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WARRIOR
10 days ago
12

A cellular telephone transmits electromagnetic waves at a frequency of 835 mhz. what is the wavelength of these waves?

Physics
1 answer:
Sav [3K]10 days ago
4 0
 Hi!

Instead of providing the answer outright, I'll assist you in grasping it using straightforward concepts.

As electromagnetic waves (consider: light) progress, they oscillate up and down. The frequency of light indicates the number of up-down oscillations occurring each second (1hz means 1 up-down oscillation per second, while 1mhz equals 1 million up-down oscillations per second). Each of these oscillations is known as a single wave.

Since we also have the information about how many meters light covers in 1 second (290,000,000 meters), it's easy to conclude that (regarding this specific question) for every '290,000,000' meters, there are '835,000,000' oscillations or waves.

Calculating the length of a single wave requires a simple division! - which refers to the wavelength of light at the specified frequency.

290,000,000 / 835,000,000 = 0.34m (or 34cm)
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inna [2995]

Answer:

Acceleration(a) = 0.75 m/s²

Explanation:

Given:

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1 month ago
A locomotive is accelerating at 1.6 m/s2. it passes through a 20.0-m-wide crossing in a time of 2.4 s. after the locomotive leav
kicyunya [3171]

Response:

Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.

Details:

  The locomotive's acceleration is 1.6 m/s^2

  The duration taken to pass the crossing is 2.4 seconds.

  We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.

  When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 m/s^2.

   32 = 0 + 1.6 * t

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  Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.

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An object is attached to a hanging unstretched ideal and massless spring and slowly lowered to its equilibrium position, a dista
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Answer:

        h = 12.8 cm

Explanation:

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distance = 6.4 cm

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        mgh = 0.5 x k x h^{2}....equation 2

  • Substituting from equation 1 into equation 2

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                y =  0.5 x h

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  • where y is 6.4, yielding the maximum elongation as

          h = 2 x 6.4 = 12.8 cm

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