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3 years ago

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A 7.1 resistor and a 4.8 resistor are connected in series with a battery. the potential difference across the 4.8 resistor is me

asured as 12 v. find the potential difference across the battery. v

1 answer:

Taya2010 [7]3 years ago

4 0

For resistors in series in a DC circuit, the potential difference is proportional to the resistance.

The voltage V across the 7.1 ohm resistor can be found from the proportion
V/7.1 = 12/4.8
V = 7.1*12/4.8 = 17.75

The potential difference at the battery terminals is the sum of the potential differences across the two series resistors.
Vbatt = 17.75 V + 12 V
Vbatt = 29.75 V

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2 years ago

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An FM radio station broadcasts electromagnetic radiation at a frequency of 93.5 MHz. The wavelength of this radiation is _______

Alborosie

Answer:

3.21

Explanation:

The relation between frequency and wavelength is shown below as:

$c=frequency\times Wavelength
$

c is the speed of light having value $3\times 10^8\ m/s$

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3 years ago

What is the frequency of a wave if the speed is 24 m/s and the wave is 2 meters

vekshin1

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3 years ago

Some expressways are curved, banked and designed to maximize ___________ at higher speeds.

Liono4ka [1.6K]

Answer:

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2 years ago

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

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

=> $z = 1 \ m$

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

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3 years ago