3.
What part of your eye is responsible for regulating the amount of light that enters your eye?

Answer:

These planets rotate around the sun in a circular path. Likewise in a heliocentric model it is believed that the sun is at the center of the universe and the planet earth along with all other planet move around it. Thus in both geocentric model and heliocentric model bodies in space move in circular orbits.

Answer:

The bodies in space move in circular orbits

Explanation:

I got it right on my test

Answer:

a)  ρ = 6.25 10⁵ μg / m³, b) ρ  = 1 10⁷ μg / m³

Explanation:

Let's analyze the exercise a little before starting, we must know the amount of pollutant in the box, that the one that enters less the one that degrades and with this value find the density or concentration.

Let's start by finding the volume of air that goes into the box

               V = Lh x

Let's find the distance of air that enters per unit of time, as it goes at constant speed

               x = v₀ t

we substitute

               V₀ = Lh v₀ t

At this same time, a quantity of pollutant is distributed

              Q₀ = r t  

the contaminant that is entering reaches the entire box, therefore the total amount of contaminant is

               Q = Qo t

we substitute

               Q = r t²

the net amount of pollutant that remains is that less enters the one that degraded in the same time, as they ask for the steady state

              [tex]Q_{net}[/tex]= Q - k t

the pollutant concentration is

              ρ = Q_net / V

              V = L L h

              ρ =[tex]\frac{r \ t^2 - k \ t}{ L^2 h}[/tex]

              ρ = [tex](r \frac{ L^2}{v_o^2} - k \frac{L}{v_o} ) \frac{1}{L^2 h}[/tex]

               ρ = [tex]\frac{r}{ v_o h} -\frac{k}{v_o L h}[/tex]

let's reduce the magnitudes to the SI system

           r = 10 kg / s

           L = 100 km = 100 10³ m

           h = 1 km = 1 10³ m

           k = dq / dt = 0.20 1/h ( 1h/3600 s) = 5.5555 10⁻⁵  1/s

           v₀ = 4 m / s

let's calculate

The volume of the box

             V = (100 100 1) 109

             V = 1 10¹³ m³

            ρ = [tex]\frac{10}{ 4^2 \ 1\ 10^3 } - \frac{5.5556 \ 10^{-5}}{ 4 \ 100 \ 10^3 1 \ 10^3}[/tex]

            ρ = [tex]6.25 10^{-4} - 1.389 ^{-13}[/tex]

            ρ = 6.25 10⁻⁴ kg / m³

let's reduce to μg / m³

               ρ = 6.25 10⁻⁻⁴ kg / m³ (10⁹ μg / 1kg)

               ρ = 6.25 10⁵ μg / m³

b) in case the air speed decreases to v₀ = 1 m / s

             ρ= \frac{10}{ 1^2 \  1\  10^3 } - \frac{5.5556 \ 10^{-5}}{ 1 \ 100 \ 10^3  1 \ 10^3}

             ρ = 1 10⁻² - 5.5556 10⁻¹³

             ρ =  1 10⁻² kg / m³

             ρ  = 1 10⁷ μg / m³

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

[tex] v = \frac{I}{nqA} [/tex]

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

[tex] n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3} [/tex]                  

Now, we can find the drift speed:

[tex]v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s[/tex]              

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!      

Answer:

A. 171.24 Ibs

Explanation:

To find the amount of salt in the tank,

Let Q = Amount of salt in the mixture

And let 100 + (3-2)t = 100 + t be the volume of mixture at anytime t.

Rate of gain - Rate of loss = dQ / dt

Concentration of salt = Q / (100+t)

For the linear differential equation,

dQ / dt = 3(2) - 2 [Q/ (100 + t)]

dQ /dt + Q [2 / (100 + t)] = 6

The general solution of the linear differential equation is:

Q (i.f) = ∫ A(t) (i.f) dt + C

Therefore,

i.f = e ^ ∫ P(t) dt

And P(t) = 2 / (100 + t)

i.f = e ^ ∫ 2 / (100 + t)

  = e ^ 2㏑ (100 + t)

     = e ^ ㏑ (100 + t) ^2 = (100 + t) ^2

Q(100 + t) ^ 2 = ∫6 (100 + t) ^2 dt + C

 Q(100 + t) ^2 = 2(100 + t) ^ 3 + C

  When t = 0, Q = 50

Therefore,

50( 100) ^2 = 2(100) ^3 + C

 C = -1.5 * 10 ^6

therefore, when t = 30,

Q (100 + 30) ^2 = 2(100 + 30) ^3 - 1.5 * 10 ^6

 Q (400) ^2 = 2(130) ^3 - 1.5 * 10 ^6

    Q = 171.24 Ibs

The amount of salt in the tank at the end of 30 minutes is 171.24 lbs.

The given parameters:

The linear differential equation of the salt solution is calculated as follows;

[tex]\frac{dx}{dt} = Gain - loss[/tex]

where;

The salt concentration at time t, is calculated as follows;

[tex]\frac{dx}{dt} = 2(3) - 2(\frac{X}{100 + t} )\\\\\frac{dx}{dt} = 6 - 2(\frac{X}{100 + t} )\\\\\frac{dx}{dt} +2(\frac{X}{100 + t} ) =6[/tex]

Apply the general solution of linear differential equation as follows;

[tex]X(f) = \int\limits {At} \, dt \ + C\\\\f = e^{\int\limits {At} \, dt}\\\\ f = e^{\int\limits {\frac{2}{100 + t} } \, dt}\\\\f = e^{2 ln(100 + t)}\\\\f = (100 + t)^2[/tex]

[tex]X(100 + t)^2 = \int\limits {6(100 + t)^2} \, dt \ + \ C\\\\ X(100 + t)^2 = 2(100 + t)^3 + C[/tex]

When t = 0 and X = 50

[tex]50(100 + 0)^2 = 2(100+ 0)^3 + C\\\\C = -1.5 \times 10^6[/tex]

When t = 30 min, the concentration is calculated as;

[tex]X (100 + 30)^2 = 2(100 + 30)^3- 1.5 \times 10^6\\\\X(130)^2 = 2(130)^3 - 1.5\times 10^6\\\\X(130)^2 = 2894000\\\\X = \frac{2894000}{130^2} \\\\X = 171.24 \ lbs[/tex]

Learn more about solution of Linear differential equation here: https://brainly.com/question/5508539

Answer:

1.8 Kg 17.6N

Explanation:

I don't know the explanation hahaha

Answer:

The gravitational force on an object increases as the object’s mass increases.

Explanation:

This is the answer on Edmentum. :)