!!! HELP!!! HELL GELP HELP HELP

LiRa [457]

Answer:

When air rises in the atmosphere it gets cooler and is under less pressure. When air cools, it's not able to hold all of the water vapor it once was. Air also can't hold as much water when air pressure drops. The vapor becomes small water droplets or ice crystals and a cloud is formed.

Explanation:

hope this helps.

8 0

3 years ago

A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit

likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density $\mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m$

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 120=753.6rad/sec$

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by $P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt$

From the relation we can see that power $P\ \propto\ A^2$

So if amplitude is halved then power will be $\frac{1}{4}$ times

So power will be equal to $\frac{0.2023}{2}=0.0505watt$

4 0

3 years ago

Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light.

pogonyaev

Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is $D = 0.59 \ m$

Explanation:

From the question we are told that

The best resolution is $\theta = 0.3 \ arcsecond$

The wavelength is $\lambda = 700 \ nm = 700 *10^{-9 } \ m$

Generally the

1 arcminute = > 60 arcseconds

=> x arcminute => 0.3 arcsecond

So

$x = \frac{0.3}{60 }$

=> $x = 0.005 \ arcminutes$

Now

60 arcminutes => 1 degree

0.005 arcminutes = > z degrees

=> $z = \frac{0.005}{60 }$

=> $z = 8.333 *10^{-5} \ degree$

Converting to radian

$\theta = z = 8.333 *10^{-5} * 0.01745 = 1.454 *10^{-6} \ radian$

Generally the resolution is mathematically represented as

$\theta = \frac{1.22 * \lambda }{ D}$

=> $D = \frac{1.22 * \lambda }{\theta }$

=> $D = \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }$

=> $D = 0.59 \ m$

4 0

3 years ago

When cleaning a storage battery you can use a solution of water and ammonia or solution of water and

Anettt [7]

Water and baking soda can be used, too.

7 0

4 years ago

If you touch the two terminals of a power supply with your two fingertips on opposite hands, the potential difference will produ

LiRa [457]

Answer:

Yes the body will receive a dangerous shock in both cases.

Explanation:

Different parts of the body has different resistance. skin has the high resistance as compared to other organs of the body.

Dry skin has high resistance than wet skin this is because water is relatively good conductor of electricity, it adds parallel path to the current flow and hence reduces skin resistance.

Dry hands body has approximately 500 kΩ resistance and if 120 V electricity supply current received will be:

I = V/R= 120/ 500*10^3

I= 0.24 mA

Even the current seems is much lower than the safe zone but this is the case in case of DC voltage in case of AC voltage the body will receive a shock this is because the skin pass more current when the voltage is changing i.e. AC.

Similarly for wet hands body resistance is 1 kΩ. so the current through the body seems to be:

I = 120 / 1000

I = 12 mA

The current is higher than safe zone so the body will receive a dangerous shock.

7 0

4 years ago