What is the velocity of the cart in these sections?
a-b
c-d
e-f
f-g

Answer:

The answer is below

Explanation:

Let x represent the number of ounce of dairy based meal and let y represent the number of vegan option in ounce.

Since the diet must contain at least 2400 mg vitamin C, therefore:

50x + 20y ≥ 2400

Since the diet must contain at least 1800 mg Calcium, therefore:

30x + 20y ≥ 1200

Since the diet must contain at least 1200 calories, therefore:

10x + 40y ≥ 1200

Therefore the constraints are:

50x + 20y ≥ 2400

30x + 20y ≥ 1200

10x + 40y ≥ 1200

x > 0, y > 0

The graph was drawn using geogebra online graphing tool, and the solution to the problem is at:

C(30, 45) and D(48, 18)

dairy-based meal costs $0.042 per ounce and the vegan option costs $0.208 per ounce. The cost equation is:

Cost = 0.042x + 0.208y

At C(30, 45);  Cost = 0.042(30) + 0.208(45) = $10.62

At C(48, 18);  Cost = 0.042(48) + 0.208(18) = $5.76

The minimum cost is at (48, 18). That is 48 dairy based meal and 18 vegan

Answer:

Iris

Explanation:

The iris seems to be the illuminated portion of the eyes which really covers the pupil. It controls the amount of light reaching the eye. The lens is indeed a translucent layer of the retina that serves to concentrate light and objects on the lens.

Answer:

I hope this helps.

Explanation:

Answer:

Option 2 both take millions of years to form

Explanation:

Both  coal and ores take millions of years to form.

Ore is a naturally occurring rock or silt that has precious minerals in it that may be extracted, processed, and sold for a profit. These minerals are usually metals. Mining is the process of removing ore from the soil. The valuable metals or minerals are then removed by treating or refining the ore, frequently through smelting.

The concentration of the desired ingredient in an ore is referred to as its grade. To decide if a rock has a high enough grade to be worth mining and is therefore regarded as an ore, the value of the metals or minerals it contains must be evaluated against the expense of extraction.

Typically, oxides, sulphides, silicates, or native metals like copper or gold are the minerals of interest. To separate the valuable components from the waste rock, ore must be treated. Numerous geological processes collectively known as ore genesis are responsible for the formation of ore deposits.

Learn more about ore here:

https://brainly.com/question/4712893

#SPJ2

Answer:

A) F = 21.134 N

B) a = 3180.76 × 10^(24) m/s²

Explanation:

A) We are given;

Mass of alpha particle; m = 4.0026 u

Now, 1u = 1.66 × 10^(-27) kg

Thus; m = 4.0026 × 1.66 × 10^(-27)

Distance apart; r = 6.60 × 10^(−15) m

Charge on the alpha particle is;

q = 2e = 2 × 1.6 × 10^(-19) C

Formula for the force between the two alpha particles is;

F = kq1.q2/r²

k = 8.99 × 10^(9) N.m²/C²

q1 = q2 = 2 × 1.6 × 10^(-19) C

F = 8.99 × 10^(9) × (2 × 1.6 × 10^(-19))²/(6.60 × 10^(−15))²

F = 21.134 N

B) acceleration is given by;

a = F/m

Thus; a = 21.134/(4.0026 × 1.66 × 10^(-27))

a = 3180.76 × 10^(24) m/s²

Answer:

I have strawberry blonde/brown hair blue eyes and a girl lol

Explanation:

Answer:

a) E = 0

b) [tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]

From which we have;

[tex]E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0[/tex]

E = 0/A = 0

E = 0

b) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]

[tex]E \cdot A = \dfrac{+q }{\varepsilon _{0}}[/tex]

[tex]E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}[/tex]

By Gauss theorem, we have;

[tex]E\oint dS = \dfrac{q}{\varepsilon _{0}}[/tex]

Therefore, we get;

[tex]E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}[/tex]

The electrical field outside the spherical shell

[tex]E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }[/tex]

[tex]k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }[/tex]

Therefore, we have;

[tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]

Answer:

A.

Explanation:

Answer: A

Explanation:

Answer:

its An uncharged object.

if its not charged the electrically wont go on it

Answer:

uncharged object

Explanation:

Answer:

yo imma so I dunno find out yourself

Explanation:

dhdhdhdnndsisijjsksskskekekekkekssisisieieieiiwiwiwieiwieidjdjddi?

