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4 years ago

What forms when moisture evaporates into the air and then cools

Physics

1 answer:

musickatia [10]4 years ago

3 0

Answer:

Clouds

Explanation:

In an area, where the temperature is relatively high, then the warm air rises upward due to its less density. As these air rises upward, the air cools due to the decreasing temperature with the increasing height. There comes such a point where the air is not capable of holding any more water vapor, and it cools below the dew point. This process eventually leads to the formation of clouds in the sky, and with further cooling, the tiny droplets are formed, resulting in the occurrence of rain.

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3 years ago

An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference

kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

$K=qV$ (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

$K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J$

The kinetic energy of the particle is also:

$K=\frac{1}{2}mv^2$ (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}$

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

$|F_B|=qvBsin(\theta)$

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

$|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N$

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

$r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm$

the radius of the trajectory of the electron is 10.1 cm

6 0

4 years ago

The electric field strength is 5.50×10^4 N/C inside a parallel-plate capacitor with a 2.50 mm spacing. A proton is released from

valentina_108 [34]

Answer:

$v=1.6\times10^{5}m/s$

Explanation:

The equation that relates electric field strength and force that a charge experiments by it is F=qE.

Newton's 2nd Law states F=ma, which for our case will mean:

$a=\frac{F}{m}=\frac{qE}{m}$

We know from accelerated motion that $v^2=v_0^2+2ad$

In our case the proton is released from rest, so $v_0=0m/s$ and we get $v=\sqrt{2ad}$

Substituting, we get our final velocity equation:

$v=\sqrt{\frac{2qEd}{m}}$

For the <em>proton </em>we know that $q=1.6\times10^{-19}C$ and $m=1.67\times10^{-27}kg$ . Writing in S.I. d=0.0025m, we obtain:

$v=\sqrt{\frac{2(1.6\times10^{-19}C)(5.5\times10^{4}N/C)(0.0025m)}{(1.67\times10^{-27}kg)}}=162318.5m/s=1.6\times10^{5}m/s$

6 0

3 years ago