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3 years ago

Which of the following statements correctly describe the various applications listed above?

Physics

1 answer:

fgiga [73]3 years ago

3 0

Answer:

The answer is "c,d,e, and g".

Explanation:

The correct choices can be defined as follows:

Higher-frequency microwaves aren't used in any of these systems.
Infrared waves aren't seen in each of these technologies.
Its shortest wavelength of all of the technologies listed is the above radiation generated by certain wireless networks.
These devices all produce waves of wavelengths ranging from 0.10 to 10.0 cm.

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A marathon runner completes a 42.238 km course in 2 h, 31 min, and 46 s . There is an uncertainty of 29 m in the distance run an

icang [17]

Answer:

The percentage uncertainty in the average speed is 0.10% (2 sig. fig.)

Explanation:

Consider the formula for average speed $\bar{v}$ .

$\displaystyle \bar{v} = \frac{s}{t}$ ,

where

$s$ is the total distance, and
$t$ is the time taken.

The percentage uncertainty of a fraction is the sum of percentage uncertainties in

the numerator, and
the denominator.

What are the percentage uncertainties in $s$ and $t$ in this question?

The unit of the absolute uncertainty in $s$ is meters. Thus, convert the unit of $s$ to meters:

$s = \rm 42.238\;km = 42.238\times 10^{3}\;m$ .

$\begin{aligned}\displaystyle \text{Percentage Uncertainty in }s &= \frac{\text{Absolute Uncertainty in } s}{\text{Measured Value of }s}\times 100\% \\ &=\rm\frac{29\; m}{42.238\times 10^{3}\;m}\times 100\%\\ &= 0.0687\%\end{aligned}$ .

The unit of the absolute uncertainty in $t$ is seconds. Convert the unit of $t$ to seconds:

$t = \rm 2\times 3600 + 31\times 60 + 46 = 9106\;s$

Similarly,

$\begin{aligned}\displaystyle \rm \text{Percentage Uncertainty in }t &= \frac{\text{Absolute Uncertainty in }t}{\text{Measured Value of }t}\times 100\% \\ &=\rm\frac{46\; s}{9106\;s}\times 100\%\\ &= 0.0329\%\end{aligned}$ .

The average speed $\bar{v}$ here is a fraction of $s$ and $t$ . Both $s$ and $t$ come with uncertainty. The percentage uncertainty in $\bar{v}$ will be the sum of percentage uncertainties in $s$ and $t$ . That is:

$\text{Percentage Uncertainty in }\bar{v}\\=(\text{Percentage Uncertainty in } s) + (\text{Percentage Uncertainty in } t)\\ = 0.0687\% + 0.0329\%\\ = 0.010\%$ .

Generally, keep

two significant figures for percentage uncertainties that are less than 2%, and
one for those that are greater than 2%.

The percentage uncertainty in $\bar{v}$ here is less than 2%. Thus, keep two significant figures. However, keep more significant figures than that in calculations to make sure that the final result is accurate.

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4 years ago

What’s the equation for the alpha decay of plutonium-244.

kvv77 [185]

Answer:

Pu-244 and 94 alpha decay = U -240 and 92

8 0

3 years ago

23) What is the magnitude of the electric field intensity at a point where a proton experiences an

Radda [10]

Answer: I am so sorry i hadn't learned this Yet~!

Explanation:

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2 years ago

The thermal efficiency of a power cycle operating in a reversible manner is found to be 50%. Assuming that the same 2 thermal re

inna [77]

Answer:

Explanation:

The thermal efficiency of a Power cycle $\eta = \dfrac{Q_H -Q_c}{Q_H}$

where;

$\eta = 50\% = 0.5$

$Q_H = Heat \ flow \ from \ higher \ temperature$

$Q_c = Heat \ flow \ from \ lower \ temperature$

$0.5 = \dfrac{Q_H -Q_c}{Q_H}$

$0.5 Q_H = Q_H - Q_c$ --- (1)

$Q_c = 0.5 Q_H$ ---- (2)

The coefficient of performance is:

$COP_R = \dfrac{Q_c}{Q_H -Q_c}$

let replace the value of $Q_c = 0.5 Q_H$ in the above equation then;

$COP_R = \dfrac{0.5Q_H}{Q_H -0.5 Q_H}$

$COP_R = \dfrac{0.5Q_H}{0.5 Q_H}$

$COP_R = 1$

The

On the other hand, the heat pump

$COP_{HP} = \dfrac{Q_H}{Q_H -Q_c}$

By replacing equation (1) into the above equation; we have:

$COP_{HP} = \dfrac{Q_H}{0.5Q_{H}}$

$COP_{HP} = \dfrac{1}{0.5}$

$COP_{HP} =2$

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3 years ago

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Aneli [31]

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4 years ago