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Lapatulllka [165]
3 years ago
5

When NASA's Skylab reentered the Earth's atmosphere on July 11, 1979, it broke into a myriad of pieces. One of the largest fragm

ents was a 1770-kg lead-lined film vault, and it landed with an estimated speed of 120 m/s.?
Physics
1 answer:
Burka [1]3 years ago
6 0

Answer:

Kinetic Enerrgy = 12744000 J or 12.744 MJ

Explanation:

Given Data:

Mass of fragment = m =1770 kg    ;

Estimated speed = v = 120 m/sec ;

Solution:

As we know that

Kinetic energy = K.E = (1/2)mv²

By putting the values, we get

              K.E = 0.5*1770*120²

              K.E = 12744000 J

                            or

              K.E = 12.744 MJ

So, the kinetic energy of lead-lined vault when it landed was 12.744MJ approximately.

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Calculate ideal work (in J) when a single stream of 1 mole of air is heated and expanded from 25 C and 1 bar to 100 C and 0.5 ba
NISA [10]

Answer:

-1786.5J

Explanation:

Temperature 1=T1=25°c

Temperature 2=T2=200°c

Pressure P1=1bar

Pressure P2=0.5bars

T=37°c+273=310k

Note number if moles=1

Recall work done =2.3026RTlogp2/P1

2.3026*8.314*310log(0.5/1)

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Leftover planetesimals that formed in the region of the solar system now occupied by the jovian planets are called
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Answer:

Comets

Explanation:

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The Jovian planets include Jupiter, Saturn, Uranus, and Neptune.

Therefore, leftovers of comets (planetesimal bodies) formed in the region of the solar system that are now occupied by the Jovian planets is due to the dusty particles  associated with the comets.

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Why is static electricity worse in winter?
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Provide an example of when momentum is conserved and explain your answer you can get 10 PTS if answered with a good explaination
dezoksy [38]

Answer:

m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s

Explanation:

<u>Conservation of Momentum </u>

The total momentum of a system of two particles is

p=m_1v_1+m_2v_2

Where m1,m2,v1, and v2 are the respective masses and velocities of the particles at a given time. Then, the two particles collide and change their velocities to v1' and v2'. The final momentum is now

p'=m_1v_1'+m_2v_2'

The momentum is conserved if no external forces are acting on the system, thus

m_1v_1+m_2v_2=m_1v_1'+m_2v_2'

Let's put some numbers in the problem and say

m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s

(8)(12)+(6)(4)=(8)(-6)+(6)(28)

96+24=-48+168

120=120

It means that when the particles collide, the first mass returns at 6 m/s and the second continues in the same direction at 28 m/s

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3 years ago
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