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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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since v is directly proportional to the force and the force remain the same, the increase in the cart speed will also remain the same.

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The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the
FrozenT [24]

Answer:

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Explanation:

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u= Free space permeability

I= Electrical current passing through the wire

a  = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

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