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3 years ago

14

When air cools, it holds _____ water than when it is warmer.

2 answers:

Doss [256]3 years ago

8 0

The answer is A. less. Hope it helps

Blizzard [7]3 years ago

6 0

It holds A. Less water than when it is warmer.

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A microcomputer that is smaller, lighter, and less powerful than a notebook, and that has a touch sensitive screen, is called a

yKpoI14uk [10]

<span>A microcomputer that is smaller, lighter, and less powerful than a notebook, and that has a touch sensitive screen, is called a tablet. Tablets are used similarly to computers in the way that information can be stored, viewed and edited on them.</span>

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3 years ago

What is the period of a wave traveling 5 m/s if its wavelength is 20 m/s

Ilia_Sergeevich [38]

Speed of wave is given as

$v = 5 m/s$

Wavelength of the wave is given as

$\lambda = 20 m$

now from the formula of wave time period we can say

$speed = \frac{wavelength}{time period}$

$5 = \frac{20}{T}$

$T = \frac{20}{5}$

$T = 4 s$

so it will have time period of T = 4 s

7 0

3 years ago

Each second, 1250 m3 of water passes over a waterfall 150 m high. Three-fourths of the kinetic energy gained by the water in fal

mote1985 [20]

Answer:

The generator produces electrical energy at a rate of 1378125000 J per second.

Explanation:

volume of water falling each second is 1250 $m^{3}$

height through which it falls, h is 150 m

mass of 1 $m^{3}$ of water is 1000 kg

⇒mass of 1250 $m^{3}$ of water, m = 1250×1000 = 1250000 kg

acceleration due to gravity, g = 9.8 $\frac{m}{sec^{2} }$

in falling through 150 m in each second, by Work-Energy Theorem:

Kinetic Energy(KE) gained by it = Potential Energy(PE) lost by it

⇒KE = mgh

= 1250000×9.8×150 J

= 1837500000 J

Electrical Energy = $\frac{3}{4}$ (KE)

= $\frac{3}{4}$ ×1837500000

= <u>1378125000 J per second</u>

8 0

3 years ago

A charging RC circuit controls the intermittent windshield wipers in a car. The emf is 12.0 V. The wipers are triggered when the

lilavasa [31]

Answer:

$R=803k\Omega$

Explanation:

We have the following information,

$V_0 = 12V\\V=10V\\c= 1.25*10^{-6}F\\t=1.8s$

We apply the equation for capacitor charging the voltage across it,

$V=V_0 (1-e^{-t/x})\\e^{-t/x}=1-(\frac{V}{V_0})\\-\frac{t}{Rc}=ln(\frac{V}{V_0})\\R=-\frac{t}{ln(\frac{V}{V_0})*c}$

Replacing values,

$R=-\frac{1.8}{ln(10/12)*1.25*10^{-6}}$

$R=803k\Omega$

3 0

3 years ago

How does electricity flow?

Irina-Kira [14]

Answer:C

Explanation:

4 0

3 years ago