Ask question

Ask question

All categories

3 years ago

12

What name is used for scientists who study the weather in an area over a long time, such as 200 years? Hydrologists weather repo

rters climatologists balloon specialists

1 answer:

andrew11 [14]3 years ago

4 0

Answer:

I think it is meteorologist because they also study of the atmosphere, atmospheric phenomena, and atmospheric effects on our weather.

You might be interested in

How do galaxies vary?

givi [52]

The universe has trillions of galaxies and counting. Astronomers give names to each galaxy base on its shape (e.i, Sombrero galaxy, Milkyway Galazy, ect,.).

Also, the size of the galaxy is taken into account and their color.

4 0

3 years ago

Under certain circumstances, potassium ions (K+) in a cell will move across the cell membrane from the inside to the outside. Th

choli [55]

Answer:

$1.368\times 10^{-20}\ J$

Explanation:

q = Charge in the potassium ion = $19e-18e$

e = Charge of electron = $1.6\times 10^{-19}\ C$

$V_2-V_1$ = Change in potential = $0-(-85.5\times 10^{-3})$

Change in electric potential is given by

$E=q(V_2-V_1)\\\Rightarrow E=(19e-18e)(0-(-85.5\times 10^{-3})\\\Rightarrow E=1.6\times 10^{-19}\times 85.5\times 10^{-3}\\\Rightarrow E=1.368\times 10^{-20}\ J$

The energy is $1.368\times 10^{-20}\ J$

3 0

3 years ago

Natasha_Volkova [10]

Answer B would be the correct answer for the questions

5 0

4 years ago

Suppose that the hatch on the side of a Mars lander is built and tested on Earth so that the internal pressure just balances the

Kaylis [27]

Answer:

The value is $F_{net} = 4444 lb$

The force will be outward

Explanation:

From the question we are told that

The diameter of the disk is $d = 50.0 \ cm = \frac{50}{100} = 0.5 \ m$

The external pressure on Mars is $P = 650 \ N/m^2$

From the question we are told that

Internal pressure = External pressure

Generally the external Force on earth is

$F_E = P_{atm} * A$

Here $P_{atm}$ is the atmospheric pressure with value $P_{atm} = 1.013*10^{5}\ Pa$

So

$F_E = 1.013 *10^{5} * \pi * \frac{d^2}{4}$

=> $F_E = 1.013 *10^{5} *3.142 * \frac{0.50 ^2}{4}$

=> $F_E = 19893 \ N$

Generally the external Force on Mars is

$F= P * A$

$F = 650 * \pi * \frac{d^2}{4}$

=> $F = 650 *3.142 * \frac{0.5^2}{4}$

=> $F = 127.6 \ N$

Net force is mathematically represented as

$F_{net} = F_E -F$

=> $F_{net} = 19893 -127.6$

=> $F_{net} = 19765.6 \ N$ converting to pounds

$F_{net} = \frac{19765.6}{4.448}$

=> $F_{net} = 4444 lb$

Given that that the value is positive then the force will be outward

7 0

3 years ago

A spherical drop of water carrying a charge of 43 pC has a potential of 540 V at its surface (with V = 0 at infinity). (a) What

almond37 [142]

Explanation:

Given that,

Charge on a spherical drop of water is 43 pC

The potential at its surface is 540 V

(a) The electric potential on the surface is given by :

$V=\dfrac{kq}{r}$

r is the radius of the drop

$r=\dfrac{kq}{V}\\\\r=\dfrac{9\times 10^9\times 43\times 10^{-12}}{540}\\\\r=7.17\times 10^{-4}\ m$

(b) Let R is the radius of the spherical drop, when two such drops of the same charge and radius combine to form a single spherical drop. ATQ,

$\dfrac{4}{3}\pi r^3+\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\pi R^3\\\\2r^3=R^3\\\\R=2^{1/3} r$

Now the charge on the new drop is 2q. New potential is given by :

$V=\dfrac{9\times 10^9\times 43\times 10^{-12}\times 2}{2^{1/3}\times 7.17\times 10^{-4}}\\\\V=856.79\ V$

Hence, the radius of the drop is $7.17\times 10^{-4}\ m$ and the potential at the surface of the new drop is 856.79 V.

5 0

3 years ago