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pantera1 [17]
3 years ago
10

A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.30mmc

os[(6.98rad/m)x + (742 rad/s)t]. Being more practical, you measure the rope to have a length of 1.35 m and a mass of 0.00338 kg. You are then asked to determine the following:
A) Amplitude
B) Frequency
C) Wavelength
D) Wave Speed
E) Direction the wave is traveling
F) Tension in the rope
G) Average power transmitted by the wave
Physics
1 answer:
MakcuM [25]3 years ago
7 0
<span>y(x,t)= 2.30mmcos[(6.98rad/m)x + (742 rad/s)t]
</span>A) Amplitude is 2.30mm<span>
B) Frequency 1/</span>2.30mm<span>
C) Wavelength is </span>6.98rad/m<span>
D) Wave Speed is </span>742 rad/s<span>
E) Direction the wave is traveling

</span>
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30 points please answer asap!
tino4ka555 [31]

Answer:

False, if its not moving there is no potential energy.

5 0
3 years ago
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Write comparison between Electrical Circuit and Magnetic Circuit.
Anarel [89]

<u>electrial</u><u> </u>

<u> </u><u>In an electrical circuit, electric current flows through the closed path</u>

<u>magnetic</u><u> </u><u>circuit</u><u> </u>

<u>In the magnetic circuit, magnetic flux flows through the closed path</u>

7 0
3 years ago
A 4-79 permalloy solenoid coil needs to produce a minimum inductance of 1.1 . If the maximum allowed current is 4 , how many tur
daser333 [38]

The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,

The magnetic field H is given as,

H = \frac{nI}{l}

Here,

n = Number of turns of the coil

I = Current that flows in the coil

l = Length of the coil

From the above equation, the number of turns of the coil is,

n = \frac{Hl}{I}

The magnetic field is again given by,

H = \frac{B}{\mu_t}

Where the minimum inductance produced by the solenoid coil is B.

We have to obtain n, that

n = \dfrac{\frac{B}{\mu_t}l}{I}

Replacing with our values we have that,

n = \dfrac{\frac{1.1Wb/m^2 }{200000}(2m)}{4mA}

n = \dfrac{(\frac{1.1Wb/m^2 }{200000})(\frac{10^4 guass}{1Wb/m^2})(2m)}{4mA(\frac{10^{-3}A}{1mA})}

n = 27.5 \approx 28

Therefore the number of turn required is 28Truns

4 0
4 years ago
A stationary horn emits a sound with a frequency of 228 Hz. A car is moving toward the horn on a straight road with constant spe
stiks02 [169]

Answer: 26.84 m/s

Explanation:

Given

Original frequency of the horn f_o=228\ Hz

Apparent frequency f'=246\ Hz

Speed of sound is V=340\ m/s

Doppler frequency is

\Rightarrow f'=f_o\left(\dfrac{v+v_o}{v-v_s}\right)

Where,

v_o=\text{Velocity of the observer}\\v_s=\text{Velocity of the source}

Insert values

\Rightarrow 246=228\left[\dfrac{340+v_o}{340-0}\right]\\\\\Rightarrow 366.84=340+v_o\\\Rightarrow v_o=26.8\ m/s

Thus, the speed of the car is 26.84\ m/s

4 0
3 years ago
When 23Na is bombarded with protons, the products are 20Ne and A. a neutron B. an alpha particle C. a deuteron D. a gamma ray pa
Vikki [24]

Answer:

The correct option is B. an alpha particle.

Explanation:

When ²³Na is bombarded with protons _{1}^{1}\textrm{H}, ²⁰Ne and one alpha particle _{2}^{4}\textrm{He} is released.

The reaction is as follows:

_{11}^{23}\textrm{Na} + _{1}^{1}\textrm{H} \rightarrow _{10}^{20}\textrm{Ne} + _{2}^{4}\textrm{He}

Therefore, an alpha particle and ²⁰Ne are released when ²³Na is bombarded with protons.

5 0
4 years ago
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