<span>this may help you
As far as the field goes, the two charges opposite each other cancel!
So E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
and since k = 8.99e9N·m²/C²,
E = 1.789e10N·m²/C² * Q / d² </span>
Answer:


Explanation:
Force of friction on M mass so that it will move down the inclined plane is given as

now if it is moving down the inclined plane at constant speed
so we will have

on other side the mass "m" will go up at constant speed
so we have

so we have

so we have

for special case when m = M
then we have

Answer:
Explanation:
Far point = 17 cm . That means he can not see beyond this distance .
He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula
1 / v - 1 / u = 1 / f
1 / - 17 - 1 / - 65 = 1 / f
= 1 / 65 - 1 / 17
= - .0434 = 1 / f
power = - 100 / f
= - 100 x .0434
= - 4.34 D .
<span>The answer is The conductance of a conductor is inversely
proportional to the cross-sectional area of the conductor.</span>
<span>Conductance is directly related to the ease offered by any material to the passage of electric current. Conductance is the opposite of resistance. The higher the conductance, the lower the resistance and vice versa, the greater the resistance, the less conductance, so both are inversely proportional</span>
Answer:
See the answers below.
Explanation:
to solve this problem we must make a free body diagram, with the forces acting on the metal rod.
i)
The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.
We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.
For the summation of forces we will take the forces upwards as positive and the negative forces downwards.
ΣF = 0

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.
ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.
ΣM = 0
![(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]](https://tex.z-dn.net/?f=%2815%2A9%29%20-%20%2818%2AW%29%20%3D%200%5C%5C135%20%3D%2018%2AW%5C%5CW%20%3D%207.5%20%5BN%5D)