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arsen [322]
1 year ago
5

The oscilloscope can be thought of as a plotting machine. What is plotted on the a axis? What is plotted on the y axis? If you t

ry to look at a 6 volt signal with the "volts/div" dial set on 0.2 you don't see anything. Why not? Should you turn the dial to 2 volts/div or to 0.02 volts/div to find the signal?
Physics
1 answer:
Feliz [49]1 year ago
5 0

Answer: The oscilloscope is not a plotting machine.

Explanation: The Oscilloscope is not a plotting machine is a device which is use to measure the frequency,period, peak to peak Voltage Vpp or any signal. That is alternating.

So, if you're such you wired your circuit whose output signal you want to measure very well and all connections and settings are done accurately, then you can reduce the volt/div below 0.2. You not seeing any signal at 0.2v/div shows that the amplitude of the signal coming into the Oscilloscope is not up to that.

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A stunt cyclist needs to make a calculation for an upcoming cycle jump. The cyclist is traveling 100 ft/sec toward an inclined r
Elena-2011 [213]

Answer:A=23.57^{\circ}

Explanation:

Given

velocity=100 ft/s

height of landing zone=10 ft

Equation of x(t)=100 t\cos (A)

y(t)=-16t^2+100t\sin (A)+10

Maximum height=35 feet

at maximum height

\frac{\mathrm{d} Y}{\mathrm{d} t}=0

\frac{\mathrm{d} x}{\mathrm{d} t}=-32t+100\sin A

t=3.125\sin A

At t=3.125\sin A

Y(t)=35=-16\times (3.125\sin A)^2+100\times 3.125\sin A\times \sin A+10

312.5\sin^2 A-156.25\sin^2 A=25

sin^2 A=0.16

A=23.57^{\circ}

8 0
1 year ago
A cat is sleeping on the floor in the middle of a 3.0-m-wide room when a barking dog enters with a speed of 1.50 m/s. as the dog
kondor19780726 [428]
Good morning.


Lets make the movement function for the dog and cat.

The cat has a start position of1.5 m(the middle of the room), with an initial speed of 0 and acceleration of 0.85 m/s².

The dog has a start position of 0, an initial speed of 1.5 m/s and acceleration of -0.1 m/s².


<u>Cat:</u>

\mathsf{X = X_0+V_0t + \dfrac{at^2}{2}}\\ \\ \mathsf{X = 1.5 + \dfrac{0.85t^2}{2}}\\ \\ \\ \mathsf{X_c = 1.5 + 0.425t^2}

<u>Dog:</u>

\mathsf{X_d= 1.5t - 0.05t^2}


Let's see if the dog reach the cat. Physically, it means \mathsf{X_d = X_c}

\mathsf{1.5t - 0.05t^2 = 1.5 + 0.425t^2}\\ \\ \mathsf{0.425t^2 + 0.05t^2 - 1.5t + 1.5 = 0}\\ \\ \mathsf{0.475t^2 - 1.5t + 1.5=0}

Now we solve for <em>t</em>:

\mathsf{\Delta = (-1.5)^2 - 4\cdot0.475\cdot1.5}\\ \\ \mathsf{\Delta = 2.25-2.85=-0.6 \ \textless \  0}

We have a negative Delta. Therefore, there is no instant t when the dog reaches the cat.
6 0
2 years ago
A bar magnet is dropped from above and falls through the loop of wire. The north pole of the bar magnet points downward towards
8_murik_8 [283]

Answer:

<em>b. The current in the loop always flows in a counterclockwise direction.</em>

<em></em>

Explanation:

When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.

3 0
1 year ago
What is the wavelength of a 100-mhz ("fm 100") radio signal?
vagabundo [1.1K]
Radio wave is about 3.10^8m/s divided by 10^8 hz is 3 nesters sound wave is 343m/s so thus Equal to approximately 0.78
7 0
1 year ago
Read 2 more answers
A uniform Rectangular Parallelepiped of mass m and edges a, b, and c is rotating with the constant angular velocity ω around an
Sonbull [250]

Answer:

(a) k = \frac{Mw^{2} }{6} (a^{2} +b^{2} )

(b)  τ = \frac{M}{3} (a^{2} +b^{2} ) ∝

Explanation:

The moment of parallel pipe rotating about it's axis is given by the formula;

I = \frac{M}{3} (a^{2} +b^{2} )   ---------------------------------1

(a) The kinetic energy of a parallel pipe is also given as;

k =\frac{1}{2} Iw^{2} --------------------------------2

Putting equation 1 into equation 2, we have;

k = \frac{M}{6} (a^{2} +b^{2} )w^{2}

k = \frac{Mw^{2} }{6} (a^{2} +b^{2} )

(b) The angular momentum is given by the formula;

τ = Iw -----------------------3

Putting equation 1 into equation 3, we have

τ = \frac{Mw}{3} (a^{2} +b^{2} )

But

τ = dτ/dt = \frac{M}{3} (a^{2} +b^{2} )\frac{dw}{dt}   ------------------4

where

dw/dt = angular acceleration =∝

Equation 4 becomes;

τ = \frac{M}{3} (a^{2} +b^{2} ) ∝

8 0
1 year ago
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