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solniwko [45]
3 years ago
7

When a particle vibrates and passes to the next vibration what happens?

Physics
1 answer:
kaheart [24]3 years ago
4 0
Im not sure how detailed your answer is supposed to be but here it is: The particle that vibrates contains energy. That exact amount of energy is passed on to the next particle. This occurs because as you must know, energy is neither created nor destroyed. The same amount of energy contained in the vibration is simply passed on.
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How long does it take an airplane to fly 1500 miles of it maintains a speed of 600 miles per hour?
prisoha [69]
2.5 hours. divide 15000 by 600




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3 years ago
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Using the equation zeff=z−s and assuming that core electrons contribute 1.00 and valence electrons contribute 0.00 to the screen
DIA [1.3K]
Thank you for posting your question here at brainly. Below is the solution. I hope the answer will help. 

<span>Cl^- 1s^2 2s^2p^6 3s^2 3p^6 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 17- 10 =7 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6; 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 19- 10 = 9 
</span>
S = 2 + 6.8 + 2.45 = 11.25 
<span>Zeff(Cl^-) = 17 – 11.25 = 5.75 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6 same S as for Cl^- but Z increases by 2 hence </span>
<span>Zeff(K^+) = 19 - 11.25 = 7.75</span>
5 0
3 years ago
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a kickball is struck with a 15.2 m/s velocity at a 63.0 degree angle. it lands on a rooftop 2.40 s later. how high is the roof?​
Serhud [2]

Refer to the attachment

3 0
3 years ago
A 5.0-kg centrifuge takes 95 s to spin up from rest to its final angular speed with constant angular acceleration. A point locat
stellarik [79]

Answer:

(a) 17.37 rad/s^2

(b) 12479

Explanation:

t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0

w = v / r = 99 / 0.06 = 1650 rad/s

(a) Use first equation of motion for rotational motion

w = w0 + α t

1650 = 0 + α x 95

α = 17.37 rad/s^2

(b) Let θ be the angular displacement

Use third equation of motion for rotational motion

w^2 = w0^2 + 2 α θ

1650^2 = 0 + 2 x 17.37 x θ

θ = 78367.87 rad

number of revolutions, n = θ / 2 π

n = 78367.87 / ( 2 x 3.14)

n = 12478.9 ≈ 12479

4 0
3 years ago
a car moving at 11 m/s crashes into an obstacle and stops in 0.26s. compute the Force that a seatbelt exerts on a 21-kg child to
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Answer:

890 N

Explanation:

Acceleration is change in velocity over change in time.

a = Δv / Δt

a = (11 m/s − 0 m/s) / 0.26 s

a = 42.3 m/s²

Force is mass times acceleration.

F = ma

F = (21 kg) (42.3 m/s²)

F ≈ 890 N

4 0
3 years ago
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