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3 years ago

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HELP PLS!!!! Light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is th

e distance light travels in one year.
State the speed in ft/s, using scientific notation.

3.069 x 101 ft/s
9.8208 x 108 ft/s
3.0 x 1016 ft/s

1 answer:

maw [93]3 years ago

6 0

You've already told us the speed in ft/s . It's right there in the question. You said that light travels about 982,080,000 ft/s.

We don't know how accurate that number is, but for purposes of THIS question, that's the number we're going with.

In scientific notation, it's written . . . <em>9.8208 x 10⁸ ft/s .</em>

We don't know where you were going with the number of seconds in a year. But to answer the question that you eventually asked, it turned out that we don't even need it.

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Answer:

ΔV=0.484mV

Explanation:

The potential difference across the end of conductor that obeys Ohms law:

ΔV=IR

Where I is current

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The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A

R=(pL)/A

Given data

Length L=3.87 cm =0.0387m

Diameter d=2.11 cm =0.0211 m

Current I=165 A

Resistivity of aluminum p=2.65×10⁻⁸ ohms

So

ΔV=IR

$=I(\frac{pL}{A})\\ =I(\frac{pL}{\pi r^{2} } )\\=I(\frac{pL}{\pi (d/2)^{2} } )\\=165A((\frac{(2.65*10^{-8})(0.0387m)}{\pi (0.0211m/2)^{2} } ))\\=4.84*10^{-4}V$

ΔV=0.484mV

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2 years ago

Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as

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Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

$h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m$

or in kilometers,

$h_2 = 47 km$

the first altitude in kilometers is

$h_1 = 155 km$

so the difference between the two altitudes is

$\Delta h = 155 km - 47 km = 108 km$

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3 years ago

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2 years ago

Read 2 more answers

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

Brut [27]

Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

Where $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

So

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

And

$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

So

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_c =0.121$

So

$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

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3 years ago

Water in a tank is pressurized by air and pressure measured using a multi-fluid manometer. Determine the gage pressure of air in

sattari [20]

Answer:

The gauge pressure of air is 110 kpa

Explanation:

Atmospheric pressure, $P_{atm}$ = 101 Kpa

$P_{gauge} + \rho_w gh_1 + \rho_o gh_2 -\rho_{Hg} gh_3 =P_{atm}$

$P_{gauge} = P_{atm} - \rho_w gh_1 - \rho_o gh_2 +\rho_{Hg} gh_3$

where;

ρw is the density of water = 1000 kg/m³

ρo is the density of oil = 800 kg/m³

ρHg is the density of mercury = 13,600 kg/m³

g is acceleration due to gravity = 9.8 m/s²

$P_{gauge} = 101,000 - (1000* 9.8*0.2) - (800* 9.8*0.3) +(13,600* 9.8*0.46)\\\\P_{gauge} = 101,000 - 1960 - 2352 + 13610.26\\\\P_{gauge} = 110,298.26 pa$

Therefore, the gauge pressure of air is 110 kpa

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3 years ago