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4 years ago

1 example of Deionization

Physics

1 answer:

xxTIMURxx [149]4 years ago

7 0

Answer:

For example, the rain water does not have any salinity, except the small quantity due to the atmosphere, while the sea water has a very high salinity.

hope this helps you friend

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The average force of a baseball is 18.9 N . It’s mass is 0.145kg fine the acceleration in m/s^2

Artyom0805 [142]

Answer:

the acceleration is 130.3m/s²

Explanation:

Given data

Force F= 18.9N

Mass of ball m= 0.145kg

Acceleration a=?

Applying the Newton's second law of motion

"The rate of change of momentum of a body is proportional to the external force".

F=ma

a= F/m

a= 18.9/0.142

a= 130.3m/s²

3 0

3 years ago

Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t

n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

$y=\frac{\lambda (m+\frac{1}{2})D}{d}$

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light used

In this problem we have:

$\lambda=683 nm = 683\cdot 10^{-9} m$ is the wavelength of the light

$d=1.1 mm = 0.0011 m$ is the width of the slit

m = 13 is the order of the minimum

$y=8.57 cm = 0.0857 m$ is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

$D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m$

6 0

3 years ago

Inside a television picture tube there is a build-up of electrons (charge of 1.602 × 10^–19 C) with an average spacing of 38.0 n

Brut [27]

The magnitude of electric field is produced by the electrons at a certain distance.

E = kQ/r²

where:
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9 N. m2 / C2

Given are the following:
Q = 1.602 × 10^–19 C
r = 38 x 10^-9 m

Substitue the given:
E = $\frac{( 1.602 x 10^{-19} )( 9.0x10_{9} )}{(38x10^{-9}) ^{2} }$

E = 998.476 kN/C

8 0

3 years ago

Read 2 more answers

A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the lo

Zepler [3.9K]

Answer:

0.5849Weber

Explanation:

The formula for calculating the magnetic flus is expressed as:

$\phi = BAcos \theta$

Given

The magnitude of the magnetic field B = 3.35T

Area of the loop = πr² = 3.14(0.24)² = 0.180864m²

angle of the wire loop θ = 15.1°

Substitute the given values into the formula:

$\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb$

Hence the magnetic flux Φ through the loop is 0.5849Weber

5 0

3 years ago

For the following types of electromagnetic radiation, how do the wavelength, frequency, and photon energy change as one goes fro

KIM [24]

Answer:

Wavelength, frequency and the photon energy changes as the one goes across the ranges of the electro-magnetic radiations.

Explanation:

Electro-magnetic radiations may be defined as the form of energy that is radiated or given by the electro-magnetic radiations. The visible light that we can see is the one of the electro-magnetic radiations. Other forms are the radio waves, gamma waves, UV rays, infrared radiations, etc.

The wavelength of the radiations decreases as we go from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.

The frequency of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.

The photon energy of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.

5 0

3 years ago