A uniformly charged spherical droplet of mercury has electric potential Vbig throughout the droplet. The droplet then breaks int

Answer: since resistance is influenced by length, the resistance will be equally distributed as the wire is cut into two equal parts, resulting in

4.5 * 10^-3 ohm.

Explanation:

The velocity of water can be decomposed into its vertical and horizontal components:


The vertical component will exhibit a parabolic trajectory due to gravity, while the horizontal component will be linear:

To determine when the water reaches the ground 2.5m away, set y= 0 and x = 2.5

vf^2 = 2ad
vf^2 = 2(9.81)(44m)
vf^2 = 863.28
vf = √863.28
vf = 29.4  - using equations of motion

ME = PE + KE
ME = mgh + 1/2mv^2
ME = (1)(9.81)(44) + 1/2(1)(3^2)
ME = 431.64 + 4.5
ME = 436.14 - applying the principle of energy conservation

hope this helps:)

<span>Let Q be the charge, thus Q = -20.0 µC.</span>
Define D as the distance between the center of the rod and the specified point. Therefore,
D=0.32 - 0.12 = 0.2 m
<span>L = 0.12 m, which represents the length of the rod
</span><span>To find the magnitude and direction of the electric field along the axis of the rod at a point 32.0 cm from its center, use the formula:
</span><span>E = K·Q/r²
</span>or<span>E = kQ/D(D+L), where k</span> is a constant equal to 8.99 x 10<span>9</span> N m
2/C2.<span>Consequently,[TAG_21]]E=(</span>8.99 x 109 N m2/C2.* (-20.0 µC))/(<span>0.2 m*0.32m)</span><span>

</span>

The new charge of the ball will amount to 8x10^8C after removing 5x10^27 electrons.

Explanation:

Initially, if the sphere is electrically neutral, its charge stands at 0C.

When an electron with a charge of (-1.6*10^-19 C) is taken away, we effectively add a positive charge, leading to:

1.6*10^-19 C as the sphere's new charge.

For a total of N electrons removed, the sphere's overall charge now becomes:

N*1.6*10^-19 C.

To calculate N when:

N*1.6*10^-19 C = 8.0x 10^8 C.

We find that N is: (8.0/1.6)x10^(8 + 19) = 5x10^27 electrons.