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3 years ago

Why does a rocket continue to rise after it runs out of fuel

Physics

2 answers:

vitfil [10]3 years ago

6 0

Answer:

In rocket flight, forces become balanced and unbalanced all the time. ... As the engines are ignited, the thrust from the rocket unbalances the forces, and the rocket travels upward. Later, when the rocket runs out of fuel, it slows down, stops at the highest point of its flight, then falls back to Earth.

Explanation:

Darina [25.2K]3 years ago

5 0

Answer: The surface of the pad pushes the rocket up while gravity tries to pull it down. As the engines are ignited, the thrust from the rocket unbalances the forces, and the rocket travels upward. Later, when the rocket runs out of fuel, it slows down, stops at the highest point of its flight, then falls back to Earth.

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Which explains the downward force produced when air flows over the winglike spoiler on a race car?

professor190 [17]

Newton's principle
newton's 2nd law of motion where F=ma

3 0

3 years ago

The weight of an astronout is 60kg on earth, find the wight of the same object on the planet where the gravitational attraction

Marianna [84]

The astronaut's weight is not 60 kg anywhere, because kg is a unit of mass, not weight.

If the astronaut's mass is 60 kg, then his weight is (60 kg)x(acceleration of gravity).

That's 588 Newtons on Earth, and 58.8 Newtons on a planet with 1/10 Earth's gravity.

The astronaut's mass of 60 kg goes with her, and doesn't depend on where she is.

3 0

3 years ago

Two children hang by their hands from the same tree branch. The branch is straight, and grows out from the tree trunk at an angl

vazorg [7]

Answer:879.29 N-m

Explanation:

Given

mass of first child $m_1=44 kg$

distance of first child from tree is $r_1=1 m$

tree is inclined at an angle of $\theta =27^{\circ}$

mass of second child $m_1=27 kg$

distance of second child from tree is $r_2=2.1 m$

Weight of first child $=m_1g=431.2 kg$

Weight of second child $=m_2g=264.6 kg$

Torque of first child weight $=m_1g\cos \theta \cdot r_1$

$T_1=44\times 9.8\times \cos 27\times 1=384.202 N-m$

Torque of second child weight $=m_2g\cos \theta \cdot r_2$

$T_2=27\times 9.8\times \cos 27\times 2.1=495.096 N-m$

Net torque $T_{net}=T_1+T_2=384.202+495.096=879.29 N-m$

6 0

4 years ago

The toy car is about 3 inches long is an example of a ?

balandron [24]

Answer:

a quantitative observation because it includes numerical data

8 0

3 years ago

An air-track glider attached to a spring oscillates between the 10 cm mark and the 60 cm mark on the track. The glider completes

Assoli18 [71]

Answer:

a) Time period is 3.3 seconds

b) The frequency is 0.3030 Hz

c) amplitude is 0.25 m

d) maximum speed is 0.476 m/s

Explanation:

Given the data in the question;

a) Period

Time Period T = Time taken for one oscillation

T = 33s / 10 = 3.3 seconds

Therefore, Time period is 3.3 seconds

b) Frequency

we know that frequency is the inverse of time period

so;

Frequency f = 1/T = 1 / 3.3 s

Frequency f = 0.3030 Hz

Therefore, The frequency is 0.3030 Hz

c) amplitude

amplitude A = $\frac{1}{2}$ ( 60 cm - 10 cm )

A = $\frac{1}{2}$ × 50 cm

A = 25 cm

A = 0.25 m

Therefore, amplitude is 0.25 m

d) maximum speed of the glider

maximum speed $V_{max}$ = ωA

and ω = 2π/T

so maximum speed $V_{max}$ = $\frac{2\pi }{T}$ A

so we substitute

so maximum speed $V_{max}$ = $\frac{2\pi }{3.3}$ × 0.25 m

so maximum speed $V_{max}$ = 0.476 m/s

Therefore, maximum speed is 0.476 m/s

5 0

3 years ago