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4 years ago

13

What is the flat shape

1 answer:

Harlamova29_29 [7]4 years ago

6 0

[C]

The reason why I say this, is because that is the only option that looks like the "flat shape." Notice that the other options have different designs that wouldn't look the same as the flat shape. The flat shape is just a rectangle you piece of cardboard with smaller rectangles at the end of each side. As you can see, shape C shows both smaller rectangles connected without any other design or cur out.

Hope This Helped! Good Luck!

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A nanometer connected to a pipe indicates a negative gauge pressure of 70mm of mercury .what is the pressure in the pipe in N/m^

prohojiy [21]

Based on the calculations, the absolute pressure in the pipe is equal to 90,670.4‬ N/m².

<h3>How to calculate the absolute pressure?</h3>

Mathematically, absolute pressure can be calculated by using this formula:

$P_{abs} = P_{atm} + P_{guage}$

<u>Scientific data:</u>

Atmospheric pressure (Patm) = 1 bar = 1 × 10⁵ N/m²

Head (h) = -70mmhg = -0.07 m hg

Acceleration due to gravity = 9.8 m/s²

Density of mercury = 13,600 kg/m³.

Next, we would determine the gauge pressure:

$P_{guage} = \rho gh\\\\P_{guage} = -13600 \times 9.8 \times 0.07\\\\P_{guage} = -9,329.6\;N/m^2$

Now, we can calculate the absolute pressure:

$P_{abs} = P_{atm} + P_{guage}\\\\P_{abs} = 1 \times 10^5 - 9329.6$

Absolute pressure = ‭90,670.4‬ N/m².

Read more on absolute pressure here: brainly.com/question/10013312

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6 0

2 years ago

Please help i give brainliest

Mazyrski [523]

Answer:

A mock-up

Explanation:

It is made of cheap and easy to access parts.

5 0

3 years ago

A jetliner flying at an altitude of 10,000 m has a Mach number of 0.5. If the jetliner has to drop down to 1000 m but still main

andreev551 [17]

Answer

Assuming

At 10000 m height temperature T = -55 C = 218 K

At 1000 m height temperature T = 0 C = 273 K

$\dfrac{V_1}{C_1} =\dfrac{V_2}{C_2} = 0.5$

R = 287 J/kg K

$C_1 = \sqrt{\gamma RT_1} = \sqrt{1.4\times 287\times 218} = 295 m/s$

$C_2 = \sqrt{\gamma RT_2} = \sqrt{1.4\times 287\times 273} = 331 m/s$

$V_2 = \dfrac{V_1}{C_1}\timesC_2$

V₂ = V₁ ×1.1222

V₁ = 0.5 × C₁ = 0.5 × 295 = 147.5 m/s

V₂ = 1.1222 × 147.5 = 165.49 m/s

so, the jetliner need to increase speed by ( V₂ -V₁ )

= 165.49 - 147.5

= 17.5 m/s

6 0

4 years ago

Alberto's mom is taking a splinter out of his hand with a pair of tweezers. The tweezers are 3 inches long. She is applying .25

Burka [1]

Answer:

ok

Explanation:

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4 years ago

Read 2 more answers

When developing a design proposal, you will need to ask: *

Liula [17]

Answer:

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3 years ago