Answer:

Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 
Explanation:
The potential energy is given by:
U=Q*V
where:
Q is the charge
V is the potential difference
Potential Difference=V=
So,

Where:
k is Coulomb Constant=
q is the charge on electron=
r is the distance=
For 3 Electrons Potential Energy or work Done is:


Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 
Answer:
V = 331.59m/s
Explanation:
First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.
S = ut + 1/2at²
Given height of the cliff S = 80m
initial velocity u = 0m/s²
a = g = 9.81m/s²
Substitute
80 = 0+1/2(9.81)t²
80 = 4.905t²
t² = 80/4.905
t² = 16.31
t = √16.31
t = 4.04s
Next is to get the vertical velocity
Vy = u + gt
Vy = 0+(9.81)(4.04)
Vy = 39.6324
Also calculate the horizontal velocity
Vx = 1330/4.04
Vx = 329.21m/s
Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.
V² = Vx²+Vy²
V² = 329.21²+39.63²
V² = 329.21²+39.63²
V² = 108,379.2241+1,570.5369
V² = 109,949.761
V = √ 109,949.761
V = 331.59m/s
Hence the speed of the shell as it hits the ground is 331.59m/s
Answer:
Dermatome. (Ans. C).
Explanation:
Dermatome is defined as the area of the human anatomy skin which is supplied by single spinal sensory nerve root. At the spinal cord these spinal sensory nerve enter the nerve root, and the branches of spinal sensory reach to the periphery of the body.
The sensory nerve which is present in the periphery of the body are the type of nerve which helps to transmit signals from sensation such as pain, temperature, etc. to the spinal cord from some specific area of the anatomy.
In 0.25h it will move in 22.5 kilometers.
Answer:
0.5 Ohms
Explanation:
We note that the node Q is also between the resistors of 1ohm and 2ohms.
We note that the node P is also between the resistors of 2ohms and 3ohms.
Thus, all these resistors are in parallel, beween nodes P and Q
1/Re=1/R1+1/R2+1/R3
1/Re=1/1+1/2+1/3=(6+3+2)/6=11/6 [ohm^(-1)]
Re=6/11=0.54ohms
Rounding to the tenth: Re=0.5 ohms