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kolbaska11 [484]
3 years ago
13

Which of the following is true about a combustion reaction?

Physics
1 answer:
Ksenya-84 [330]3 years ago
7 0

Answer:

b

Explanation:

this is because inoder. for combustion or burning to take place oxygen must be available

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What happens to jetstream’s as they get closer to the equator
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Answer:They stop because jet streams follow boundaries between hot and cold air.

Explanation:

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The distance traveled by a wave during a single period is
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I think it's wavelength
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How did astronomers precisely determine the length of an Astronomical Unit in the 1960s?
zaharov [31]

Answer:

Use of telemetry and radar astronomy

Explanation:

An astronomical Unit (AU) is a unit of measuring distances in outer space, which is based on the approximate distance between the earth and the Sun.

After several years of trying to approximate the distance between the Sun and the Earth using several methods based on geometry and some other calculations, advancements in technology made available the presence of special motoring equipment, which can be placed in outer space to remotely monitor and measure the position of the sun.

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5 0
3 years ago
Atoms of two different elements must have different
Tresset [83]

Answer:

Number of neautrons

Explanation:

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4 0
3 years ago
Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al
Citrus2011 [14]

To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire(m) = 1.10g = 1.10*10^{-3} kg

Density (\rho)= 8.92*10^3kg/m^3

Resistively of copper (\gamma) = 1.7*10^{-8}\Omega \cdot m

Resistance (R) = 0.390\Omega

Volume is defined as,

V= lA \text{ and }\frac{m}{\rho}

lA= \frac{1.10*10^{-3}}{8.92*10^3}

lA = 1.233*10^{-7} m^3 (1)

We know that,

\frac{l}{A} = \frac{R}{\gamma}

\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}

\frac{l}{A} = 2.2941*10^7 m^{-1} (2)

Multiplying equation we have

l^2 = (1.233*10^{-7})( 2.2941*10^7)

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l =\sqrt{2.8286m^2}

l = 1.68m

Therefore the length of the wire is 1.68m

6 0
3 years ago
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