Answer:
I'm pretty sure it's 3.
Explanation:
Because if you look at your options the only that would be relevant to tick marks would be either 4 or 3. And it said in the question that we're looking for the one for the dependent variable. And the dependent variable is on the Y- Axis and the 3 is the tick marks for the y-axis. So your answer is 3.
Answer:
21.3 V, 1.2 A
Explanation:
1.
These resistors are in series, so the net resistance is:
R = R₁ + R₂ + R₃
R = 20 + 30 + 45
R = 95
So the current is:
V = IR
45 = I (95)
I = 9/19
So the voltage drop across R₃ is:
V = IR
V = (9/19) (45)
V ≈ 21.3 V
2.
First, we need to find the equivalent resistance of R₂ and R₃, which are in parallel:
1/R₂₃ = 1/R₂ + 1/R₃
1/R₂₃ = 1/10 + 1/10
R₂₃ = 5
Now we find the overall resistance by adding the resistors in series:
R = R₁ + R₂₃ + R₄
R = 10 + 5 + 10
R = 25
So the current through R₁ is:
V = IR
30 = I (25)
I = 1.2 A
Answer:
Q=185.84C
Explanation:
We have to take into account the integral
In this case we have a superficial density in coordinate system.
Hence, we have for R: x2 + y2 ≤ 4
but, for symmetry:
![Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C](https://tex.z-dn.net/?f=Q%3D4%5Cint_0%5E2%5Cint_0%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%5Crho%20dydx%5C%5C%5C%5CQ%3D4%5Cint_0%5E2%5Cint_0%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%284x%2B4y%2B4x%5E2%2B4y%5E2%29%20dydx%5C%5C%5C%5CQ%3D4%5Cint_0%5E%7B2%7D%5B4x%5Csqrt%7B4-x%5E2%7D%2B2%284-x%5E2%29%2B4x%5E2%5Csqrt%7B4-x%5E2%7D%2B%5Cfrac%7B4%7D%7B3%7D%284-x%5E2%29%5E%7B3%2F2%7D%5Ddx%5C%5C%5C%5CQ%3D4%5B46.46%5D%3D185.84C)
HOPE THIS HELPS!!
