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OverLord2011 [107]
3 years ago
14

How are the electric field lines around a positive charge affected when a second positive charge is near it?

Physics
1 answer:
Liono4ka [1.6K]3 years ago
8 0

Answer:Because it likes to repel and attract opposites

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Does the thickness of the wire affect the strength of an electroscope
TiliK225 [7]
It probably does. I'm not 100% sure about it, but a thicker wire would increase the number of positive and negative charges in it.
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a closed tank is partially filled with glycerin. if the air pressure in the tank is 6 lb/in.2 and the depth of glycerin is 10 ft
vlabodo [156]

Answer:

<u><em>note:</em></u>

<u><em>solution is attached due to error in mathematical equation. please find the attachment</em></u>

4 0
3 years ago
An object in the shape of a thin ring has radius a and mass M. A uniform sphere with mass m and radius R is placed with its cent
madreJ [45]

Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

dF = GmdM/L²

Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³    ∫dM = M

F = GmMx/[√(a² + x²)]³  

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

3 0
3 years ago
What is a group of individuals of the same species living in the same geographic area
balu736 [363]
A group of individuals  living in a particular geographic area is termed population.
8 0
3 years ago
A satellite of mass m is moving in a circular orbit around the earth at a constant speed v and at an altitude h above the earth'
qwelly [4]

The satellite executes a rotation motion around the earth, because Earth's force of attraction plays the role of centripetal force:

Fa=Fcp=>k*Mp*m/(Rp+r)²=mv²/(Rp+r)=>v=√(k*Mp/(Rp+r))=√(6.67*10⁻¹¹*5.98*10²⁴/(6371*10³+1000*10³))=√(39.88*10¹³/(7371*10³))=√(5.41*10⁷)=7355.53 m/s


Check the calculations again !


7 0
3 years ago
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