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e-lub [12.9K]
3 years ago
11

Wireless Internet networks, including many used in homes, often make use of high-frequency radio waves. High-frequency waves are

useful because they can carry a lot of information. However, high-frequency waves are less capable of passing through objects than are low-frequency waves. As a result, waves traveling from a person's wireless laptop computer, for example, could be interrupted by objects between the computer and the modem.
Due to this limitation of high frequency waves, which of the following statements best explains why digital waves are commonly used in high-frequency wireless networks instead of analog waves?
A.
Analog waves cannot reach high enough amplitudes to be used in wireless networks.
B.
Analog waves cannot reach large enough wavelengths to be used in wireless networks.
C.
It is easier to correct small disruptions in digital waves than in analog waves.
D.
Small disruptions in analog waves are more easily corrected than in digital waves.
Physics
2 answers:
IgorLugansk [536]3 years ago
7 0

Answer:

The CORRECT answer is c

lutik1710 [3]3 years ago
4 0
The answer is d , small disruptions
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Answer

a) For the rock

\dfrac{v_t^2sin 2\theta}{g} = \dfrac{v_t^2sin^2\theta}{2g}

2sin\thetacos\theta = \dfrac{sin^2\theta}{2}

2cos\theta = \dfrac{sin\theta}{2}

tan\theta = 4

\theta = tan^{-1} 4

\theta = 76^0

b) \theta = 45^0 for maximum range

\dfrac{d_{max}}{d}=\dfrac{(v_tcos 45^0)(2v_tsin 45^0)g}{(v_tcos 76^0)(2v_tsin 76^0)g}

\dfrac{d_{max}}{d}=\dfrac{0.707\times 0.707)}{0.97\times 0.242}

\dfrac{d_{max}}{d}=2.129

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c) The value of θ is the same on every planet as g divides out.

5 0
3 years ago
Physics double pivot question​
andriy [413]

Explanation:

Assuming the wall is frictionless, there are four forces acting on the ladder.

Weight pulling down at the center of the ladder (mg).

Reaction force pushing to the left at the wall (Rw).

Reaction force pushing up at the foot of the ladder (Rf).

Friction force pushing to the right at the foot of the ladder (Ff).

(a) Calculate the reaction force at the wall.

Take the sum of the moments about the foot of the ladder.

∑τ = Iα

Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0

Rw (3.0 sin 60°) = mg (1.5 cos 60°)

Rw = mg / (2 tan 60°)

Rw = (10 kg) (9.8 m/s²) / (2√3)

Rw = 28 N

(b) State the friction at the foot of the ladder.

Take the sum of the forces in the x direction.

∑F = ma

Ff − Rw = 0

Ff = Rw

Ff = 28 N

(c) State the reaction at the foot of the ladder.

Take the sum of the forces in the y direction.

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Rf = mg

Rf = 98 N

3 0
3 years ago
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option D pregnant women

4 0
3 years ago
A parallel-plate capacitor is connected to a battery and becomes fully charged. The capacitor is then disconnected, and the sepa
AnnZ [28]

Answer:

Increases

Explanation:

The expression for the capacitance is as follows as;

C=\frac{\epsilon _{0}A}{d}

Here, C is the capacitance, \epsilon _{0} is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.

It can be concluded from the above expression, the capacitance is inversely proportional to the distance. According to the given problem, the capacitor is disconnected from the battery and the distance between the plates is increased. Then, the capacitance of the given capacitor will decrease in this case.

The expression for the energy stored in the parallel plate capacitor is as follows;

E=\frac{Q^{2}}{2C}

Here, E is the energy stored in the capacitor, C is the capacitance and Q is the charge.

Energy stored in the given capacitor is inversely proportional to the capacitor. The charge on the capacitor is constant. In the given problem, as the distance between the parallel plates is being separated, the energy stored in this capacitor increases.

Therefore, the option (c) is correct.

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Which source of information is generally considered credible?


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