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vfiekz [6]
2 years ago
7

Oscilloscope time base problem. You are asked to display a 1 kHzsine wave signal on the screen. Since the period T is related to

the frequency f by the formula T = 1 / f, the period of this waveis 1/(103/s)=1ms, so a reasonable setting for the"Time/div" dial is 1 ms/div. But suppose you select aninappropriate setting for the Time/div dial by mistake. What wouldyou see if you set the dial on 1 μ s/div?100 ms/div?A. Asine wave shifted to the left on 1 μ s/div and to the right on100 ms/divB. A(nearly) horizontal line on 1 μ s/div andmany (nearly) vertical lines of 100 ms/divC.Nothing on either (no trigger)D. Anormal sine wave in both casesE.None of these
Physics
1 answer:
Dmitrij [34]2 years ago
5 0

Answer:

B.

Explanation:

If we need to show on the screen a sine wave of 1 Khz, this means that the period (time to complete one cycle), will be 1/f = 1 ms.

So, in order to show some cycles on the screen at the same time, it is reasonable to choose a 1ms/div setting for "Time/div" dial.

If we by mistake choose a much much shorter setting (like 1 μs/div), what we will look at the screen will be very different, as we will be looking at 1/1000 of a cycle per div, so we wil see a nearly horizontal line.

If instead we choose 100 ms/div, this means that each division  will show 100 cycles of the sine wave, that they will look at if they were a lot of nearly vertical lines very closely spaced.

So, the statement B). is the one that is true.

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Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
netineya [11]

Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{2}} and C is a constant

6 0
2 years ago
a marine biologist wants to know the total vertical distance a dolphin travel during a jump with the surface of the water being
marin [14]

From the starting depth to the surface, the vertical distance is 35 ft.

From the surface to the peak of the jump, the vertical distance is 27 ft.

From the peak of the jump to the surface, the vertical distance is 27 ft.

From the surface to the ending depth, the vertical distance is 18 ft.

Then the total vertical distance is ...

  35 ft + 27 ft + 27 ft + 18 ft = 107 ft

3 0
2 years ago
Read 2 more answers
A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen
saw5 [17]

Answer:

The electric field strength is 4.5\times 10^{4} N/C

Solution:

As per the question:

Area of the electrode, A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}

Charge, q = 50 nC = 50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}

Now, the electric field strength of the electrode is:

\vec{E} = \frac{\sigma}{2\epsilon_{o}}

where

\epsilon_{o} = 8.85\times 10^{- 12} F/m

\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}

\vec{E} = 4.5\times 10^{4} N/C

7 0
2 years ago
A child pushes her toy across a level floor at a steady velocity of 0.50 m/s using an applied force of 2.0 N. If the weight of t
choli [55]
Since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.

Rate of speed that toy is moving is irelevant.

childs force is:
Fc = 2N
Fc = Ff  (Ff -friction force)

Ff = a*Q

where Q is weight of the toy and a is friction

if we express a we get
a = F/Q = 2/8 = 0.25
8 0
2 years ago
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For a Physics course containing 10 students, the maximum point total for the quarter was 200. The point totals for the 10 studen
Talja [164]

Answer:

130.5

Explanation:

According to the stemplot attached (Which I think it is, and if not, then you only need to replace the procedure with your data and you should be fine), you need to calculate first the points of all ten students. In that plot, we can easily calculate the points.

The first number in the colum represents the centen of the point, while the numbers of the second column, represents the units of that centen, for example if you see:

16 | 8 5 6

This means that the point for the students are 168, 165 and 166. Three students, three points.

If you watch the stemplot, the points for the students are:

116, 118, 121, 124, 128, 133, 137, 142, 146 and 179.

The median can be calculated using the mean between the two values in the middle of the sequence.

In this case, half of ten is 5, so, the numbers from the middle in this sequence are 128 and 133, therefore:

Median = 128 + 133 / 2 = 130.5

5 0
2 years ago
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