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labwork [276]
4 years ago
11

I am the particle that is so small that scientists treat me as though I have no mass. Scientists also figure that I probably spe

nd most of my time in the cloud outside the nucleus. Give my name? can yall plz help me I'm in 8th grade and I need to know what this is because this is due tomorrow ​
Physics
1 answer:
tatyana61 [14]4 years ago
8 0

Answer:

its electrons

Explanation:

only electrons stays outside the nucleus unlike protons and neutrons and it has little to no mass

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What is the total moved without regard to direction​
soldi70 [24.7K]

The entire motion of an object, regardless of direction.

In physics, motion is that the phenomenon in which an object changes its position with respect to time. Motion is mathematically described in terms of displacement, distance, velocity, acceleration, speed, and frame of regard to an observer and measuring the change in position of the body relative to that frame with change in time. The branch of physics describing the motion of objects without regard to their cause is called kinematics, while the branch studying forces and their effect on motion are named dynamics.

If an object isn't changing relative to a given frame of reference,  the thing is said to be at rest, motionless, immobile, stationary, or to possess a constant or time-invariant position with reference to its surroundings. Modern physics holds that, as  there's no absolute frame of reference, Newton's concept of absolute motion can't be  determined

To  learn more about motion from the given link:

brainly.com/question/22810476

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6 0
2 years ago
Why is there a difference between potential and kinetic energy?
inn [45]

Potential energy is stored energy because it has the potential to do something which laters turns into kinetic energy which is the moving energy.

3 0
3 years ago
Jody drives 120 miles south from Kingston, NY to Princeton, NJ. If she drives 60 mph, how long will it take her to make the trip
Bas_tet [7]

sorry if its wrong but 2 hours??

6 0
3 years ago
Read 2 more answers
Is there any induced current in the pickup coil when the polarity of the bar magnet is flipped?
stiks02 [169]

Answer:

Faraday's law , he direction of the magnetic field changes by 180º, in the polarity inversion processes, induces a voltage.  

Explanation:

For this exercise let's use Faraday's law

          E = - dФ / dt

          Ф = B.A = B A cos θ

where B is the magnetic field, A is the area and θ is the angle between the field line and the normal to the area.

We can see that an electromotive force (voltage) is indexed when there is a variation of the field B, a variation of the area and change of the angle or when there is a combinational of them.

In this case, the magnitude of the field is constant, as the wire is rigid metal, the area is constant, but the direction of the magnetic field changes by 180º, in the polarity inversion processes, for which reason each change induces a voltage.

If a voltage is created in the ring, which has a resistance, a current is also generated in it.

Therefore the answer is If a current is created in the hoop

8 0
3 years ago
On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s
Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2}  } \\

F_{m} =G*\frac{m_{m}*m_{a}  }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]

When we match these equations the masses cancel out as the universal gravitational constant

G*\frac{m_{e} *m_{a} }{r_{e}^{2}  } = G*\frac{m_{m} *m_{a} }{r_{m}^{2}  }\\\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2}  }

To solve this equation we have to replace the first equation of related with the distances.

\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m}  )^{2}  } = \frac{7.36*10^{22}  }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17}  \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c }  }{2*a}\\  where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) }  }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

6 0
3 years ago
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