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MariettaO [177]
3 years ago
14

A wave is described by y(x,t) = 0.1 sin(3x + 10t), where x is in meters, y is in centimeters and t is in seconds. The angular wa

ve number is:
Physics
1 answer:
Mila [183]3 years ago
5 0

Answer: 3 radians/meter.

Explanation:

The general sinusoidal function will be something like:

y = A*sin(k*x - ω*t) + C

Where:

A is the amplitude.

k is the wave number.

x is the spatial variable

ω is the angular frequency

t is the time variable.

C is the mid-value.

The rule that we can use to solve this problem, is that the argument of the sin( ) function must be in radians (or in degrees)

Then if x is in meters, the wave-number must be in radians/meters, so when these numbers multiply the "meters" part is canceled.

Then for the case of the function:

y(x,t) = 0.1 sin(3x + 10t)

Where x is in meters, the units of the wave number (the 3) must be in radians/meters. Then the angular wave number is 3 radians/meter.

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Motor branch circuit
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3 years ago
A metal ring 4.60 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend
telo118 [61]

Answer:

A. Ein = 8.05*10^-4 V/m

B. Clockwise sense

Explanation:

A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:

\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}            (1)

Ein: induced electric field

ds: differential of a path of the ring

ФB: magnetic flux in the ring

The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):

\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}   (2)

dB/dt = -0.280T/s     (it is decreasing)

A: area of the ring = π(r/2)^2= (π/4) r^2

r: radius of the ring = 4.60/2 = 2.30 cm

Then, you replace the values of all variables in the equation (2):

E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}

hence, the induced electric field is 8.05*10^-4 V/m

B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.

6 0
4 years ago
Probably the deadliest aspect of a thunderstorm is _____. rain thunder lightning hail
max2010maxim [7]
The answer to the given question is lightning.
Lightning is most probably the deadliest aspect of a thunderstorm.  It is <span> a sudden </span>electrostatic discharge<span> during an </span>electrical storm. It happens <span> when there is an electrostatic discharge between </span>electrically charged<span> regions of a </span>cloud, which is referred as the i<span>ntra-cloud lightning or IC, between that cloud and another cloud or the CC lightning, or between a cloud and the ground (CG lightning).</span>
8 0
3 years ago
Read 2 more answers
I NEED HELP ASAP! BRAINIEST TO THE CORRECT ANSWER. HELP ME NOW!
sergij07 [2.7K]

Answer:

<h3>a)</h3>

\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}

  • V = 12 V
  • P = 24 W

\implies \mathsf{24=\frac{12^2}{R} }

\implies \mathsf{24R=12^2 }

\implies \mathsf{24R=144 }

<u>=> R= 6 Ohms(Ω)</u>

<h3>b)</h3>

\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}

  • Power (P)= 100 W

<em>these lights operate at the usual 240 volts direct from the main electricity supply. Therefore,</em>

  • V = 240 V

\implies \mathsf{100=\frac{240^2}{R} }

<em>R and 100 can interchange places</em>

\implies \mathsf{R=\frac{240^2}{100} }

\implies \mathsf{R=\frac{57600}{100} }

<u>=> R = 576 Ω</u>

<u></u>

By Ohm's Law:

\boxed{\mathsf{Voltage(V)=Current(I) \times Resistance(R)}}

=> 240 = I × 576

=>

=> I = 0.417 A

<h3 /><h3>c)</h3>

I don't know it's resistance,... so sorry

<h3>d)</h3>

The brightness of the bulb in series is <em><u>less than</u></em> when they're placed individually.

For bulbs in series their resistance gets added to form the equivalent resistance of the two bulbs.

Their resistances are nothing but mere numbers and the sum of two numbers(positive of course) is greater than the numbers.

So, the effective resistance of some bulbs in series <u>is more</u> than the individual resistance.

And

<em>Brightness, i. e., Power</em>

\boxed{\mathfrak{Power \propto  \frac{1}{Resistance} }}

If resistance increases, Power decreases.

Here, the effective resistance was for sure larger, therefore resistance was increasing, hence power decreased taking brightness along with it.

3 0
3 years ago
2 Points
Tom [10]
The answers are A and E!
6 0
3 years ago
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