Fe=K Q1/Q2/d2
Q1 is the first charge
Q2 is the second charge
d is the distance
K= 9x10^9 NM^2/C2
Now let’s plug the numbers
Fe=9x10^9NM^2/C2 (2x10^-4C)(8x10^-4C) / (0.3m^2) you notice we took away the negative charges when we plugged the charges
Ok now we notice that we have C2 which is C to the power 2 we can write it as C^2 and we have two CSU’s beside each one of the charges we can get rid of them all by curtailment
And we can curtailment the M^2and the other M^2
Now we left with only 9x10^9N (2x10^-4)(8x10^-4)/ 0.3
Let’s multiply the (9)(2)(8)=144
And add the exponents (9)+(-4)+(-4)=1
So now we got 144x10N divide by the distance which is 0.3
144x10N / 0.3 = 4800N
Hope it helps u understand :)
There is no thing here I can answer so imma just say all of the above because I have no idea what else to say so thx for the points
Answer:
resistor R₂ has the lowest current density
Explanation:
The current density is
j = I / A
now let's analyze each case
a) R₂ has an area 2A₀ and a length L₀ that R₁
b) R₃ has an area Ao and a length 3L₀ what R₁
we can see that all the area is given in relation to the resistance R₁
the current density in R₁ is
j₁ = I / A₀
the current density in R₂
j₂ = I / 2A₀
j₂ 2 = ½ I/A₀
the current density in R₃
j₃ = I / A₀
j₂ < j₁ = j₃
therefore resistor R₂ has the lowest current density
Answer:
= 0.0298V
Explanation:
Electric field between two conductor = σ / ε₀
σ = 33pC/m²
= 33 × 10⁻¹²C/m²
ε₀ = 8.85 × 10 ⁻¹²C²/Nm²
E = 33 × 10⁻¹² / 8.85 × 10 ⁻¹²
= 3.7288N/C
potential difference between two conductors = Ed
where d = 8mm = 8 × 10 ⁻³m
V = 3.7288 × 8 × 10 ⁻³
= 29.83 × 10 ⁻³
= 0.0298V
Use a net with not too large holes, yet not too small holes.
Hence, the small beefs with fall out, and the large ones will remain in the sieve/net.