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bogdanovich [222]
3 years ago
6

Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8

mg drops each have a charge of +29 pC. The centers of the droplets are at the same height and 0.36 cm apart.
Physics
1 answer:
Blababa [14]3 years ago
4 0

Answer:

F = 5.83 10⁻¹⁷ N

Explanation:

The electric force is given by

    F = k q₁  q₂ / r²

With Gauss's law electric field flow is equal to the charge inside the Gaussian surface, if we make a spherical surface around each drop, the force independent of small deformations due to air resistance

   q₁ = q₂

   F = 8.99 10⁹ (29 10⁻¹²)² / (0.36 10⁻²)²

   F = 5.83 10⁻¹⁷ N

As the two drops have a charge of the same sign they repel

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An uncharged conductor has a hollow cavity inside of it. Within this cavity there is a charge of +10 µC that does not touch the
aliina [53]

Answer:

Explanation:

we have to make charge inside the conductor zero because we know that electric field inside the conductor should be zero

so,  the outer surface of the conductor should contain + 10 uC of charge and the inner surface contains -10 uC

8 0
3 years ago
A kayaker needs to paddle north across a 100-m-wide harbor. The tide is going out, creating a tidal current that flows to the ea
Savatey [412]

Answer:

41.81^{\circ}

Explanation:

The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.

Let Vector \overrightarrow{OA} is the tidal current velocity as shown in the diagram.

In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, \vec {R} must be in the north direction.

Let \overrightarrow{AB} is the speed of the kayaker having angle \theta measured north of east as shown in the figure.

For the resultant velocity in the north direction, the tail of the vector \overrightarrow {OA} and head of the vector \overrightarrow{AB} must lie on the north-south line.

Now, for this condition, from the triangle OAB

|\overrightarrow{AB}|\sin \theta=|\overrightarrow{OA}|

\Rightarrow \sin\theta=\frac{|\overrightarrow{OA}|}{|\overrightarrow{AB}|}=\frac 2 3

\Rightarrow \theta=\sin^{-1}\frac23

\Rightarrow \theta=41.81^{\circ}

Hence, the kayaker must paddle in the direction of 41.81^{\circ}  in the north of east direction.

3 0
4 years ago
Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa
Lorico [155]

Answer:r_2=11.81 cm

Explanation:

Given

m_1=2.2\times 10^{-8} kg

m_2=4.8\times 10^{-8} kg

same charge on both masses

potential Energy due to Magnetic Field =Kinetic Energy of Particle

qV=\frac{mv^2}{2}

v=\sqrt{\frac{2qV}{m}}

and we know

Force due to magnetic field will Provide centripetal Force

qvB=\frac{mv^2}{r}

B=\frac{\sqrt{\frac{2Vm}{q}}}{r}

and B is equal for both particles

thus \frac{m}{r^2}=constant

\frac{m_1}{r_1^2}=\frac{m_2}{r_2^2}

r_2^2=\frac{4.8}{2.2}\times 8^2

r_2=11.81 cm

4 0
3 years ago
What do we call the bright, sphere-shaped region of stars that occupies the central few thousand light-years of the milky way ga
Law Incorporation [45]

Answer: The galaxy's bulge

Explanation:

4 0
2 years ago
The amount of the lighted side of the moon you can see is the same during
attashe74 [19]

<u>Answer:</u>

The amount of the lighted side of the moon you can see is the same during "how much of the sunlit side of the moon faces Earth".

<u>Explanation:</u>

The Moon is in sequential rotation with Earth, and thus displays the Sun, the close side, always on the same side. Thanks to libration, Earth can display slightly greater than half (nearly 59 per cent) of the entire lunar surface.

The side of the Moon facing Earth is considered the near side, and the far side is called the reverse. The far side is often referred to as the "dark side" inaccurately but it is actually highlighted as often as the near side: once every 29.5 Earth days. During the New Moon the near side becomes blurred.

5 0
3 years ago
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