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ludmilkaskok [199]
2 years ago
8

A radio wave has a frequency of 5.5 × 104 hertz and travels at a speed of 3.0 × 108 meters/second. What is its wavelength?

Physics
3 answers:
tigry1 [53]2 years ago
3 0

Answer:5.45X10^3m

Explanation:So use the formula,v= fλ

3X10^8=5.5X10^4λ what Im saying is divide both and u should get 5454.54m but do sig figs to get answer

Vera_Pavlovna [14]2 years ago
3 0

Answer:

5.45X10^3m

Explanation:

Mushroom2 years ago
0 0

4.3 × 1014 hertz

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Current Flow and Ohm's Law

Ohm's law is the most important, basic law of electricity. It defines the relationship between the three fundamental electrical quantities: current, voltage, and resistance. When a voltage is applied to a circuit containing only resistive elements (i.e. no coils), current flows according to Ohm's Law, which is shown below.

I = V / R 

Where: 

I =

Electrical Current (Amperes)

V =

Voltage (Voltage)

R =

Resistance (Ohms)

    

Ohm's law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). Therefore, if the voltage is increased, the current will increase provided the resistance of the circuit does not change. Similarly, increasing the resistance of the circuit will lower the current flow if the voltage is not changed. The formula can be reorganized so that the relationship can easily be seen for all of the three variables.

The Java applet below allows the user to vary each of these three parameters in Ohm's Law and see the effect on the other two parameters. Values may be input into the dialog boxes, or the resistance and voltage may also be varied by moving the arrows in the applet. Current and voltage are shown as they would be displayed on an oscilloscope with the X-axis being time and the Y-axis being the amplitude of the current or voltage. Ohm's Law is valid for both direct current (DC) and alternating current (AC). Note that in AC circuits consisting of purely resistive elements, the current and voltage are always in phase with each other.

Exercise: Use the interactive applet below to investigate the relationship of the variables in Ohm's law. Vary the voltage in the circuit by clicking and dragging the head of the arrow, which is marked with the V. The resistance in the circuit can be increased by dragging the arrow head under the variable resister, which is marked R. Please note that the vertical scale of the oscilloscope screen automatically adjusts to reflect the value of the current.

See what happens to the voltage and current as the resistance in the circuit is increased. What happens if there is not enough resistance in a circuit? If the resistance is increased, what must happen in order to maintain the same level of current flow?


4 0
3 years ago
Please help me anything will help
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Your answer is C) The speed of sound is higher in solids than in liquids. 
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Describe how the properties of sound waves change as they spread out in a spherical pattern.
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They go up and down
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The observation of a "wet spot" on a hot highway in the summer is caused by:_________.
maxonik [38]

Answer: Refraction

Explanation:

5 0
3 years ago
A concave mirror with a radius of curvature of 20 cm has a focal length of
xxTIMURxx [149]

Answer:

A concave mirror has a radius of curvature of 20 cm. What is it's focal length? If an object is placed 15 cm in front of it, where would the image be formed? What is it's magnification?

The focal length is of 10 cm, object distance is 30 cm and magnification is -2.

Explanation:

Given:

A concave mirror:

Radius of curvature of the mirror, as C = 20 cm

Object distance in-front of the mirror = 15 cm

a.

Focal length:

Focal length is half of the radius of curvature.

Focal length of the mirror =  \frac{C}{2} = 10 cm

According to the sign convention we will put the mirror on (0,0) point, of the Cartesian coordinate open towards the negative x-axis.

Object and the focal length are also on the negative x-axis where focal length and image distance will be negative numerically.

b.

We have to find the object distance:

Formula to be use:

⇒ \frac{1}{focal\ length}= \frac{1}{image\ distance} + \frac{1}{object\ distance}

⇒ Plugging the values.

⇒ \frac{1}{-10} =\frac{1}{image\ distance}+\frac{1}{-15}

⇒ \frac{1}{-10} -\frac{1}{-15}=\frac{1}{image\ distance}

⇒ \frac{1}{-10} + \frac{1}{15}=\frac{1}{image\ distance}

⇒ \frac{-3+2}{30} =\frac{1}{image\ distance}

⇒ \frac{-1}{30} =\frac{1}{image\ distance}

⇒ -30\ cm=image\ distance

Image will be formed towards negative x-axis 30 cm away from the pole.

c.

Magnification (m) is the negative ratio of mage distance and object distance:

⇒ m=-\frac{image\ distance}{object\ distance}

⇒ m=-\frac{(-30)}{(-15)}

⇒ m=-2

The focal length of the concave mirror, is of 10 cm, object distance is 30 cm and magnification is -2.

5 0
3 years ago
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