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NARA [144]
3 years ago
13

Astronaut Mark Kelly and his crew traveled to the International Space Station in the space shuttle. The shuttle burned off all r

emaining fuel before entering Earth's atmosphere. When the shuttle returned to Earth it landed at a speed of approximately 354 km/hr or 220 mi/hr, about 20 times the landing speed of a commercial aircraft. What accounts for the shuttle's speed upon landing?
Physics
2 answers:
Snowcat [4.5K]3 years ago
7 0

A) Earth's atmosphere  

B) initial thrust of take-off  

C) angle of approach upon landing  

D) gravitational potential energy

It would be D. Gravitational potential energy As the shuttle prepares to return to Earth it reduces its speed so that gravity pulls it out of its orbit and toward Earth. As it approaches the earth the gravitational potential energy is converted to kinetic energy. The drag of Earth's atmosphere slows the shuttle's descent and causes some kinetic energy to convert to heat energy.

Nitella [24]3 years ago
6 0

A couple of things, if the shuttle burned all of it's fuel before entering Earth's atmosphere then that means that the shuttle was accelerating towards Earth until it ran out of fuel.  At that point, there is little to no air resistance (friction) by the lack of an atmosphere so it keeps accelerating  due to Earth's gravitational force.  The closer the shuttle gets to Earth the stronger the gravitational pull the shuttle experiences.  Note that, once the shuttle reaches Earth's atmosphere it will cause significant amount of friction and thus will cause the shuttle to slow down.

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Desde una altura de 120 m se deja caer un cuerpo. Calcular a los 2,5 s i) la rapidez que lleva; ii) cuánto ha descendido; iii) c
stira [4]

Answer:

i) 24.5 m/s

ii) 30,656 m

iii) 89,344 m

Explanation:

Desde una altura de 120 m se deja caer un cuerpo. Calcule a 2.5 s i) la velocidad que toma; ii) cuánto ha disminuido; iii) cuánto queda por hacer

i) Los parámetros dados son;

Altura inicial, s = 120 m

El tiempo en caída libre = 2.5 s

De la ecuación de caída libre, tenemos;

v = u + gt

Dónde:

u = Velocidad inicial = 0 m / s

g = Aceleración debida a la gravedad = 9.81 m / s²

t = Tiempo de caída libre = 2.5 s

Por lo tanto;

v = 0 + 9.8 × 2.5 = 24.5 m / s

ii) El nivel que el cuerpo ha alcanzado en 2.5 segundos está dado por la relación

s = u · t + 1/2 · g · t²

= 0 × 2.5 + 1/2 × 9.81 × 2.5² = 30.656 m

iii) La altura restante = 120 - 30.656 = 89.344 m.

6 0
3 years ago
3.
ratelena [41]

Answer:

1.84 kJ  (kilojoules)

Explanation:

A specific heat of 0.46 J/g Cº means that it takes 0.46 Joules of energy to raise the temperature of 1 gram of iron by 1 Cº.

If we want to heat 50 g of iron from 20° C to 100° C, we can make the following calculation:

Heat = (specific heat)*(mass)*(temp change)

Heat = (0.46 J/g Cº)*(50g)*(100° C -  20° C)

[Note how the units cancel to yield just Joules]

Heat = 1840 Joules, or 1.84 kJ

[Note that the number is positive:  Energy is added to the system.  If we used cold iron to cool 50g of 100° C water, the temperature change would be (Final - Initial) or (20° C - 100° C).  The number is -1.84 kJ:  the negative means heat was removed from the system (the iron).

8 0
2 years ago
Magnetic pole reversal worksheet
Likurg_2 [28]
??????????????????????
8 0
3 years ago
57:07
Reptile [31]

Answer:

v (speed) = S / t = 4 * 400 m / (6 * 60 sec) = 4.4 m/s

The average velocity  is zero because there is no net vector displacement.

5 0
3 years ago
A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are
igor_vitrenko [27]

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

T = I\alpha = 0.303*0.6454 = 0.196 Nm

So the force acting on the other end to generate this torque mush be:

F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N

4 0
3 years ago
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