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sattari [20]
3 years ago
11

Officials begin to release water from a full man-made lake at a rate that would empty the lake in 4 weeks, but a river that can

fill the lake in 15 weeks is replenishing the lake at the same time. How many weeks does it take to empty the lake? Express your answer as a fraction reduced to lowest terms, if needed.
Physics
1 answer:
iris [78.8K]3 years ago
6 0

Answer:

5 weeks and 5 days is required to empty the lake

Explanation:

Officials begin the remove water from a full man made lake

The lake can be emptied in 4 weeks

= -1/4

A river can fill the lake up in 15 weeks

= 1/15

Let t represent the number of weeks that is required to empty the lake

= -1/t

Therefore the number of weeks it takes to empty the lake can be calculated as follows

-1/t= -1/4 + 1/15

-1/t= -11/60

Cross multiply

-11×t= -1×60

-11t= -60

t = 60/11

t= 5 5/11

Hence it takes 5 weeks and 5 days to empty the lake

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M (b) Calculate the speed of an electron that is accelerated through the same electric potential difference.
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The speed of a electron that is accelerated from rest through an electric potential difference of 120 V  is 6.49\times10^6m/s

<h3>How to calculate the speed of the electron?</h3>

We know, that the  energy of the system is always conserved.

Using the Law of Conservation of energy,

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Now, substituting the formula of U and K, we get:

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Let v_f and v_i represent the final and initial speed.

Here, v_i =0

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4 0
1 year ago
Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti
Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

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b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

c = \lambda \cdot \nu  (1)

Where c is the speed of light, \lambda is the wavelength and \nu is the frequency.  

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, \lambda can be isolated from equation 1:

\lambda = \frac{c}{\nu} (2)

since the value of c is 3x10^{8}m/s. It is necessary to express the frequency in units of hertz.

\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 100200000Hz

But 1Hz = s^{-1}

\nu = 100200000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}

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Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

<em> </em>

\nu = 1070kHz . \frac{1000Hz}{1kHz} ⇒ 1070000Hz

But  1Hz = s^{-1}

\nu = 1070000s^{-1}

Finally, equation 2 can be used:

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\lambda = 280.3 m

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 835600000Hz

But  1Hz = s^{-1}

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Finally, equation 2 can be used:

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Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

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