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3 years ago

What process allows radios to receive signals even when they are not in line of a tower?

1 answer:

3 0

The answer is A. Refraction

Refraction is a change in direction of propagation of a wave due to a change in its transmission medium, by maintaining its frequency while changing the wave velocity at the same time will allows radios to receieve signals even though when they're not in line

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Our eyes can detect light only within a range of ____________ called visible light. frequencies speeds mediums periods

ahrayia [7]

Our eyes detect light only within a range of frequencies called visible light.

3 0

3 years ago

A leaky 12-kg bucket is lifted from the ground to a height of 15 m at a constant speed with a rope that weights 0.75 kg/m. Initi

Scrat [10]

Answer:

Approximately 5.5 kJ.

Explanation:

The mass of water in the bucket is $\rm 40\; kg$ at a height of 0 meters above the ground.
At a height of 15 meters, the mass of the water would be $\rm 0\; kg$ .

The bucket goes up at a constant rate. Also, the water leaks at a constant rate. As a result, the mass of water in the bucket changes at a constant rate as height increases.

Find the rate at which the mass of water decreases over height:

$\begin{aligned} &\text{Rate at which water leaks}\cr &= \frac{\text{Change in mass of Water}}{\text{Change in height}} \cr &= \rm \frac{40 \; kg - 0\; kg}{15\; m} \cr &= \rm \frac{40}{15}\; kg \cdot m^{-1}\end{aligned}$ .

In other words, $\displaystyle \rm \frac{40}{15}\; kg$ of water leaks out of the bucket everytime its height increases by $\rm 1\; m$ .

Hence, the mass of water at $h\; \rm m$ would be $\displaystyle 40 - \frac{40}{15}\, h$ $\rm kg$ . Write that as a function of height:

$\displaystyle m_\text{water}(h) = 40 - \frac{40}{15}\, h = \frac{40}{15} \, (15 - h)$ .

On the other hand, $(40 - h)\; \rm m$ of rope is still in the air while the bucket is at a height of $h\; \rm m$ . The mass of the rope that needs to be lifted at would thus be $0.75\; (15 - h)\; \rm kg$ .

Write that also as a function of the height of the bucket:

$\displaystyle m_\text{rope}(h) = 0.75\; (15 - h)$ .

Add the mass of the bucket:

$\displaystyle m(h) = \frac{40}{15}\, (15 - h) + 0.75\; (15 - h) + 12$ .

Simplify this expression:

$\begin{aligned} m(h) &= \frac{40}{15}\, (15 - h) + 0.75\; (40 - h) + 12 \cr &= \left(\frac{40}{15} + 0.75\right) (15 - h) + 12\end{aligned}$ .

The only two forces on the bucket are gravitational force and tension force from the rope. Since the bucket is moving at constant speed, the net force on it should be zero. The two forces must balance each other. In other words, they must act in opposite directions. Most importantly, the must be of equal magnitude. That is:

$\begin{aligned} F(h) &= W(h) \cr &= m(h) \cdot g \cr &= 9.81 \times \left(\left(\frac{40}{15} + 0.75\right) (15 - h) + 12\right)\cr &= 9.81 \times \left(\frac{40}{15} + 0.75\right) (15 - h) + 9.81 \times 12 \cr &\approx 502.763 - 33.5175\; h + 117.72 \cr &= 620.483 - 33.5175\; h\end{aligned}$ .

Integrate force over distance travelled to find work done. In this case, since the bucket is moving upwards, distance travelled is equal to height.

$\begin{aligned}W &= \int \limits_{0}^{15} (620.483 - 33.5175\; h) \, dh \cr &= \left[620.483\; h - \frac{1}{2}\times 33.5175\; h^2\right]^{15}_{0} \cr &= \left(620.483 \times 15 - \frac{1}{2} \times 33.5175 \times 15^2\right) - 0 \cr &\approx 5.5\times 10^{3}\; \rm J \cr &= 5.5\;\rm kJ \end{aligned}$ .

7 0

3 years ago

A wave with a period of 1⁄3 second has a frequency of A. 67 Hz. B. 33 Hz. C. 300 Hz. D. 3 Hz.

Triss [41]

A wave with a period of 1⁄3 second has a frequency of D. 3 Hz. To calculate this we will use the formula that represents the correlation between a frequency (f) and a time period (T): T = 1/f. Or: f = 1/T. The unit for the time period is second "s" while the unit for frequency is Hertz "Hz" (=1/s). We know that T = 1/3 s. That means that f = 1/(1/3s) = 3 1/s = 3 Hz.

3 0

4 years ago

A car slowed down from 70 mils to 30 mi/s on the highway in 20

Arlecino [84]

It was 40 I hope it helps

4 0

4 years ago

How much KCI can be dissolved in 100 g of water at 30°C?

asambeis [7]

Answer:

Explanation:

The problem provides you with thesolubility of potassium chloride, KCl , in water at 20∘C , which is said to be equal to 34 g / 100 g H2O . This means that at 20∘C , a saturated solution of potassium chloride will contain 34 g of dissolved salt for every100 g of water.

6 0

3 years ago

Read 2 more answers