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Oksana_A [137]
3 years ago
14

What is population education?​

Physics
2 answers:
Naily [24]3 years ago
6 0

Answer:

Population is derived from French word "populous" meaning 'people' so population is study of people living together and education is life long knowledge.

Explanation:

sana makatulong

Leviafan [203]3 years ago
3 0

Answer:

Is all about people – how many of us there are, how we shape the world around us, and how we interact with each other.

Explanation:

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A vertically polarized beam of light of intensity 100 W/m2 passes through two ideal polarizers. The transmission axis of the fir
TEA [102]

To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:

I=I_0 cos^2\theta

Where,

I_ {0} indicates the intensity of the light before passing through the polarizer,

I is the resulting intensity, and

\theta indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

Since we have two objects the law would be,

I=I_0cos^2\theta_1*cos^2(\theta_2-\theta_1)

Replacing the values,

I=100*cos^2(20)*cos^2(40-20)

I=100*cos^4(20)

I=77.91W/m^2

Therefore the intesity of the light after it has passes through both polarizers is 77.91W/m^2

7 0
3 years ago
1. The mass defect of iron-56 is 0.52875 amu. What is the energy equivalent of this mass?
Delvig [45]

Answer:

Hello! Your answer is BELOW

Explanation:

1.About 91.754% of all iron is iron-56. Of all nuclides, iron-56 has the lowest mass per nucleon. With 8.8 MeV binding energy per nucleon, iron-56 is one of the most tightly bound nuclei.

2.The atomic weight of lead is quite variable in nature because the three heaviest isotopes are the stable end-products of the radioactive decay of uranium (238U to 206Pb and 235U to 207Pb) and thorium (232Th to 208Pb).

3.Mass defect for uranium-238 is 3.983 × 10-25 kg.

4.Energy and Mass Are Relative

The equation E = mc^2 states that the amount of energy possessed by an object is equal to its mass multiplied by the square of the speed of light.

Hope I helped! Ask me anything if you have any questions. Brainiest plz!♥ Hope you make a 100%. Have a nice morning! -Amelia♥

7 0
3 years ago
The bottom of cooking utensils are often kept black. Give reason​
stepan [7]

Answer:

absorb more heat and get heated quickly

Explanation:

7 0
3 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
2 years ago
A racing car travels on a circular track of radius 158 m, moving with a constant linear speed of 19.1 m/s. Find its angular spee
SOVA2 [1]

Answer:

\omega=0.12\frac{rad}{s}

Explanation:

In a uniform circular motion, since a complete revolution represents 2π radians, the angular velocity, which is defined as the angle rotated by a unit of time, is given by:

\omega=\frac{2\pi}{T}(1)

Here T is the period, that is, the time taken to complete onee revolution:

T=\frac{2\pi r}{v}(2)

Replacing (2) in (1):

\omega=\frac{2\pi}{\frac{2\pi r}{v}}=\frac{v}{r}\\\omega=\frac{19.1\frac{m}{s}}{158m}\\\omega=0.12\frac{rad}{s}

3 0
3 years ago
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