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3 years ago

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The Periodic Table Question 4 of 10 If an element forms a 2+ ion, in which group of the periodic table would you expect to find

it? D A. 2 B. 17 O C. 1 D. 18 7

1 answer:

zvonat [6]3 years ago

3 0

Answer:

B is the difference قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *قصدي لا انا اعرف بشير بس ما اعرف اخوه يعني *

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The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. it is a good approximation to assume tha

Rufina [12.5K]

This assumption is correct. The emissivity of a substance is the ratio of the energy radiated from a body to the energy radiated from a black body (perfect emitter). Because these stars reflect negligible radiation compared to the amount they emit.

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3 years ago

Light travels at a speed of 3.00 x 10^11 cm/s. What is the speed of light in kilometers/hour?

rewona [7]

As 1 km = 1000 m = 1000,00 cm,

So, 1 cm = (1/1000,00) km

1 hour = 60 × 60 s = 3600 s

So, 1 s = (1 / 3600) hour

The light travels at a speed of $3.00 \times 10^{11} \ cm/s$ .

In kilometer/hour,

$3.00 \times 10^{11} \ cm/s = 3.00 \times 10^{11} \frac{(1/1000,00) km}{ (1 / 3600) hour} = 108 \times 10^8 \ km/hour$

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4 years ago

A train travels 94 kilometers in 5 hours, and then 84 kilometers in 3 hours. What is its average speed?

diamong [38]

Average speed equal to = ,total distance traveled/total time taken
so it would be = 94+84/8
= 22.25 kmph

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3 years ago

Two loudspeakers are located 4.965 m apart on an outdoor stage. A listener is 12.608 m from one and 18.368 m from the other. Dur

leonid [27]

Answer:

Approximately $30.0\; \rm Hz$ .

Explanation:

Look up the speed of sounds in the air at $25\; ^\circ\rm C$ : $v \approx 346\; \rm m \cdot s^{-1}$ .

Let the frequency of this tune be $f\; \rm Hz$ . The wavelength of the tune would be $\displaystyle \lambda = \frac{v}{f} \approx \frac{346}{f}$ .

The distance between the first speaker and the listener is $12.608\; \rm m$ . How many wavelengths can fit into that distance?

$\displaystyle \frac{12.608}{\lambda} \approx \frac{12.608}{346 / f} = \frac{12.608}{346} \cdot f$ .

Similarly, the distance between the second speaker and the listener is $18.368\; \rm m$ . Number of wavelengths in that distance:

$\displaystyle \frac{18.368}{\lambda} \approx \frac{18.368}{346 / f} = \frac{18.368}{346} \cdot f$ .

Difference between these two numbers:

$\begin{aligned} &\frac{18.368}{346} \cdot f - \frac{12.608}{346} \cdot f = \frac{5.76}{346}\cdot f\end{aligned}$ .

For destructive interference to occur, that difference should be equal to $\displaystyle \frac{1}{2}$ , $\displaystyle 1 + \frac{1}{2}= \frac{3}{2}$ , $\cdots$ , or $\displaystyle \left(k+ \frac{1}{2}\right)$ in general ( $k$ can be any non-negative whole number.)

Let $\begin{aligned} \frac{5.76}{346}\cdot f = k + \frac{1}{2}\end{aligned}$ , and solve for $f$ .

$\begin{aligned} f&= \left.\left(k + \frac{1}{2}\right) \right/\frac{5.76}{346}\\ &= \frac{346}{5.76} \, k + \frac{1}{2} \times \frac{346}{5.76} \\ &\approx 60.1\, k + 30.0 \end{aligned}$ .

The next step is to find the values of $k$ that ensure $20 \le f \le 20\times 10^{3}$ (frequency is between $20\; \rm Hz$ and $20\, \rm kHz$ .) It turns out that $k = 0$ (the smallest $k$ value possible) would be sufficient. In that case, the frequency is approximately $30.0\; \rm Hz$ . Using a larger $k$ would only increase the frequency.

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3 years ago

Ella applies for a car loan, but is not approved for the loan. Ella checks her credit report and finds several unauthorized line

Inessa05 [86]

She could have given it to somebody, or the thief could have stolen it from her.

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3 years ago