Answer:

i think 692m/s2 is the correct answer

Answer:

d

Explanation:

Answer:

a)  W = 1.63 10⁻²⁸ J,  b)  W = 1.407 10⁻²⁷ J, c) W = 1.68 10⁻²⁸ J,

d)  W = - 4.93 10⁻²⁸ J

Explanation:

a) In this problem we have an electron at the origin, work is requested to carry another electron from infinity to the point x₂ = 0, y₂ = 2.00m

If we use the law of conservation of energy, work is the change in energy of the system

          W = ΔU = U_∞ -U

the potential energy for point charges is

           U =k [tex]\sum \frac{q_i q_j}{r_{ij} }[/tex]

in this case we only have two particles

           U = k [tex]\frac{q_1q_2}{r_{12} }[/tex]

the distance is

           r₁₂ = [tex]\sqrt{(x_2-x_1)^2 + ( y_2-y_1)^2 }[/tex]

           r₁₂ =[tex]\sqrt{ 0 + ( 2-0)^2}[/tex]Ra 0 + (2-0)

           r₁₂ = √2= 1.4142 m

we substitute

           W = k \sum \frac{q_i q_j}{r_{ij} }

let's calculate

            W = [tex]\frac{ 9 \ 10^9 (1.6 \ 10^{-19})^2 }{1.4142}[/tex] 9 109 1.6 10-19 1.6 10-19 / 1.4142

            W = 1.63 10⁻²⁸ J

b) the two electrons are fixed, what is the work to bring another electron to x₃ = 3.00 m y₃ = 0

in this case we have two fixed electrons

            U = k [tex]( \frac{q_1q_3}{r_{13} } + \frac{q_2q_3}{r_{23} } )[/tex]

in this case all charges are electrons

             q₁ = q₂ = q₃ = q

             W = U = k q² [tex]( \frac{1}{r_{13} } + \frac{1}{r_{23} } )[/tex]

the distances are

            r₁₃ = [tex]\sqrt{(3-0)^2 + 0}[/tex]RA (3.00 -0) 2 + 0

            r₁₃ = 3

            r₂₃ = [tex]\sqrt{ 3^2 + 2^2}[/tex]Ra (3 0) 2 + (2 0) 2

            r₂₃ = √13

            r₂₃ = 3.606 m

let's look for the job

            W = U

let's calculate

            W =[tex]{9 \ 10^3 ( 1.6 10^{-19})^2 }({\frac{1}{3} + \frac{1}{3.606} } )[/tex]

            W = 1.407 10⁻²⁷ J

c) the three electrons are fixed, we bring the four electron to x₄ = 3.00m,

y₄ = 4.00 m

             W = U = k [tex]( \frac{q_1q_4}{r_{14 }} + \frac{q_2q_4}{r_{24} } + \frac{q_3q_4}{r_{34} } )[/tex]

all charges are equal q₁ = q₂ = q₃ = q₄ = q

             W = k q² [tex](\frac{1}{r_{14} } + \frac{1}{r_{24} } + \frac{1}{r_{34} } )[/tex]

let's look for the distances

             r₁₄ = [tex]\sqrt{3^2 +4^2}[/tex]

             r₁₄ = 5 m

             r₂₄ = [tex]\sqrt{3^2 + ( 4-2)^2}[/tex]

             r₂₄ = √13 = 3.606 m

             r₃₄ = [tex]\sqrt{(3-3)^2 + (4-0)^2}[/tex]

            r₃₄ = 4 m

we calculate

           W = 9 10⁹ (1.6 10⁻¹⁹)²  [tex]( \frac{1}{5} + \frac{1}{3.606} + \frac{1}{4} )[/tex]

           W = 1.68 10⁻²⁸ J

d) we take the proton to the location x5 = 1m y5 = 1m

            W = U = k [tex]( \frac{q_1q_5}{r_{15} } + \frac{q_2q_5}{r_{25} } + \frac{q_3q_5}{r_{35} } + \frac{q_4q_5}{r_{45} } )[/tex]

in this case the charges have the same values ​​but charge 5 is positive and the others negative, so the products of the charges give a negative value

            W = - k q² [tex]( \frac{1}{r_{15} } + \frac{1}{r_{25} } + \frac{1}{r_{35} } + \frac{1}{r_{45} } )[/tex]

we look for distances

            r₁₅ = [tex]\sqrt{ 1^2 +1^2}[/tex]Ra (1-0) 2 + (1-0) 2

            r₁₅ = √ 2 = 1.4142 m

            r₂₅ = [tex]\sqrt{ (2-1)^2 +1^2}[/tex]

            r₂₅ = √2 = 1.4142 m

            r₃₅ = [tex]\sqrt{ ( 3-1)^2 +1^2}[/tex]

            r₃₅ = √5 = 2.236 m

            r₄₅ = [tex]\sqrt{ (3-1)^2 + (4-1)^2}[/tex]

            r₄₅ = √13 = 3.606 m

we calculate

           W = - 9 10⁹ (1.6 10⁻¹⁹)² [tex]( \frac{1}{1.4142} +\frac{1}{1.4142} + \frac{1}{2.236} + \frac{1}{3.606} )[/tex]

            W = - 4.93 10⁻²⁸ J

Answer:

a. of liquid at the time bubbles first emerge slowly from the liquid.

