3 years ago

Question 1 of 10

Physics

1 answer:

skad [1K]3 years ago

5 0

Answer:

A is the right answer of the following statement

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local AM radio station broadcasts at a frequency of 685.9 kHz. Calculate the wavelength at which they are broadcasting. Waveleng

natta225 [31]

Answer:

λ = 437 m.

Explanation:

Since a radio wave is a electromagnetic type of wave, it propagates approximately at the same speed of light in vacuum, 3*10⁸ m/s.
As in any wave, there exist a fixed relationship between the propagation speed, the wavelength and the frequency of the radio wave, as follows:

$c = \lambda* f (1)$

Replacing by the values of the speed and the frequency, and solving for the wavelength λ, we get:

$\lambda = \frac{c}{f} =\frac{3e8 m/s}{685.9e31/s} =437 m (2)$

8 0

2 years ago

A student practicing for a track meet ran 263 m in 30 sec. What was her average speed?

alexira [117]

Answer:

8.3 m/s

Explanation:

8 0

3 years ago

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How would you know if a force is acting on an object?

Artemon [7]

Answer:

microscope

Explanation:

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2 years ago

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In a certain process a gas ends in its original thermodynamic state. Of the following, which is possible as the net result of th

drek231 [11]

Answer:

E. The gas absorbs 50 of energy as heat and does 50」ot work

Explanation:

This is following the law of thermodynamics that energy is neither created nor destroyed

3 0

3 years ago

A person stands on the ball of one foot. The normal force due to the ground pushing up on the ball of the foot has magnitude 750

White raven [17]

Answer:

the force exerted on the foot by the tibia would be 2975 N

Explanation:

Given the data in the question;

To maintain equilibrium between the foot and the ball vertically, the addition normal normal force $N^>$ (750 N) and the tension in the Achilles tendon $F^>_{Achilles}$ (2225 N) must be equal to the force exerted on the foot by the tibia;

| $N^>$ | + | $F^>_{Achilles}$ | = | $F^>_{Tibia}$ |

so force exerted on the foot by the tibia will be;

| $F^>_{Tibia}$ | = | $N^>$ | + | $F^>_{Achilles}$ |

so we substitute IN OUR VALUES

| $F^>_{Tibia}$ | = 750 N + 2225 N

| $F^>_{Tibia}$ | = 2975 N

Therefore, the force exerted on the foot by the tibia would be 2975 N

3 0

3 years ago