Find the surface area and volume of the prism.
1 answer:
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Question:
A number cube with faces labeled from 1 to 6 will be rolled once. The number rolled will be recorded as the outcome.
Give the sample space describing all possible outcomes. Then give all of the outcomes for the event of rolling a number greater than 2.
If there is more than one element in the set, separate them with commas.
Answer:
Step-by-step explanation:
Given
A roll of a 6 sided number cube
Solving (a): The sample space
This implies that we list out all number on the number cube.
So:
Solving (b): Outcomes greater than 2
This implies that we list out all number on the number cube greater than 2 i.e. 3 to 6.
So:
Two uniformly charged, infinite, nonconducting planes are parallel to a yz plane and positioned at x = -50 cm and x = +50 cm. The charge densities on the planes are -50 nC/m² and +25 nC/m², respectively. What is the magnitude of the potential difference between the origin and the point on the x axis at x = +80 cm? (Hint:Use Gauss’ law.)
Answer:
ΔV = 2520 V
Step-by-step explanation:
We are given;
Charge density 1; σ1 = -50 nC/m² = -50 × 10^(-9) nC/m²
Charge density 2; σ2 = +25 nC/m² = +25 × 10^(-9) nC/m²
Now, formula for electric field strength is;
E = -(½ε)(|σ1| ± |σ2|)
ε is vacuum permittivity with a constant value as 8.85 × 10^(−12) C²/N.m²
|σ1| and |σ2| means we are taking the absolute values and would therefore not use the negative sign attached to them.
Now, in between x = 0 cm(0 m) and 50 cm (0.5 m), electric field generated by both charge densities would be in the same direction and thus;
E_in = -(½ε)(|σ1| + |σ2|)
E_in = -(½ × 8.85 × 10^(−12)) × [(50 × 10^(-9)) + (25 × 10^(-9))]
E_in = -4200 V/m
In contrast, when x ≥ 50 cm (0.5 m), electric field generated by both charge densities would be in opposite directions and thus;
E_out = -(½ε)(|σ1| - |σ2|)
E_out = -(½ × 8.85 × 10^(−12)) × [(50 × 10^(-9)) - (25 × 10^(-9))]
E_out = 1400 V/m
The magnitude of the potential difference from the origin to x = 80 cm(0.8 m) is calculated as attached;
From the attachment, we see that;
ΔV = 2520 V
