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3 years ago

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Suppose you are asked to find the area of a rectangle that is 2.1-cm wide by 5.6-cm long. Your calculator answer would be 11.76

cm2 . Now suppose you are asked to enter the answer to two significant figures. (Note that if you do not round your answer to two significant figures, your answer will fall outside of the grading tolerance and be graded as incorrect.)

1 answer:

lianna [129]3 years ago

3 0

The area of a rectangle whose length is 5.6 cm and the width is 2.1 cm to two significant figures is 12 cm ².

<h2>Further Explanation </h2><h3>Area </h3>

Area is a measure of how much space is occupied by a given shape.
Area of a substance is determined by the type of shape in question.

For example;

Area of a rectangle is given by; Length multiplied by width

Area of a triangle = 1/2 x base x height

Area of a circle = πr². where r is the radius of a circle,

Area of a square = S², Where s is the side of the square.etc.

Perimeter

Perimeter is defined as the distance along a two dimension shape. Perimeter of different shapes is given by different formulas

For example

The perimeter of a rectangle = 2(length+width)

The perimeter of a triangle = a+b+c; where a, b and c are the sides of the triangle. etc.

In this question

The rectangle has a;

Length = 5.6 cm

Width = 2.1 cm

Area of a rectangle = length × width

= 2.1 cm × 5.6 cm

= 11.76 cm²

The area of the rectangle to two significant figures is 12 cm²

Keywords; Area, Area of a rectangle

<h3>Learn more about;</h3>

Area: brainly.com/question/1322653

Perimeter: brainly.com/question/1322653
Area of a rectangle: brainly.com/question/1322653

Level: Middle school

Subject; Mathematics

Topic: Area and Perimeter

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The specific heat capacity of concrete is $3.0 J/(g^{\circ}C)$

Explanation:

When a certain amount of energy Q is supplied/given off to/from a sample of substance with mass m, the temperature of the substance increases/decreases by an amount $\Delta T$ , according to the equation

$Q=mC_s \Delta T$

where

m is the mass of the substance

$C_s$ is the specific heat capacity of the substance

$\Delta T$ is the change in temperature of the substance

In this problem, we have:

m = 1 kg = 1000 g is the mass of the concrete slab

$Q = -12,000 J$ is the amount of energy lost by the slab

$\Delta T = 30-26= -4^{\circ}C$ is the change in temperature of the slab

Solving the equation for $C_s$ , we find the specific heat capacity of concrete:

$C_s = \frac{Q}{m \Delta T}=\frac{-12,000}{(1000)(-4)}=3.0 J/(g^{\circ}C)$

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3 years ago

According to the nebular theory, what early event eventually led to the formation of our solar system?

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2 years ago

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.4 ft/s,

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Answer:

$\dfrac{d\theta}{dt} =-0.233\ rad/s$

Explanation:

given,

length of ladder = 10 ft

let x be the distance of the bottom and y be the distance of the top of ladder.

x² + y² = 100

differentiating with respect to time we get

$2 x\dfrac{dx}{dt}+2y\dfrac{dy}{dt} = 0$ ..............(1)

when x = 8 and y = 6 and when \dfrac{dx}{dt} = 1.4ft/s

from equation (1)

now,

$16\times 1.4 + 12\dfrac{dy}{dt} = 0$

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let the angle between the ladders be θ

$tan\theta = \dfrac{y}{x}$

y = xtan θ

$\dfrac{dy}{dt} =\dfrac{dy}{dt} tan\theta + x sec^2\theta\dfrac{d\theta}{dt}$

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3 years ago

spaceship of mass m travels from the Earth to the Moon along a line that passes through the center of the Earth and the center o

satela [25.4K]

Answer:

the correct result is r = 3.71 10⁸ m

Explanation:

For this exercise we will use the law of universal gravitation

F = $- \frac{m_{1} m_{2} }{r^2}$

We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m

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F₁ = - \frac{m' M }{r^2}

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F₂ = - \frac{m' m }{(d-r)^2}

in the problem ask for what point the force has the relation

2 F₁ = F₂

let's substitute

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(d-r) ² = $\frac{m}{2M}$ r²

d² - 2rd + r² = \frac{m}{2M} r²

r² (1 -\frac{m}{2M}) - 2rd + d² = 0

Let's solve this quadratic equation to find the distance r, let's call

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a = 0.99385

a r² - 2d r + d² = 0

r = $\frac {2d \frac{+}{-} \sqrt{4d^2 - 4 a d^2}} {2a}$

r = [2d ± 2d $\sqrt{1-a}$ ] / 2a

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r₁ = \frac{d}{a} 1.04

r₂ = \frac{d}{a} 0.96

let's calculate

r₁ = $\frac{3.84 10^8}{0.99385}$ 1.04

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r₂ = 3.71 10⁸ m

therefore the correct result is r = 3.71 10⁸ m

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2 years ago

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cestrela7 [59]

Answer:

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