B) Smartphone, iPad, and tablet are also a kind of computer.

What type of conversion is taking place when natural gas is used to heat water?

frutty [35]

The type of conversion that is taking place when natural gas is

burned to heat water is referred to as chemical conversion. In this

conversion thermal energy is produced.

6 0

3 years ago

HELP PLZZZZZ NOW!!!

Aneli [31]

Answer:Climatic processes affect the dynamics of Earth's ice sheets and glaciers, and along ... by abrupt events and by continuous reshaping of Earth's surface from surface ... Forecasting natural disasters, including the timing and size of earthquakes, the . Last, human activity has a profound impact on water resources, landscape

Explanation:

step by step

3 0

3 years ago

Read 2 more answers

What would MOST LIKELY happen if the amount of incoming solar energy decreased without a change in the amount of reflection or o

777dan777 [17]

<h2><u><em>Well, you see, that depends. </em></u></h2><h2><u><em>The firsy thing we have to tak intp account is the angle at witch the sun's rays hit the earth, and that fact can make all the difference, seeing as it does discriminate against seasons. It's more likely that i the winter, a more drastic effect would talk.</em></u></h2><h2 /><h2 /><h2 /><h2>oωo</h2>

8 0

3 years ago

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A commuter airplane starts from an airport and takes theroute. The plane first flies to city A, located 175 km away in a directi

bearhunter [10]

Answer:

245.45km in a direction 21.45° west of north from city A

Explanation:

Let's place the origin of a coordinate system at city A.

The final position of the airplane is given by:

rf = ra + rb + rc where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:

rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km

rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km

The module of this position is:

$rf = \sqrt{rfX^2+rfY^2} = 245.45km$

And the angle measure from the y-axis is:

$\alpha =atan(rfX/rfY) = 21.45\°$

So the answer is 245.45km in a direction 21.45° west of north from city A

6 0

3 years ago

You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int

uysha [10]

In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
$Q(t) = Q_0 (1-e^{-t/\tau})$
where t is the time, $Q_0 = CV$ is the maximum charge on the capacitor at voltage V, and $\tau = RC$ is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using $R=39.1 \Omega$ and $C= 65.7 mF=65.7\cdot 10^{-3}F$ , the time constant of the circuit is:
$\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s$

2) To find the charge on the capacitor at time $t=1.95 \tau$ , we must find before the maximum charge on the capacitor, which is
$Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C$
And then, the charge at time $t=1.95 \tau$ is equal to
$Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C$

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be $Q_0 = 0.59 C$ . We can see this also from the previous formule, by using $t=\infty$ :
$Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C$

4 0

3 years ago