><em><u> ki</u></em><em><u> nahi</u></em><em><u> kuch</u></em><em><u> nahi</u></em><em><u> kiya</u></em><em><u> tha</u></em><em><u> aur</u></em><em><u> bhajo</u></em><em><u> ham</u></em><em><u> nahi</u></em><em><u> sun</u></em><em><u> and</u></em><em><u> unlike</u></em><em><u> decimals</u></em><em><u> the</u></em><em><u> poem</u></em><em><u> </u></em><em><u>y</u></em><em><u> and</u></em><em><u> hai</u></em><em><u> </u></em><em><u>ki</u></em><em><u> </u></em><em><u>ueugdhdhrh</u></em><em><u> </u></em><em><u>Hur</u></em><em><u> and</u></em><em><u> hai</u></em><em><u> to</u></em><em><u> bola</u></em><em><u> ke</u></em><em><u> arey</u></em><em><u> yaar</u></em><em><u> hai</u></em><em><u> ki</u></em><em><u> nahi</u></em><em><u> kuch</u></em><em><u> nahi</u></em><em><u> kiya</u></em><em><u> hai</u></em><em><u> GSV</u></em><em><u> moers</u></em><em><u> VWSA</u></em><em><u> the</u></em><em><u> poem</u></em><em><u> by</u></em><em><u> kho</u></em><em><u> in</u></em><em><u> the</u></em><em><u> poem</u></em><em><u> by</u></em><em><u> kho</u></em><em><u> in</u></em><em><u> the</u></em><em><u> family</u></em><em><u> of</u></em><em><u> the</u></em><em><u> poem</u></em><em><u> by</u></em><em><u> kho</u></em><em><u> in</u></em><em><u> the</u></em><em><u> family</u></em><em><u> of</u></em><em><u> the</u></em><em><u> poem</u></em><em><u> by</u></em><em><u> Sunil</u></em><em><u> Kumar</u></em><em><u> Sharma</u></em><em><u> and</u></em><em><u> </u></em><em><u>i</u></em><em><u>t</u></em><em><u> with</u></em><em><u> a</u></em><em><u> few</u></em><em><u> more</u></em><em><u> things</u></em><em><u> are</u></em><em><u> you</u></em></h3>

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What unit of measure would you use for mass?

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grams, pounds, kilos, etc

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An astronaut sitting on the launch pad on Earth's surface is 6,400 kilometers from Earth's center and weighs 400 newtons. Calcul

PIT_PIT [208]

Answer:

weight at height = 100 N .

Explanation:

The problem relates to variation of weight due to change in height .

Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .

At the surface :

Applying Newton's law of gravitation

mg₀ = G Mm / R²

At height h from centre

mg₁ = G Mm /h²

Given mg₀ = 400 N

400 = G Mm / R²

400 = G Mm / (6400 x 10³ )²

G Mm = 400 x (6400 x 10³ )²

At height h from centre

mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²

= 400 / 4

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weight at height = 100 N

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3 years ago

What is the definition of motivation?

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3 years ago

A 20.0-kg block is initially at rest on a horizontal surface. A horizontal force of 77.0 N is required to set the block in motio

saveliy_v [14]

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3 years ago

Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate

masha68 [24]

Answer:

The heat loss per unit length is $\frac{Q}{L} = 2981 W/m$

Explanation:

From the question we are told that

The outer diameter of the pipe is $d = 104mm = \frac{104}{1000} = 0.104 m$

The thickness is $D = 2mm = \frac{2}{1000} = 0.002m$

The temperature of water is $T = 90^oC = 90 + 273 = 363K$

The outside air temperature is $T_a = -10^oC = -10 +273 = 263K$

The water side heat transfer coefficient is $z_1 = 300 W/ m^2 \cdot K$

The heat transfer coefficient is $z_2 = 20 W/m^2 \cdot K$

The heat lost per unit length is mathematically represented as

$\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}$

Substituting values

$\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}$

$\frac{Q}{L} = \frac{628}{0.2107}$

$\frac{Q}{L} = 2981 W/m$

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3 years ago