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3 years ago

If the water cycle is effective why are some areas in the world always short on water?

1 answer:

6 0

They are short on water because the environment is not the same everywhere some are more dry so not water is there at all or just to hot like a dessert

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I’ll give u BRAINLIEST PLEASE!! HURRY

zalisa [80]

Answer:

Explanation:

6 0

2 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

3 years ago

A race car moving along a circular track has a centripetal acceleration of 15.4 m/s? If the car has

Helen [10]

Answer:

r = 58.44 [m]

Explanation:

To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.

a = v²/r

where:

a = centripetal acceleration = 15.4 [m/s²]

v = tangential speed = 30 [m/s]

r = radius or distance [m]

r = v²/a

r = 30²/15.4

r = 58.44 [m]

3 0

2 years ago

What is a real life example of something impermeable? (please help now!!) (due tomorrow!!)

Nutka1998 [239]

Impermeable is a substance that is water proofed e.g glass, aluminium e.t.c <span />

6 0

3 years ago

A 55.4 g sample of water at 99.61 °C is placed in a constant pressure calorimeter. Then, 23.4 g of zinc metal at 21.6 °C is adde

Zolol [24]

Answer:

The specific heat capacity of the zinc metal measured in this experiment is 0.427 J/g.°C

Explanation:

From the experimental data, the water loses heat because its initial temperature is greater than the final temperature of the mixture. On the other hand, the zinc metal gains heat because its initial temperature is less than the final temperature of the mixture

Heat loss by water = Heat gain by zinc metal

m1C1(T1 - T3) = m2C2(T3 - T2)

m1 is mass of water = 55.4 g

C1 is specific heat capacity of water = 4.2 J/g.°C

m2 is mass of zinc metal = 23.4 g

C2 is specific heat capacity of zinc metal

T1 is the initial temperature of water = 99.61 °C

T2 is the initial temperature of zinc metal = 21.6 °C

T3 is the final temperature of the mixture = 96.4 °C

55.4×4.2(99.61 - 96.4) = 23.4×C2(96.4 - 21.6)

746.9028 = 1750.32C2

C2 = 746.9028/1750.32 = 0.427 J/g.°C

3 0

3 years ago