Explanation:

Boiling point of liquid happens due to heat energy. This is an exothermic reaction as heat is released in to the environment. The initial boiling vapors slowly move away from the liquid and as the temperature increases the vapors start moving quickly.

Answer:

4.55

Explanation:

The terminal speed of a diver is given by:

[tex]v_t=\sqrt{\frac{2mg}{C\rho A} } \\\\Where\ m=mass\ of \ driver,d=acceleration\ due\ to\ gravity,C=drag\ \\coefficient,A=cross\ sectional\ Area.\\\\Therefore:\\\\A=\frac{2mg}{C \rho v_t^2} \\\\For\ area\ with\ terminal\ speed\ in\ spread\ angle\ position(v_s):\\\\A_s=\frac{2mg}{C \rho v_s^2} \\\\For\ area\ with\ terminal\ speed\ in\ nose\ dive\ position(v_n):\\\\A_n=\frac{2mg}{C \rho v_n^2}\\\\Therefore\ since\ g,m,C,\rho\ are\ constant:\\\\[/tex]

[tex]\frac{A_s}{A_n}= \frac{\frac{2mg}{C \rho v_s^2}}{\frac{2mg}{C \rho v_n^2}}\\\\\frac{A_s}{A_n}= \frac{v_n}{v_s} \\\\v_n=320\ km/h,v_s=150\ km/h\\\\\frac{A_s}{A_n}=\frac{320^2}{150^2} =4.55[/tex]

Answer:

L = 8.35 m

Explanation:

The lenght of a wire L can be calculated using the following expression:

L = R A/ρ  (1)

Where:

R: resistance of the wire

A: Cross section area of the wire

ρ: resistivity of the copper wire.

With this expression we realize that we do not have the area of the cross section, and the resistance of the wire either.

To calculate the area we can use the following expression:

A = πr²    (2)

If the diameter is 0.25 mm, then the radius is half, 0.125 mm. Converting this in meter it will have to be:

0.125 /1000 = 0.000125 m

Replacing we have:

A = π(0.000125)²

A = 4.91x10⁻⁸ m²

The reported resistivity of a copper wire is 1.68x10⁻⁸ Ω.m, so we just need to determine the resistance, which can be found using Ohm's law:

R = V/I  (3)

Replacing (3) into (1) we have:

L = (V * A) / (I * ρ) (4)

So finally, the length of the copper wire will be:

L = (12 * 4.91x10⁻⁸) / (4.2 * 1.68x10⁻⁸)

Hope this helps

Answer:

a. volume of gas:  (decreases)

b. temperature of gas:  (same)

c. internal energy of gas: (same)

d. pressure of gas: (increases)

Explanation:

We have a gas (let's suppose that is ideal) in a piston with a fixed volume V.

Then we put in a reservoir at 0°C (the mixture of water and ice)

remember that the state equation for an ideal gas is:

P*V = n*R*T

and:

U = c*n*R*T

where:

P = pressure

V = volume

n = number of mols

R = constant

c = constant

T = temperature.

Now, we have equilibrium at T = 0°C, then we can assume that T is also a constant.

Then in the equation:

P*V = n*R*T

all the terms in the left side are constants.

P*V = constant

And knowing that:

U = c*n*R*T

then:

n*R*T = U/c

We can replace it in the other equation to get:

P*V = U/c = constant.

Now, the piston is (slowly) moving inwards, then:

a) Volume of the gas: as the piston moves inwards, the volume where the gas can be is smaller, then the volume of the gas decreases.

b) temperature of the gas: we know that the gas is a thermal equilibrium with the mixture (this happens because we are in a slow process) then the temperature of the gas does not change.

c) Internal energy of the gas:

we have:

P*V = n*R*T = constant

and:

P*V = U/c = constant.

Then:

U = c*Constant

This means that the internal energy does not change.

d) Pressure of the gas:

Here we can use the relation:

P*V = constant

then:

P = (constant)/V

Now, if V decreases, the denominator in that equation will be smaller. We know that if we decrease the value of the denominator, the value of the quotient increases.

And the quotient is equal to P.

Then if the volume decreases, we will see that the pressure increases.

