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Blizzard [7]
3 years ago
13

What are the divisions within a shell called

Physics
1 answer:
blsea [12.9K]3 years ago
5 0
The divisions within an atom's shell are called subshells. This means that each shell consists of several subshells that are made up of orbitals. Each orbital consists of 1 or 2 electrons. The outermost shell of an atom is what we call the valence electrons, and they are what participate in chemical bonding.
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A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a rope that goes over an ideal pulley. If the masses are
irga5000 [103]

Answer:

The time taken will be 0.553 seconds.

Explanation:

We should start off by finding the force exerted by the rope on the 3kg weight in this case.

Weight of 5kg mass = 5 * 9.81 = 49.05 N

Weight of 3kg mass = 3 * 9.81 = 29.43 N

The force acting upward on the 3kg mass will equal the weight of the 5kg mass. Thus the resultant force acting on the 3kg mass is:

Total force = 49.05 - 29.43 = 19.62 N (upwards)

We can now find the acceleration:

F = m * a

19.62 = 3 * a

a = 6.54 m/s^2

We now use the following equation of motion to get the time taken to travel 1 meter:

s=u*t+\frac{1}{2} (a*t^2)

1=0*t+\frac{1}{2} (6.54*t^2)

t = 0.553 seconds

4 0
3 years ago
The gauge pressure in the tires of your car is 210 kPa (30.5 psi) when the temperature is 25°C (77 °F). Several days later it is
DaniilM [7]

Under the assumption that the tires do not change in volume, apply Gay-Lussac's law:

P/T = const.

P = pressure, T = temperature, the quotient of P/T must stay constant.

Initial P and T values:

P = 210kPa + 101.325kPa

P = 311.325kPa (add 101.325 to change gauge pressure to absolute pressure)

T = 25°C = 298.15K

Final P and T values:

P = ?, T = 0°C = 273.15K

Set the initial and final P/T values equal to each other and solve for the final P:

311.325/298.15 = P/273.15

P = 285.220kPa

Subtract 101.325kPa to find the final gauge pressure:

285.220kPa - 101.325kPa = 183.895271kPa

The final gauge pressure is 184kPa or 26.7psi.

8 0
3 years ago
Why might exposing astronauts to weightlessness be harmful?
love history [14]
The body doesn't have to work as hard when there's no gravity for it to work against, so it becomes accustomed to a much lower work load on every level. It leads to lower bone mass and weaker muscles, including the heart, leading to a drop in blood pressure that can eventually build up to create problems with cognitive function. After so long, minor accidents can lead to major, even life threatening problems. A simple bump that would do little more than leave a bruise on you and I can result in a broken femur bone or broken neck on an astronaut who has been exposed to a weightless environment for too long.

This is one of the several hurdles that must be overcome in order for a manned mission to Mars to succeed. Exposure to a weightless environment on the order of roughly two years for a manned Mars mission would be so degrading to the body that the rough, turbulent re-entry into Earth's atmosphere might prove to be too violent for an astronaut to survive.

The problem is bones.

On Earth, every time you do something with "impact" (like walking), there are microcracks in your bones. Calcium is used by the body to fix these cracks... and that is how the bones grow and become strong.

No weight = no impact = no cracks = no "repairs" being done by the body = the body gets rid of un-neede calcium and bones become brittle and weak.

There are some other operations in the body that require gravity as a "director", or resistance to movement as a driver of change (think of muscles in the legs, when there is no need to walk).

The organ themelves are (generally) OK since many things can work in any orientation.
5 0
3 years ago
A pilot claims to have seen a UFO moving initially at a speed of about 351 m/s in an easterly direction and then, in a time inte
serious [3.7K]

Answer:

Explanation:

This problem can be solved easily if we represent velocity in the form of vector.

The velocity of 351 was towards easterly direction so

V₁ = 351 i

The velocity of 351 was towards south west making - 48° with east or + ve x direction.

V₂ = 351 Cos 48 i - 351 sin 48 j

V₂ = 234.86 i - 260.84 j

Change in velocity

= V₂ - V₁ = 234.86 i - 260.84 j - 351 i

= -116.14 i - 260.84 j

acceleration

= change in velocity / time

(-116.14 i - 260.84 j )/ 1

= -116.14 i - 260.84 j

magnitude = 285.53 ms⁻²

Direction

Tan θ = 260.84 / 116.14 = 2.246

θ = 66 degree south of west .

3 0
3 years ago
Vocabulary should be taught:
lapo4ka [179]

Answer:

The third one

Explanation:

it is better for vocab to be taught through natural learning and lessons, instead of one or the other.

7 0
3 years ago
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