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3 years ago

When did they start using the word "Scrubbed" for launches?

Physics

1 answer:

VladimirAG [237]3 years ago

6 0

Answer:

Many speculation exists for why the term 'scrubbed' is used in relation to launches. One of these is that during world war II, when the Air force planned bombing raids. These Air force missions had their briefings done with illustration of the details of the launch and and flight drawn on chalkboards. When some of mission flight takeoffs were canceled (due to bad weather or a malfunction or other reasons) the details were then scrubbed off of the chalkboard. This idea of the mission being scrubbed meaning that it is no longer to be followed through with, proceeded to this modern day rocket and aviation launches.

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A race car completes race in 10 minutes. Its average speed is 200km/hr. How long is the racetrack?

tigry1 [53]

Answer:

33.34km

Explanation:

S=Vxt

0.167hr x 200km/hr

ans=33.34km

8 0

2 years ago

What is the wavenumber of the stretching vibrational mode for the CO molecule, given that the force constant of the bond is 680

Gnesinka [82]

Answer:

1.10134 * 10⁻⁹m⁻¹

Explanation:

K = 680Nm⁻¹

μ = ?

μ = (m₁ + m₂) / m₁m₂

compound = CO

C = 12.0 g/mol = 0.012kg/mol

O = 16.0g/mol = 0.016kg/mol

μ = (m₁ + m₂) / m₁m₂

μ = (0.012 + 0.016) / (0.012*0.016) = 145.83

v = 1/2πc * √(k/μ)

ν = 1/ 2*3.142* 3.0*10⁸ * √(630/145.83)

v = 5.30*10⁻¹⁰ * 2.078

v = 1.10134*10⁻⁹m⁻¹

8 0

3 years ago

Suppose that the electric field in the Earth's atmosphere is E = 1.16 102 N/C, pointing downward. Determine the electric charge

butalik [34]

Answer:

Electric charge in the earth will be $Q=5.231\times 10^5C$

Explanation:

We have given that E = 116 N/C

Radius of the earth R = 6371 km = 6371000 m

We have to find the electric charge in the earth '

We know that electric field due to charge is given by $E=\frac{1}{4\pi \epsilon _0}\frac{Q}{R^20}=\frac{KQ}{R^2}$ . here K is coulomb's constant

So $116=\frac{9\times 10^9\times Q}{(6371000)^2}$

$Q=5.231\times 10^5C$

So electric charge in the earth will be $Q=5.231\times 10^5C$

5 0

3 years ago

If the moment acting on the cross section is M=630N⋅m, determine the maximum bending stress in the beam. Express your answer to

-BARSIC- [3]

Answer:

2.17 Mpa

Explanation:

The location of neutral axis from the top will be

$\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm$

Moment of inertia from neutral axis will be given by $\frac {bd^{3}}{12}+ ay^{2}$

Therefore, moment of inertia will be

$\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}$