Answer:

8

Explanation:

tsijtsiztuztuistizrizturzurz

Answer:

8

Explanation:

Mercury, Venus, earth , Mars, jupiter, saturn , Uranus,Neptune

Answer:

In the collision, the strength of the force exerted on the pod is greater than the strength of the force exerted on the space station, but those forces are exerted in opposite directions.

Explanation:

Answer:

Liquids

Explanation:

Diffusion occurs fastest in liquids.

Answer:

swimming, cycling, jump rope, brisk walking, gardening, jogging

Answer:

1. cell membrane

2. golgi body

3. mitochondrion

4. cytoplasm

5. nucleolus

6. nucleus

Explanation:

The correct answer to this question is Option A; 6.

Why?

In a plant cell, the nucleus surrounds the nucleolous, which would be number 5. Therefore, number 6 would be your correct answer.

~Thank you~

Answer:

The electric potential at the center of the meter stick is 54 KV.

Explanation:

Electric potential (V) is given as:

i.e V = [tex]\frac{kq}{r}[/tex]

Where: k is the Coulomb constant, q is the charge and r is the distance.

Given: q = 3.0 μC = 3.0 x [tex]10^{-6}[/tex] C, r = 0.5 m

So that,

V = [tex]\frac{9*10^{9}*3.0*10^{-6} }{0.5}[/tex]

   = [tex]\frac{2.7*10^{4} }{0.5}[/tex]

V = 54000

  = 54 000 volts

The electric potential at the center of the meter stick is 54 KV.

Answer:

Explanation:

Total heat gain by the block ΔQ = 2 x 12 kJ = 24 kJ .

Gain of potential energy = mgh = 12 x 9.8 x 32 = 3.763 kJ

Decrease in kinetic energy KE = 1/2 x 12 ( 40² - 7² )

= 9.306 kJ

increase in internal energy ΔE = 70 kJ

ΔQ =  ΔE + PE - KE + W , W is work done by the gas

Putting the values

24 = 70 + 3.763 - 9.306 + W

W = - 40.457 kJ .

Since W is negative that means work is done on the system .

Answer:

0.179 N

Explanation:

What is the magnitude of the horizontal part of the contact force on the shaker by the turntable?

The horizontal part of the constant force of the turntable on the shaker is the centripetal force of the turntable on the shaker, F.

So, F = mv²/r where m  = mass of shaker = 79 g = 0.079 kg, v = speed of shaker = 0.7222 m/s and r = radius of turntable = 0.23 m

So, substituting the values of the variables into the equation, we have

F = mv²/r

F = 0.079 kg (0.7222 m/s)²/0.23 m

F = 0.0412 kgm/s² ÷ 0.23 m

F = 0.179 kgm/s²

F = 0.179 N

Answer:

Explanation:

The relation between electric field and potential difference is as follows

E = - dV / dr

That means if dV is positive , E is negative . In other words , if potential increases , E is negative or in opposite direction in which potential increases .

Here the electric potential increases uniformly from east to west , that means electric field is from west to east . Since potential is uniformly increasing that means

dV / dr = constant

E = constant

Electric field is constant .

So the option which is correct is

" points east and does not vary with position " .

Answer:

h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2  \  t_o^2 =0

The correct result is that of a positive height

Explanation:

For this exercise we use the kinematic relations, let's start by finding the time it takes for the sound to reach the man

             v_s = y / t

             t = [tex]\frac{y}{ v_s}[/tex]

this height is y = h

             t =  \frac{h}{ v_s}

the man has a response time of t = t₀, therefore

time to move is

             t' =  t - t₀

the initial height of flower pot is

           y = y₀ + v₀ t' - ½ g  t'²

when it reaches the floor the height is zero y = 0 and as the pot is dropped its initial velocity is zero v₀ = 0

            0 = y₀ +0 - ½ g (t -t₀)²

if the initial height is i = h,

            h = ½ g ([tex]\frac{h}{v_s}[/tex] - t₀)²2

            [tex]\frac{2}{g} h[/tex] = [tex]\frac{h^2}{v_s^2}[/tex] - [tex]\frac{2t_o }{v_s} h[/tex] + t₀²

            [tex]\frac{h^2}{v_s^2} - ( \frac{2t_o}{v_s} + \frac{2}{g} ) h + t_o^2 = 0[/tex]h2 / vs2 - (2nd / vs + 2 / g) h + to2 - = 0

            [tex]h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2 \ t_o^2 =0[/tex]

To know the height, you must solve the second degree equation, it is much easier with numerical values.

             The correct result is that of a positive height

Answer:

Explanation:

FBEDAC

Answer:

1g/cm3

Explanation:

volume of block is 3 cubed which is 27 cm3

we know density is m/v so d= 27g/27cm3

which is 1g/cm3

if my answer helps please mark as brainliest