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gayaneshka [121]
4 years ago
15

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

Physics
1 answer:
VladimirAG [237]4 years ago
6 0

Answer:

Many speculation exists for why the term 'scrubbed' is used in relation to launches. <em>One of these is that during world war II, when the Air force planned bombing raids</em>. These  Air force missions had their briefings done with illustration of the details of the launch and and flight drawn on chalkboards. <em>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</em>. This idea of the mission <em>being scrubbed meaning that it is no longer to be followed through with</em>, proceeded to this modern day rocket and aviation launches.

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Fofino [41]
<span>Hooke's law is F=-kx, which means the elastic force contained by the spring is a product of the distance it stretches and its spring constant, but the direction of the force is opposite that of the displacement. We calculate as follows:

</span><span>(3 kg)(9.8 m/s^2) = -k(-0.38 m)
</span>k =<span> 77.4

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7 0
3 years ago
A motorcycle running on gasoline wastes a large amount of energy mainly as A) heat energy and sound energy. B) light energy and
vlada-n [284]

A motorcycle mainly wastes energy as heat <u>energy</u> and <u>sound</u> energy. In the engine, chemical energy is transformed into mechanical energy. However, the engine is inefficient and much of the chemical energy is lost as heat energy. Also, some of the energy is transformed to sound energy. This explains why the motorcycle is noisy and has an exhaust pipe.

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4 years ago
One end of a 7-cm-long spring is attached to the ceiling. When a 5.4 kg mass is hung from the other end, the spring is stretched
mash [69]

Answer:

2.63 cm

Explanation:

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Making c the subject of the formula then

c=\frac {F}{y}

Since F is gm but taking the given mass to be F

c=\frac {5.4 kg}{4.3 cm}=1.2558139534883720930232558139534883720930

By substitution now considering F to be 3.3 kg

y=\frac {3.3 kg}{1.2558139534883720930232558139534883720930}=2.6277777777777 cm\approx 2.63 cm

8 0
3 years ago
A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the 7) magnitude of the change
IRISSAK [1]

Answer:

Change in momentum will be -4.4 kgm/sec

So option (A) is correct option

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Mass of the ball is given m = 0.10 kg

Initial velocity of ball v_1=25m/sec

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We have to find the change in momentum

So change in momentum is equal to =m(v_2-v_1)=0.1\times (-19-25)=-4.4kgm/sec ( here negative sign shows only direction )

So option (A) will be correct answer

5 0
3 years ago
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Answer:

Just find the formula then you can do it

4 0
3 years ago
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