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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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Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo
djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for  the force of static friction:

f_s = \mu_s N --- (1)

where;

f_s = static friction force

\mu_s = coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

N = ma_{min}

From equation (1)

f_s = \mu_s ma_{min}

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

F = f_s

mg = \mu_s ma_{min}

a_{min} = \dfrac{mg }{ \mu_s m}

a_{min} = \dfrac{g }{ \mu_s }

where;

\mu_s = 0.383 and g = 9.8 m/s²

a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }

\mathbf{a_{min}= 25.59 \ m/s^2}

3 0
2 years ago
You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp
GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d)539.55\mu

(e) 0

Explanation:

The period for 1 circle 2\pi of the merry go around is 9.5s. It means the angular speed is:

\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s

(a)The speed is

v = \omega * R = 0.661 * 17 = 11.24 m/s

(b) Centripetal acceleration:

a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2

(c) Magnitude of the force that keeps you go around at this acceleration

F = ma = 55 * 7.44 = 409 N

(d) let the coefficient of friction by \mu. The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

F_f = mg\mu = 55*9.81\mu = 539.55\mu

3 0
2 years ago
<img src="https://tex.z-dn.net/?f=%5Cbold%7BWhat%20%5C%3A%20is%20%5C%3A%20the%20%5C%3A%20speed%3F%20%7D" id="TexFormula1" title=
scoray [572]

Refer to the above attachment

8 0
2 years ago
Read 2 more answers
A solid cylinder has a diameter of 17.4 mm and a length of 50.3mm. It's mass is 49g . What is its density of the cylinder in met
Tamiku [17]

Answer:

4 tonne/m³

Explanation:

ρ = m / V

ρ = 49 g / (π (17.4 mm / 2)² (50.3 mm))

ρ = 0.0041 g/mm³

Converting to tonnes/m³:

ρ = 0.0041 g/mm³ (1 kg / 1000 g) (1 tonne / 1000 kg) (1000 mm / m)³

ρ = 4.1 tonne/m³

Rounding to one significant figure, the density is 4 tonne/m³.

6 0
2 years ago
A 20kg object acceleration by a force of 200N with coefficient of kineticfriction is 0.4 what is acceleration of the object?​
Schach [20]

Answer:

<u>Given</u><em> </em><em>-</em><em> </em><u>M</u><u> </u><u>=</u><u> </u>20 kg

k = 0.4

F = 200 N

<u>To </u><u>find </u><u>-</u><u> </u> acceleration

<u>Solution </u><u>-</u><u> </u>

F= kMA

200 = 0.4 * 20 * acceleration

200 = 8 * a

a = 8/200

a = 0.04 m s²

<h3>a = 0.04 m s²</h3>
5 0
1 year ago
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