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FinnZ [79.3K]
4 years ago
12

A rocket firing its engine and accelerating in outer space (no gravity, no air resistance) suddenly runs out of fuel. Which best

describes its motion after burnout?
A. It continues to accelerate but at a gradually decreasing rate, until it reaches a constant velocity.


B. It continues to accelerate at the same rate.


C. It immediately begins slowing down and gradually approaches zero velocity.


D. It immediately stops accelerating, and continues moving at the velocity it had when burnout occurred.
Physics
1 answer:
notsponge [240]4 years ago
4 0

the correct choice is

D. It immediately stops accelerating, and continues moving at the velocity it had when burnout occurred.

as soon as the rocket runs out of fuel, its engine stops firing gases which is mainly responsible for the acceleration of the rocket. hence the rocket stops accelerating. also, since there is no gravity or air resistance , there is no net force on the rocket after it runs out of the fuel. hence

a = acceleration = 0

we know that , at no acceleration , object moves at constant velocity. hence the rocket keeps moving at the velocity it had when burnout occured.

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a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car
Vaselesa [24]

The momentum of the 155 kg car afterwards is 469.7 kg m/s to the right

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum of the system is conserved before and after the collision, so we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 282 kg is the mass of the bumper car

u_1 = 3.50 m/s is the initial velocity of the bumper car (we take the right as positive direction)

v_1 = 0.800 m/s is the final velocity of the bumper car

m_2 = 155 kg is the mass of the second bumper car

u_2 = -1.88 m/s is the initial velocity of the second car (moving to the left)

v_2 is the final velocity of the second car

Solving for v_2,

v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(282)(3.50)+(155)(-1.88)-(282)(0.800)}{155}=3.03 m/s

where the positive sign means the direction is to the right.

And now we can find the momentum of the 155 kg afterwards, which is

p_2 = m_2 v_2 = (155)(3.03)=469.7 kg m/s (to the right)

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

7 0
3 years ago
Material A is less optically dense than material B. How can you tell from the diagram above?
zimovet [89]
You can tell because the line bends and the closer it is to horizontal or past horizontal it is more dense
7 0
3 years ago
Below is an oscilloscope trace from an AC supply. Calculate the frequency of the supply if each horizontal division represents 0
GuDViN [60]

Answer:

20 hertz of frequency produced.

Explanation:

\boxed{frequency = \frac{1}{period} }

Here we will find frequency and period should be in second, here given: 0.05 seconds

using the formula:

\boxed{frequency = \frac{1}{0.05} }

\boxed{frequency =20}

8 0
3 years ago
To ancient peoples, why were planets special?
coldgirl [10]

Answer:

Planets were like gods.

Explanation:

To the people of many ancient civilizations, the planets were thought to be deities. Our names for the planets are the Roman names for these deities. For example, Mars was the god of war and Venus the goddess of love.

7 0
3 years ago
How do I convert 14.8 cm into MEters?
stellarik [79]
Multiply it by a fraction equal to ' 1 ', like this:

(14.8 cm) x (1 meter/100 cm) = 14.8/100 = 0.148 meter
6 0
3 years ago
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