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IRINA_888 [86]
3 years ago
6

Newton’s first law relates motion to balanced and unbalanced forces.

Physics
2 answers:
BlackZzzverrR [31]3 years ago
8 0
True, an object at rest stays and rest and an object in motion stays in motion
gulaghasi [49]3 years ago
8 0

YES !  

True and certainly.  

Uh huh, uh huh.

That's a fact.  

More intense veracity is seldom encountered.

Did you have a question to ask ?

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A space vehicle approaches a space station in orbit. The intent of the engineers is to have the vehicle slowly approach, reducin
N76 [4]

Answer: The total momentum before the docking maneuver is mV_{1}+MV_{2} and after the docking maneuver is (m+M) U

Explanation:

Linear momentum p (generally just called momentum) is defined as mass in motion and is given by the following equation:  

p=m.v  

Where m is the mass of the object and v its velocity.

According to the conservation of momentum law:

<em>"If two objects or bodies are in a closed system and both collide, the total momentum of these two objects before the collision </em>p_{i} <em>will be the same as the total momentum of these same two objects after the collision </em>p_{f}<em>". </em>

<em />

p_{i}=p_{f}

This means, that although the momentum of each object may change after the collision, the total momentum of the system does not change.

Now, the docking of a space vehicle with the space station is an inelastic collision, which means both objects remain together after the collision.

Hence, the<u> initial momentum</u> is:

p_{i}=mV_{1}+MV_{2}

Where:

m is the mass of the vehicle

V_{1} is the velocity of th vehicle

M is the mass of the space station

V_{2} is the velocity of the space station

And the <u>final momentum</u> is:

p_{f}=(m+M)U

Where:

U is the velocity of the vehicle and space station docked

6 0
3 years ago
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Which of the three recording stations was closest to the epicenter of the earthquake, based on the seismograms shown below?
Nastasia [14]

Answer:C

Explanation:took quiz

3 0
3 years ago
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You hang a heavy ball with a mass of 30 kg from a tungsten rod 2.8 m long by 1.5 mm by 2.6 mm. You measure the stretch of the ro
guajiro [1.7K]

Answer:

Young's modulus (Y) = 3.56×10^11 N/m^2

The speed of sound in tungsten = 6166.4 m/s

Explanation:

Young's modulus (Y) = stress/strain

Stress = force/area

Force = mg = 30×9.8 = 294 N

Area = 1.5 × 2.6 = 3.9 mm^2 = 3.9/10^6 = 3.9×10^-6 m^2

Stress = 294/3.9×10^-6 = 7.54×10^7 N/m^2

Strain = extension/length

Extension = 0.000594 m

Length = 2.8 m

Strain = 0.000594/2.8 = 2.12×10^-4

Y = 7.54×10^7/2.12×10^-4 = 3.56×10^11 N/m^2

Y = h × rho × g

rho = 18.7 g/cm^3 = 18.7 g/cm^3 × 1 kg/1000 g × (100 cm/1 m)^3 = 18,700 kg/m^3

h = 3.56×10^11/(18,700×9.8) = 1.94×10^6 m

From the equations of motion

v^2 = u^2 + 2gh =

Initial speed (u) = 0 m/s

v = sqrt (2×9.8×1.94×10^6)

v = 6166.4 m/s

7 0
3 years ago
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If a pickup is placed 16.25 cm from one of the fixed ends of a 65.00-cm-long string, which of the harmonics from n=1 to n=12 wil
Lina20 [59]

Answer:

The answer to this question can be defined as follows:

Explanation:

Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.

Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.

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The electric current running through the wire coil in an electric motor exerts force directly onto A) the battery. B) an aluminu
antiseptic1488 [7]
<span>C)<span>a powerful magnet.

</span></span>
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