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Sever21 [200]
3 years ago
8

The maximum number of eletrons that can occupy a 3p sublevel is

Physics
1 answer:
nadezda [96]3 years ago
3 0
The maximum number is 6
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If a sound travels 343m/s through air and has a frequency of 800Hz, what would the wavelength be of the same sound traveling thr
VARVARA [1.3K]

Answer:

0.625m

Explanation:

Now Velocity of a wave ,V = frequency × wavelength

Wavelength =velocity /frequency

=500/800 =0.625m

7 0
3 years ago
Define measurements.​
Reptile [31]

Answer:

act or process of measuring

Explanation:

6 0
3 years ago
Scientific observations should be reported with bias as long as they benefit a scientist's employer
Ira Lisetskai [31]
False, there are always pros and cons to research
8 0
3 years ago
in a human cannonball a person is shot from a cannon with a barrel that is 3.05 m long. 16700 J of work are done to accelerate t
ankoles [38]
The work done to accelerate the acrobat is given by
W=Fd
where F is the force applied and d the distance of application of the force.
If the barrel is 3.05 m long, then d=3.05 m. Therefore we can find the force:
F= \frac{W}{d} = \frac{17600 J}{3.05 m} =5475 N
6 0
4 years ago
A day on a distant planet observed orbiting a nearby star is 21.5 hr. Also, a year on the planet lasts 69.3 Earth days. In other
serg [7]

Answer:

Part A

The angular speed of rotation of the plane is 8.11781 × 10⁻⁵ rad/s

Part B

The angular speed of orbit of the planet is 1.04938 × 10⁻⁶ rad/s

Explanation:

The parameters of the planet are;

The duration of a day on the distant planet = 21.5 hr.

The duration of a year on the distant planet = 69.3 Earth days

Part A

The duration of a day = The time to make one complete revolution of 2·π radians

∴ The average angular speed about its axis, \omega_{rotation} = Angle turned/Time

∴ \omega_{rotation}  = 2·π/(21.5 × 60 × 60) s ≈ 8.11781 × 10⁻⁵ rad/s

The average angular speed of the planet about its own axis, \omega_{rotation}  = 8.11781 × 10⁻⁵ rad/s

The angular speed of rotation of the plane \omega_{rotation}  = 8.11781 × 10⁻⁵ rad/s

Part B

The time it takes the planet to revolve round the neighboring star once = 69.3 Earth days

Therefore, the average angular speed of the planet around its neighboring star, \omega _{Star}, is given as follows;

\omega _{Orbit}  = 2·π/((69.3 × 24 × 60 × 60) s) = 1.04938 × 10⁻⁶ rad/s

The average angular speed of orbit, \omega _{Orbit} = 1.04938 × 10⁻⁶ rad/s

The angular speed of orbit of the planet, \omega _{Orbit} = 1.04938 × 10⁻⁶ rad/s.

3 0
3 years ago
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