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4 years ago

11

What tern is used to describe the measurement of a wave equal to half the distance from crest to trough?

1 answer:

Papessa [141]4 years ago

3 0

Are you referring the term Amplitude.

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Through which of the following will a light wave travel the fastest

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B. Outer Space.

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Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid

Korolek [52]

Answer:

a

$z= 2.5 \ m$

b

$z = (1 \ m , 4 \ m )$

Explanation:

From the question we are told that

Their distance apart is $d = 5.00 \ m$

The wavelength of each source wave $\lambda = 6.0 \ m$

Let the distance from source A where the construct interference occurred be z

Generally the path difference for constructive interference is

$z - (d-z) = m \lambda$

Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0

so

$z - (5-z) = 0$

=> $2 z - 5 = 0$

=> $z= 2.5 \ m$

Generally the path difference for destructive interference is

$|z-(d-z)| = (2m + 1)\frac{\lambda}{2}$

=> $|2z - d |= (0 + 1)\frac{\lambda}{2}$

=> $|2z - d| =\frac{\lambda}{2}$

substituting values

$|2z - 5| =\frac{6}{2}$

=> $z = \frac{5 \pm 3}{2}$

So

$z = \frac{5 + 3}{2}$

$z = 4\ m$

and

$z = \frac{ 5 -3 }{2}$

=> $z = 1 \ m$

=> $z = (1 \ m , 4 \ m )$

7 0

3 years ago

A 5 m3 tank containing 5kg of an unknown ideal gas at 500 kPa is connected through a valve to another tank containing 10 kg of t

Ivan

Answer:

a) $V_{T} = 9\,m^{2}$ , b) $m_{T} = 15\,kg$ , c) $P_{T} = 416.667\,kPa$

Explanation:

a) The equation of state for ideal gas is:

$P \cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

Given the existence of an isothermal process, the following relation is derived:

$\frac{P_{1}\cdot V_{1}}{m_{1}} = \frac{P_{2}\cdot V_{2}}{m_{2}}$

The volume of the other tank is:

$V_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}}\right)\cdot V_{1}$

$V_{2} = \left(\frac{10\,kg}{5\,kg} \right)\cdot \left(\frac{200\,kPa}{500\,kPa}\right)\cdot (5\,m^{3})$

$V_{2} = 4\,m^{3}$

The total volume is:

$V_{T} = V_{1} + V_{2}$

$V_{T} = 5\,m^{3} + 4\,m^{3}$

$V_{T} = 9\,m^{2}$

b) The total mass is:

$m_{T} = m_{1} + m_{2}$

$m_{T} = 5\,kg + 10\,kg$

$m_{T} = 15\,kg$

c) The pressure of the gas in the two tanks is:

$P_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{V_{1}}{V_{2}}\right)\cdot P_{1}$

$P_{T} = \left(\frac{15\,kg}{5\,kg}\right)\cdot \left(\frac{5\,m^{2}}{9\,m^{2}} \right)\cdot (500\,kPa)$

$P_{T} = 416.667\,kPa$

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4 years ago