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Svet_ta [14]
4 years ago
14

Nitrogen (Z = 7) has three electrons in the 2p level (in addition to two electrons each in the 1s and 2s levels).

Physics
1 answer:
solmaris [256]4 years ago
8 0
Based on electrons would be answer b.
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kozerog [31]

Answer:

The answer is C

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5 0
3 years ago
A throttle position sensor waveform is going to be observed. At what setting should the volts per division be set to see the ent
Mars2501 [29]

Answer:

1 V / div

Explanation:

Solution:

- The vertical scale has eight divisions.  

- If each division is set to equal 1 volt, the display will show 0 to 8 volts.  

- This is okay in a 0 to 5 volt variable sensor such as a throttle position (TP) sensor.  

- The volts per division (V/div) should be set so that the entire anticipated waveform can be viewed.

3 0
3 years ago
However, sometimes levers are designed to increase the input force needed to move an object. Which of the following is most like
Tems11 [23]

What are the answer choices, if there are any?

6 0
4 years ago
Independent measurements of two lengths are made. The results are x=95 +- 0.1 and y=2.30+-0.02 cm
notsponge [240]
95-2.30+0.1+0.02cm =0.00089186
100m
3 0
3 years ago
A uniformly dense solid disk with a mass of 4 kg and a radius of 4 m is free to rotate around an axis that passes through the ce
Dmitry [639]

Answer:

3.44 rad

Explanation:

The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk

Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is  ΔK = 21 J/s × 3.3 s = 69.3 J

So, ΔK = 1/2I(ω² - ω₀²)

Since ω₀ = 0 rad/s

ΔK = 1/2I(ω² - 0)

ΔK = 1/2Iω²

ΔK = 1/2(MR²/2)ω²

ΔK = MR²ω²/4

ω² = (4ΔK/MR²)

ω = √(4ΔK/MR²)

ω = 2√(ΔK/MR²)

Substituting the values of the variables into the equation, we have

ω = 2√(ΔK/MR²)

ω = 2√(69.3 J/( 4 kg × (4 m)²))

ω = 2√(69.3 J/[ 4 kg × 16 m²])

ω = 2√(69.3 J/64 kgm²)

ω = 2√(1.083 J/kgm²)

ω = 2 × 1.041 rad/s

ω = 2.082 rad/s

The angular displacement θ is gotten from

θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²

Substituting the values of the variables into the equation, we have

θ = ω₀t + 1/2αt²

θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²

θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²

θ = 1/2 × 6.87159 rad

θ = 3.436 rad

θ ≅ 3.44 rad

6 0
3 years ago
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