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poizon [28]
3 years ago
13

examine the following graph. a) What is the amplitude of the oscillation? b) What is the period of the oscillation?​

Physics
1 answer:
Aleks [24]3 years ago
6 0

Answer:

Explanation:

The amplitude looks like about 0.5

period = 2/3

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An iron nail goes through a chemical change when iron in the nail combines with oxygen in the air to form iron oxide. This compo
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It is like that, except most nails are steel or stainless steel, slowing to rusting process to about 5 years.
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3 years ago
An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 ​7 ​​ m/s over
stich3 [128]

Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :  

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron  

• a is the acceleration of the electron  

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.  

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m  

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

          a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

  = 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m  

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .  

xf =xi+vi*t+1/2*a*t

xf =xi+vi*t+1/2*a*t

  t=√2(xf-xi)/a

 t=3.765×10^-10 s

now we can use the power P Eq.  

 P=W/Δt => ΔK/Δt  

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0
3 years ago
1. Boyle's law relates the pressure of a gas to its
KIM [24]

Answer:

volume is the correct answer

Explanation:

7 0
2 years ago
A truck is traveling at 2.0 m/s. It slows to a stop at a constant rate over 3.00 s. How far does the car travel during those 3.0
earnstyle [38]

Answer:

During those 3.00 seconds before stopping, the car travels a distance of 6 m.

Explanation:

The simple rule of three is a tool that is used to quickly solve problems, where three pieces of information must be known, and one of them operates as an unknown to be known.

Two magnitudes are directly proportional if one magnitude increases the other also does it, and if the magnitude decreases the other in the same way.

Being a, b and c known data and x the unknown, the value that we want to know, the rule of three when the magnitudes are directly proportional is applied as follows:

a ⇒ b

c ⇒ x

So: x=\frac{c*b}{a}

In this case, knowing that a truck travels at 2 m/s, the rule of three applies as follows: if in 1 second the truck travels 2 m, in 3 seconds how much distance does it travel?

distance=\frac{3 s*2 m}{1 s}

distance= 6 m

<u><em> During those 3.00 seconds before stopping, the car travels a distance of 6 m.</em></u>

8 0
3 years ago
Why is Pluto no longer a planet?
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The International Astronomical Union (IAU) downgraded the status of Pluto to that of a dwarf planet because it did not meet the three criteria the IAU uses to define a full-sized planet. Essentially Pluto meets all the criteria except one—it “has not cleared its neighboring region of other objects.
5 0
3 years ago
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