The answer is 24 because u have multiple it
Answer:
+5.20 µC
Explanation:
Given that
Charge =−5.20 µC and radius= 5.40 cm.
Insert a +10.4- µC charge at the center of the sphere
As we know that according to Gauss's theorem electric field inside the metal is zero.
When +10.4 µC charge insert but the total charge on the sphere should be −5.20 µC.So the charge on the outer surface of the sphere will be +5.20 µC.
Answer:
um where is the question? im confused
Step-by-step explanation:
Answer:
B. 5070π in³
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Geometry
</u>
Volume of a Cylinder Formula: V = πr²h
- <em>r</em> is radius
- <em>h</em> is height
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
<em>r</em> = 13 in
<em>h</em> = 30 in
<u>Step 2: Find Volume</u>
- Substitute in variables [Volume of a Cylinder Formula]: V = π(13 in)²(30 in)
- Evaluate exponents: V = π(169 in²)(30 in)
- Multiply: V = 5070π in³
