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scZoUnD [109]
2 years ago
11

astronauts brought back 500 lb of rock samples from the moon. how many kilograms did they bring back? 1 kg

Physics
1 answer:
Nadya [2.5K]2 years ago
8 0

Astronauts brought back 500 lbs rock samples from the moon. They brought back 227 kilograms of sample.

An Astronaut is a person who has been educated, prepared, and put into space as part of a human spaceflight mission to serve as a commander or crew member. The word is frequently used to refer to scientists as well as professional space explorers, while being usually reserved for all of those who travel into space on a daily basis.

According to the given question, Astronaut bought back 550 lbs sample,

We know that,

2.20lbs = 1 kg.

1 lbs = 1/2.20 kg

1 lbs = 0.45 kg

Hence, 500 lbs equals 227kgs according to the given relation.

So, Astronauts brought back 227 kilograms of sample.

Learn more about Astronaut here, brainly.com/question/11244838

#SPJ4

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Answer:

151.9 N

Explanation:

Force = mass x acceleration

Acceleration due to gravity is 9.8 m/s^2 (you should memorize this number).

F = ma

F = (15.5)(9.8)

F = 151.9

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Studying neutrinos helped to explain how our Sun works but led to changes in theories of particle physics, how is this process c
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Explanation:

Yes,  evidently, the process is consistent with the scientific process because in scientific process falsification and modification are two very important traits. So this new concept have modified the existing theories.

Through the modification a theory is adjusted without undermining other discovering made through the theory's prediction.

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Read 2 more answers
In a physics lab, a 0.500-kg cart (Cart A) moving with a speed of 129 cm/s encounters a magnetic collision with a 1.50-kg cart (
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Answer:

58 cm/s

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7 0
3 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
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