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nignag [31]
3 years ago
11

An object with a positive electric charge will be attracted to what? *

Physics
1 answer:
Liono4ka [1.6K]3 years ago
7 0
I think it’s A but go wit what yu think
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An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a ra
lukranit [14]

Answer:

\frac{dA}{dt} = 28800 \ m^2/year

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

<u>Geometry</u>

  • Area of a Rectangle: A = lw

<u>Algebra I</u>

  • Exponential Property: w^n \cdot w^m = w^{n + m}

<u>Calculus</u>

Derivatives

Differentiating with respect to time

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Explanation:

<u>Step 1: Define</u>

Area is A = lw

2w = l

w = 300 m

\frac{dw}{dt} = 24 \ m/year

<u>Step 2: Rewrite Equation</u>

  1. Substitute in <em>l</em>:                    A = (2w)w
  2. Multiply:                              A = 2w²

<u>Step 3: Differentiate</u>

<em>Differentiate the new area formula with respect to time.</em>

  1. Differentiate [Basic Power Rule]:                                                                   \frac{dA}{dt} = 2 \cdot 2w^{2-1}\frac{dw}{dt}
  2. Simplify:                                                                                                           \frac{dA}{dt} = 4w\frac{dw}{dt}

<u>Step 4: Find Rate</u>

<em>Use defined variables</em>

  1. Substitute:                    \frac{dA}{dt} = 4(300 \ m)(24 \ m/year)
  2. Multiply:                        \frac{dA}{dt} = (1200 \ m)(24 \ m/year)
  3. Multiply:                        \frac{dA}{dt} = 28800 \ m^2/year
3 0
3 years ago
Read 2 more answers
The temperature at the surface of the Sun is approximately 5,300 K, and the temperature at the surface of the Earth is approxima
N76 [4]

Answer:

The entropy change of the Universe that occurs is 19.346 J/K

Explanation:

Given;

temperature of the sun, T_s = 5,300 K

temperature of the Earth, T_E = 293 K

radiation energy transferred by the sun to the earth, E = 6000 J

The sun loses Q of heat and therefore decreases its entropy by the amount

\delta S_{sun} = \frac{-Q}{T_s}

The earth gains Q of heat and therefore increases its entropy by the amount

\delta S_{Earth} = \frac{-Q}{T_E}

The total entropy change is:

\delta S_{Earth} + \delta S_{sun} = \frac{Q}{T_E} -\frac{Q}{T_S} \\\\                                                      = Q(\frac{1}{T_E} -\frac{1}{T_S} )\\\\= 6000(\frac{1}{293} -\frac{1}{5300} )\\\\=6000(0.0032243)\\\\= 19.346 \ J/K

Therefore, the entropy change of the Universe that occurs is 19.346 J/K

4 0
3 years ago
Help pls you don't have to answer all​
miskamm [114]

Answer:

5. The 2 features we need to harness the energy from tides are when tides come in and when tides come out.

6. Tidal energy is the most predictable type of renewable energy is because we know the exact locations of the sun and moon all year round.

6 0
4 years ago
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The jet stream flows from east to west across the United States.<br><br><br> True False
velikii [3]
This is false. they flow west to east
8 0
4 years ago
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PLEASE HELP. I NEED THIS FOR TOMMOROW. WILL GIVE BRAINLIEST ​
Fantom [35]

Ansthethethethethe

the is the is

Explanation:

3 0
2 years ago
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