Free falling objects accelerate at 9.8 m/s because the force of

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:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Geometry

Area of a Rectangle: A = lw

Algebra I

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

Calculus

Derivatives

Differentiating with respect to time

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Explanation:

Step 1: Define

Area is A = lw

2w = l

w = 300 m

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

Step 2: Rewrite Equation

Substitute in l: A = (2w)w
Multiply: A = 2w²

Step 3: Differentiate

Differentiate the new area formula with respect to time.

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

Step 4: Find Rate

Use defined variables

Substitute: $\frac{dA}{dt} = 4(300 \ m)(24 \ m/year)$
Multiply: $\frac{dA}{dt} = (1200 \ m)(24 \ m/year)$
Multiply: $\frac{dA}{dt} = 28800 \ m^2/year$

3 0

3 years ago

Read 2 more answers

An electron is in a vacuum near Earth's surface and located at y = 0 m on a vertical y axis. At what value of y should a group o

Bingel [31]

Answer:

the bunch of 23 electrons must be placed on y axis at coordinate y = - 24.35 m.

Explanation:

As we know that the gravitational force on electron at y = 0 is counter-balanced by the weight of the electron

So we have

$\frac{kq_1q_2}{r^2} = mg$

here we have

$q_1 = e$

$q_2 = 23 e$

$m = 9.11 \times 10^{-31} kg$

also we know that

$e = 1.6 \times 10^{-19} C$

so we will have

$\frac{(9\times 10^9)(1.6 \times 10^{-19})(23\times 1.6 \times 10^{-19})}{r^2} = (9.11 \times 10^{-31})(9.81)$

$\frac{5.3 \times 10^{-27}}{r^2} = 8.94 \times 10^{-30}$

$r = 24.35 m$

so the bunch of 23 electrons must be placed on y axis at coordinate y = - 24.35 m.

7 0

3 years ago

A ball rolls off a level table with a speed of 2.2 m/s. The table is 1.2 meters high. How far away from the base of the table do

r-ruslan [8.4K]

The base of table is "1.08856 m" far away from the ball land.

Given:

Distance travelled by ball,

S = 1.2 m

Initial velocity,

u = 0 m/s

Acceleration,

a = 9.8 m/s²

Constant speed,

2.2 m/s

As we know the formula,

→ $S = ut+\frac{1}{2}at^2$

By substituting the values, we get

→ $1.2=\frac{1}{2}\times 9.8t^2$

→ $1.2 = 4.9 t^2$

→ $t = \sqrt{\frac{1.2}{4.9} }$

→ $= 0.4948 \ seconds$

Let,

Horizontal distance will be "x".
Time = t

→ $x = 2.2\times t$

$= 2.2\times 0.4948$

$= 1.08856 \ m$

Thus the solution above is right.

Learn more:

brainly.com/question/19170664

8 0

3 years ago

URGENT HELP IT IS DUE BY TONIGHT PLEASE HELP

Law Incorporation [45]

Answer:

45.8

Explanation:

becuse 9.8+36=45.8 simple

5 0

3 years ago

Definition: A command economy is where a central government makes all economic decisions. The government or a collective owns th

denis-greek [22]

A command economy is a system where the government, rather than the free market, determines what goods should be produced, how much should be produced and the price at which the goods are offered for sale.

7 0

3 years ago