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gizmo_the_mogwai [7]
2 years ago
7

Uma volta inteira na Terra corresponde a uma distância 2πr≅40.000 km. Se um satélite leva 85 minutos para dar uma volta completa

ao redor da Terra, qual a sua velocidade média em m/s (metros por segundo)
Mathematics
1 answer:
Shalnov [3]2 years ago
4 0

Answer:

7.8. 10 ^ 3 m / s

Step-by-step explanation:

The computation of the average speed in m/s is shown below:

The average speed equation is

= ΔS ÷ Δt;

As We know that

an entire revolution would be40,000 km

and a satellite takes 85 minutes to complete a revolution,  

So,

ΔS = 40,000 km = 40,000,000 m

Δt = 85 min = 5 100 s

Now the average speed is

= 40 000 000 ÷ 5 ÷ 5 100

 = 7843.14 m / s

= 7.8. 10 ^ 3 m / s

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Step-by-step explanation:

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PtichkaEL [24]
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What is (8-√2)(2+√8)?
siniylev [52]

Answer:

(8 - \sqrt{2}) \cdot (2 + \sqrt{8}) = 12 + 14\; \sqrt{2}.

Step-by-step explanation:

Step One: Simplify the square roots.

8 = 4 \times 2 = 2^2 \times 2.

The square root of 8 \sqrt{8} can be simplified as the product of an integer and the square root of 2:

\sqrt{8} = \sqrt{4 \times 2} = \sqrt{2^2 \times 2} = \sqrt{2^{2}} \times \sqrt{2} = 2 \; \sqrt{2}.

As a result,

(8 - \sqrt{2}) \cdot (2 + \sqrt{8}) = (8 - \sqrt{2}) \cdot (2 + 2\; \sqrt{2}).

Step Two: Expand the product of the two binomials.

(8 - \sqrt{2}) \cdot (2 + 2\; \sqrt{2}) is the product of two binomials:

  • Binomial One: 8 - \sqrt{2}.
  • Binomial Two: 2 + 2\; \sqrt{2}

Start by applying the distributive law to the first binomial. Multiply each term in the first binomial (without brackets) with the second binomial (with brackets)

({\bf 8} - {\bf \sqrt{2}}) \cdot {(2 + 2\; \sqrt{2})}\\= [{\bf 8} \cdot {(2 + 2\; \sqrt{2})}] - [{\bf \sqrt{2}} \cdot {(2 + 2\; \sqrt{2})}]

Now, apply the distributive law once again to terms in the second binomial.

[8 \cdot ({\bf 2} + {\bf 2\; \sqrt{2}})] - [\sqrt{2} \cdot ({\bf 2} + {\bf 2\; \sqrt{2}})]\\= [8 \times {\bf 2} + 8 \times {\bf 2\;\sqrt{2}}] - [\sqrt{2} \times {\bf 2} + \sqrt{2} \times {\bf 2\; \sqrt{2}}].

Step Three: Simplify the expression.

The square of a square root is the same as the number under the square root. For example, \sqrt{2} \times \sqrt{2} = (\sqrt{2})^{2} = 2.

[8 \times 2 + 8 \times 2\;\sqrt{2}] - [\sqrt{2} \times 2 + 2 \sqrt{2} \times \sqrt{2}]\\ =[16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 4]\\= 16 + 16\;\sqrt{2} - 2\; \sqrt{2} - 4.

Combine the terms with the square root of two and those without the square root of two:

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Factor the square root of two out of the second term:

(16 - 4) + (16 \; \sqrt{2} - 2\; \sqrt{2})\\= (16 - 4) + (16 - 2) \; \sqrt{2} \\= 12 - 14 \; \sqrt{2}.

Combining the steps:

(8 - \sqrt{2}) \cdot (2 + 2\; \sqrt{2})\\= [8 \cdot (2 + 2\; \sqrt{2})] - [\sqrt{2} \cdot (2 + 2\; \sqrt{2})]\\= [8 \times 2 + 8 \times 2\;\sqrt{2}] - [\sqrt{2} \times 2 + \sqrt{2} \times 2 \;\sqrt{2}]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 2 \times (\sqrt{2})^{2}]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 2 \times 2]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 4]\\= 16 + 16\;\sqrt{2} - 2\; \sqrt{2} - 4\\= (16 - 4) + (16 \; \sqrt{2} - 2\; \sqrt{2})\\

= (16 - 4) + (16 - 2) \; \sqrt{2}\\= 12 - 14 \; \sqrt{2}.

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Digiron [165]
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