A wave has a wavelength of 10 mm and a frequency of 14 hertz. What is its speed?

A gas-turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 and 1600 kPa. The working flui

anzhelika [568]

Answer:

a) 6498.84 kW

b) 0.51

c) 0.379

Explanation:

See the attached picture below for the solution

7 0

3 years ago

Calculate the location xcm of the center of mass of the Earth-Moon system. Use a coordinate system in which the center of the Ea

Jet001 [13]

Answer:

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

Explanation:

Let suppose that planet and satellite can be treated as particles. The masses of Earth and Moon ( $m_{E}$ , $m_{M}$ ) are $5.972\times 10^{24}\,kg$ and $7.349\times 10^{22}\,kg$ , respectively. The distance between centers is 384,403 kilometers. The location of the center of mass can be found by using weighted averages:

$\bar x = \frac{x_{E}\cdot m_{E}+x_{M}\cdot m_{M}}{m_{E}+m_{M}}$

If $x_{E} = 0\,km$ and $x_{M} = 384,403\,km$ , then:

$\bar x = \frac{(0\,km)\cdot (5.972\times 10^{24}\,kg)+(384,403\,km)\cdot (7.349\times 10^{22}\,kg)}{5.972\times 10^{24}\,kg+7.349\times 10^{22}\,kg}$

$\bar x = 4.673\,km$

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

8 0

2 years ago

the half-life of carbon-14 is 5,730 years. After 11,460 year, how much of original carbon-14 remains?

Inessa [10]

Via the half-life equation:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}$

Where the time elapse is 11,460 year and the half-life is 5,730 years.

$A_{final}=A_{initial}(\frac{1}{2})^\frac{11460}{5730} \\\\A_{final}=A_{initial}\frac{1}{4} \\\\A_{final}=\frac{1}{4}A_{initial}$

Therefore after 11,460 years the amount of carbon-14 is one fourth (1/4) of the original amount.

6 0

2 years ago

A man goes for a walk, starting from the origin of an xyz coordinate system, with the xy plane horizontal and the x axis eastwar

tekilochka [14]

<h2>a) Displacement of penny = 1300 i + 2400 j - 640 k</h2><h2>b) Magnitude of his displacement = 2729.47 m</h2>

Explanation:

a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.

1300 m east = 1300 i

2400 m north = 2400 j

Drops the penny from a cliff 640 m high = -640 k

Displacement of penny = 1300 i + 2400 j - 640 k

b) Displacement of man for return trip = -1300 i - 2400 j

$\texttt{Magnitude = }\sqrt{(-1300)^2+(-2400)^2}=2729.47m$

Magnitude of his displacement = 2729.47 m

3 0

2 years ago

Read 2 more answers

Why is the curve between 1950 and 1980 relatively flat and centered around zero degrees difference from the baseline? (Hint: how

zimovet [89]

Look at the title of the graph, in small print under it.

Each point is "compared to 1950-1980 baseline". So the set of data for those years is being compared to itself. No wonder it matches up pretty close !

3 0

3 years ago