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egoroff_w [7]
3 years ago
5

Oscilloscope amplitude and frequency problem. Study the above graph. The volts/div dial is set to 2 volts/div and the time/div d

ial is set to 5 msec/div. What is the peak-to-peak amplitude of the displayed signal? What is the frequency in khz?

Physics
1 answer:
denis-greek [22]3 years ago
8 0

Answer:

Amplitude = 8 Volts

Frequency = 0.067 kHz

Explanation:

Note: The missing picture in question is attached for your review.

Given:

Volts/Div = 2 V/div

Time/Div = 5 msec/div

Finding Amplitude:

Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,

Amplitude = 4 div/volts * 2 volts/ div )\\Amplitude = 8 Volts

(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)

Finding Frequency:

As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,

Time Period = 3 div * 5msec /div\\Time Perod = 15 msec

Since,

Frequency = \frac{1}{Time Period}\\Frequency = \frac{1}{15m}\\Frequency = 0.067 kHz

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A ball is thrown vertically upward from the top of a building 80 feet tall with an initial velocity of 64 feet per second. The d
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Answer:

a) t=6.37s

b) t=3.3333s

Explanation:

The knowable variables are the initial hight and initial velocity

s_{o}=80ft

v_{os}=64ft/s

The equation that describes the motion of the ball is:

s=80+64t-16t^{2}

If we want to know the time that takes the ball to hit the ground, we need to calculate it by doing s=0 that is the final hight.

0=80+64t-12t^{2}

a) Solving for t, we are going to have two answers

t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}

a=-16

b=64

c=80

t=-1.045 s or t=6.378s

<em><u>Since time can not be negative the answer is t=6.378s </u></em>

b) To find the time that takes the ball to pass the top of the building on its way down, we must find how much does it move too

First of all, we need to find the maximum hight and how much time does it take to reach it:

v_{y}=v_{o}+gt

at maximum point the velocity is 0

0=64-32.2t

Solving for t

t=1.9875 s

Now, we must know how much distance does it take to reach maximum point

s=0+64t-16t^{2} =64(1.9875)-12(1.9875)^{2} =80ft

So, the ball pass the top of the building on its way down at 160 ft

160=80+64t-16t^{2}

Solving for t

t=2s or t=3.333s

Since the time that the ball reaches maximum point is almost t=2s that answer can not be possible, so the answer is t=3.333s for the ball to go up and down, passing the top of the building

4 0
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Answer:

W_A_B=-1.6\times 10^{-18} J

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V_B_A=\frac{W_A_B}{q}

Therefore, isolating W_A_B and replacing the data provided:

W_A_B=V_B_A *q=-2*(8\times 10^{-19}) =-1.6\times 10^{-18}J

6 0
4 years ago
A person pushes a 50kg box to the right with 100N of force at a constant velocity of 5m/s. Calculate the coefficient of friction
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3 years ago
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Answer:

B

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Momentum is the product or multiplication of a body's mass and its speed.

Since all options have the mass and speed in the same units, there is no need for conversion.

A. 20 x 500 = 10000,  B. 200 x 60 = 12000

The same goes for the rest!

6 0
3 years ago
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