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vlada-n [284]
3 years ago
7

Assume the space shuttle's main engines produce 764,576 newtons of thrust, and the shuttle has a mass of 78,018 kg. Why does the

shuttle need the two solid rocket boosters (in addition to the main engines)?
A. The boosters get the shuttle to space quicker.
B. The boosters are not needed at launch and are used later after achieving orbit.
C. The boosters are not needed, but they are there in the event one of the main engines does not work.
D. The thrust produced by the main engines equals the shuttle's weight, so the additional thrust from the boosters is needed to lift the shuttle off the launch pad.
Physics
1 answer:
Nady [450]3 years ago
4 0

Weight of anything = (mass) x (gravity in the place where the thing is)

Weight of anything on Earth = (mass) x (9.81 m/s²)

Weight of the shuttle = (78,018 kg) x (9.81 m/s²)

Weight of the shuttle, on Earth = 765,357 Newtons

Thrust of main engines = 764,576 Newtons

Are you starting to see the problem yet ?

The weight of the whole thing standing on the launch pad is 751 Newtons more than the maximum thrust of the main engines, and the engines can't lift it !  Even with all throttles wide open, the main engines alone would need about 175 <em>more</em> pounds of thrust to budge that load off the ground.  Even with the pedal to the metal, with flame and smoke belching out and covering the whole launch complex, the shuttle would just sit there and never leave the pad.

Well, no.  That's not exactly what would happen.  As the fuel in the main monster fuel tank is burned, the weight decreases.  So it would actually happen like this:  After the man announced "Zero !  We have ignition !  All engine running !", the ship would just sit there on the pad ... at first.  It would go nowhere and not even wiggle, <em>UNTIL</em> the first 175 pounds of fuel got burned without accomplishing anything.  The ship would then be 175 pounds lighter.  At that point, the weight would be exactly equal to the thrust of the main engines, and the vertical forces on the ship would be balanced.  Then, as MORE fuel continued to be wasted and the weight continued to decrease, the main engines could just begin to lift the ship off the pad.

So the correct answer is <em>choice-D</em> .  It tells the whole story, quicker than I can tell it.

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Who represents the worst of the iron age in his attempt to feed jupiter once living flesh?
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Lycaon represents the worst of the iron age in his attempt to feed jupiter once living flesh.

<h3>What is iron age?</h3>

The iron Age is the final epoch of the three-age division of the prehistory and protohistory of humanity.

It was preceded by the Stone Age, and the Bronze Age.

Who is lycaon?

In Greek mythology, Lycaon was a king of Arcadia who, in the most popular version of the myth, tested Zeus' omniscience by serving him the roasted flesh of Lycaon's own son Nyctimus, in order to see whether Zeus was truly all-knowing.

Thus, Lycaon represents the worst of the iron age in his attempt to feed jupiter once living flesh.

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I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
Masteriza [31]

Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
  • After it has traveled through 1 cm : od( 1\ cm) =  0.2606
  • After it has traveled through 2 cm :  od( 2\ cm) =  0.5211

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = 0.3012 \ I_0

<h2>b</h2>

The optical density od is given by:

od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

So, after travelling 1 cm:

od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

od( 1\ cm) = - (  - 0.2606)

od( 1\ cm) =  0.2606

After travelling 2 cm:

od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

3 0
3 years ago
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