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liq [111]
3 years ago
13

A rocket is sitting on the launch pad. The engines ignite, and the rocket begins to rise straight upward, picking up speed as it

goes. At about 1000 m above the ground the engines shut down, but the rocket continues straight upward, losing speed as it goes. It reaches the top of its flight path and then falls back to earth. Ignoring air resistance, decide which one of the following statements is true.
a. Only part of the rocket's motion, from just after the engines shut down until just before it lands, is free-fall.
b. Only the rocket's motion, while the engines are firing, is free-fall.
c. Only part of the rocket's motion, from just after the engines shut down until it reaches the top of its flight path, is free-fall.
d. All of the rocket's motion, from the moment the engines ignite until just before the rocket lands, is free-fall.
e. Only the rocket's motion from the top of its flight path until just before landing is free-fall.
Physics
1 answer:
grandymaker [24]3 years ago
4 0

Answer:

e. Only the rocket's motion from the top of its flight path until just before landing is free-fall.

Explanation:

A free-fall  is a fall just under force of gravity. The rocket;s upward motion is result of engine push - even if it was shut down  - and rocket free of engine push effect when it reaches it's maximum height after shutting down of engine. Then rockets stops at it's maximum height for a moment and rtuens back as free fall with only force of gravitation pulling it back to ground with acceleration 'g'.

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How many seconds will it take for a the International Space Station to travel 450 km at a rate of 100 m/s?
SVEN [57.7K]

Time = (distance) / (speed)

<em></em>

Time = (450 km) / (100 m/s)

Time = (450,000 m) / (100 m/s)

Time = <em>4500 seconds </em>(that's 75 minutes)

Note:

This is about HALF the speed of the passenger jet you fly in when you go to visit Grandma for Christmas.

If the International Space Station flew at this speed, it would immediately go ker-PLUNK into the ocean.

The speed of the International Space Station in its orbit is more like 3,100 m/s, not 100 m/s.

8 0
3 years ago
An airplane in a holding pattern flies at constant altitude along a circular path of radius 3.38 km. If the airplane rounds half
Andrej [43]

Answer:

a) 6076 m

b) 43.33 m/s

c) 68 m/s

Explanation:

(a) If the airplane rounds half the circle in 156s, its displacement is the circle diameter in 156s, or twice the circle's radius

s = 2r = 2* 3.38km = 6.76 km or 6760 m

(b) The average velocity would be displacement over unit of time

v = s/t = 6760 / 156 = 43.33 m/s

(c) The length of the chord it's swept in 156s is half of the circle perimeter

c = πr = π3.38 = 10.62 km or 10620 m

The airplane average speed is its chord length over a unit of time

c / t = c / 156 = 68 m/s

4 0
3 years ago
Tides occurs in oceans but not in lakes.why?​
-Dominant- [34]

Answer:

Explanation:

Tides occur in the ocean but nit in lakes because an ocean is a free flowing body of water that can travel a large area of the globe while a lake or pond only covers a small area of the earth so it is not affected by gravity as violently and in turn, prevents the formation of tides.

4 0
3 years ago
¿Cuál es la aceleración centrípeta de un móvil que recorre una
Lady bird [3.3K]

Answer:

a)  a = 4.57 m/s², b)  a = 6.48 m / s² , c)  a = 1.42 m / s²,d)   r = 82.3 m

 

Explanation:

The centripetal acceleration is the acceleration responsible for the change of direction of the acceleration vector and occurs in circular movements, the expression is

           a = v² / r

let's apply this precaution to our cases

a) let's calculate

          a = 8²/14

         a = 4.57 m/s²

b) an automobile at v = 65 km / h (1000 m / 1km) (1 h / 3600 s) =18,055 m/s

let's reduce feet to meters

          1 ft = 0.3048 m

           r = 165 ft (0.3048 m / 1 ft) = 50.292 m

          a = 18,055 2 / 50,292

           a = 6.48 m / s²

c) we calculate

          a = 1.25²2 / 1.1

          a = 1.42 m / s²

d) we look for the radius

          a = v² / r

          r = v² / a

we reduce

          v = 80 km / h (1000 m / 1km) (1h / 3600s) = 22.22  ms

          r = 22.22²/6

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e) the cenripeta acceleration is used to take the curves on the highway,

    Used in centrifuges to separate compounds

       It is used in the games of the park of atraccio

     Used in CD players and computer hard drives

5 0
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Natali5045456 [20]

<span> </span>For any prism-shaped geometry, the volume (V) is assumed by the product of cross-sectional area (A) and height (h). 

<span> V = Ah </span>

<span>
Distinguishing with respect to time gives the relationship between the rates. 
dV/dt = A*dh/dt</span>

<span> in the meantime the area is not altering </span>

<span>
dV/dt = π*(1 ft)^2*(-0.5 ft/min) </span>

<span>
dV/dt = -π/2 ft^3/min ≈ -1.571 ft^3/min 

Water is draining from the tank at the rate of π/2 ft^3/min.</span>

5 0
4 years ago
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