Which of the following is often found in individuals who are active and eating a healthy diet?

The question is incomplete. Here is the complete question.

Calculate the maximum deceleration  of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.

Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²

Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:

[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]

The [tex]W_x[/tex] is a x-component of force due to gravity (W) and, in this case, is given by: [tex]W_x[/tex] = W.sin(14)

W is described as: W = m.g

Force due to friction ([tex]f_s[/tex]) is given by: [tex]f_s[/tex] = μs.N

N is the normal force and, in the system, is equivalent of [tex]W_y[/tex], so:

[tex]W_y[/tex] = m.g.cos(14)

Therefore, the formula will be:

[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]

m.a = - (m.g.sin14) - (μs.mg.cos14)

a = - g (sin14 + μscos 14)

a) For dry concrete, μs = 1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 1.cos14)

a = - 11.05 m/s²

b) For wet concrete, μs = 0.7:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin 14 + 0.7.cos14)

a = - 10.64 m/s²

c) For ice, μs = 0.1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 0.1cos14)

a = - 9.84 m/s²

Answer:

1A,2D,3B

Explanation:

hope this helps

Answer:

SO THAT

Answer:

since the bed is a cuboid, we find the volume by L×W×H

4.50 × 2.50 × 1.50 = 16.875m³

HOPE THIS HELPS

Answer:

v = 7.67 m/s

Explanation:

The equation for apparent weight in the situation of weightlessness is given as:

Apparent Weight = m(g - a)

where,

Apparent Weight = 360 N

m = mass passenger = 61.2 kg

a = acceleration of roller coaster

g = acceleration due to gravity = 9.8 m/s²

Therefore,

360 N = (61.2 kg)(9.8 m/s² - a)

9.8 m/s² - a = 360 N/61.2 kg

a = 9.8 m/s² - 5.88 m/s²

a = 3.92 m/s²

Since, this acceleration is due to the change in direction of velocity on a circular path. Therefore, it can b represented by centripetal acceleration and its formula is given as:

a = v²/r

where,

a = centripetal acceleration = 3.92 m/s²

v = speed of roller coaster = ?

r = radius of circular rise = 15 m

Therefore,

3.92 m/s² = v²/15 m

v² = (3.92 m.s²)(15 m)

v = √(58.8 m²/s²)

v = 7.67 m/s

Answer:  arrows

Explanation:

A vector quantity is usually represented by an arrow, with the direction of the vector being the direction in which the arrow points and the length of the arrow representing the vector's magnitude.

What is the vector quantity unit?

The meter is the only fundamental SI unit that is a vector. The rest are all scalars. Derived quantities might be scalar or vector, but all vector quantities require meters as part of their definition and measurement.

The term "vector quantities" refers to physical quantities whose magnitude and direction are well specified.

Arrows are used to depict vectors. An arrow has a direction and a magnitude (how long it is) (the direction in which it points).

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Answer:

Speed v = 2.04 m/s

the speed at which gasoline moves through the fuel line is 2.04 m/s

Explanation:

Given;

Mass transfer rate m = 5.37x10^-2 kg/s.

Density d = 739 kg/m3

radius of pipe r = 3.37x10^-3 m

We know that;

Density = mass/volume

Volume = mass/density

Volumetric flow rate V = mass transfer rate/density

V = m/d

V = 5.37x10^-2 kg/s ÷ 739 kg/m3

V = 0.00007266576454 m^3/s

V = 7.267 × 10^-5 m^3/s

V = cross sectional area × speed

V = Av

Area A = πr^2

V = πr^2 × v

v = V/πr^2

Substituting the given values;

v = 7.267 × 10^-5 m^3/s/(π×(3.37x10^-3 m)^2))

v = 0.203678639672 × 10 m/s

v = 2.04 m/s

the speed at which gasoline moves through the fuel line is 2.04 m/s

Answer:

λ1 = 0.0129m = 1.29cm

λ2 = 0.00923m = 0.92 cm

Explanation:

To find the distance between the first order bright fringe and the central peak, can be calculated by using the following formula:

[tex]y_m=\frac{m\lambda D}{d}[/tex]    (1)

m: order of the bright fringe = 1

λ: wavelength of the light = 660 nm, 470 nm

D: distance from the screen = 5.50 m

d: distance between slits = 0.280mm = 0.280 *10^⁻3 m

ym: height of the m-th fringe

You replace the values of the variables in the equation (1) for each wavelength:

For λ = 660 nm = 660*10^-9 m

[tex]y_1=\frac{(1)(660*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.0129m=1.29cm[/tex]

For λ = 470 nm = 470*10^-9 m

[tex]y_1=\frac{(1)(470*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.00923m=0.92cm[/tex]

Answer:

The lifetime of the particle is  [tex]\Delta t = 2.6*10^{-25} \ s[/tex]

Explanation:

From the question we are told that

    The average rest energy is [tex]E = 91.19 \ GeV = 91.19GeV * \frac{1.60 *10^{-10} J }{1GeV} = 1.46 *10^{-8}J[/tex]

    The intrinsic width is  [tex]\Delta E =2.5eV = 2.5GeV * \frac{1.60 *10^{-10}J }{1GeV} = 4*10^{-10} J[/tex]

The lifetime is mathematically represented as

     [tex]\Delta t = \frac{h}{\Delta E}[/tex]

Where h is the Planck's constant with a value of  [tex]1.055*10^{-34} \ J\cdot s[/tex]  

substituting values

    [tex]\Delta t = \frac{1.055*10^{-34}}{4 *10^{-10}}[/tex]

     [tex]\Delta t = 2.6*10^{-25} \ s[/tex]

Complete question is;

Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery with power rating, 12 V, 50 Ampere-minutes.

Answer:

Amount of water required to charge the battery = 7.35 m³

Explanation:

The formula for Potential energy of the water at that height = mgh

Where;

m = mass of the water

g = acceleration due to gravity = 9.8 m/s²

h = height of water = 50 cm = 0.5 m

We know that in density, m = ρV

Where;

ρ = density of water = 1000 kg/m³

V = volume of water

So, potential energy is now given as;

Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J

Now, formula for energy of the battery is given as;

E = qV

We are given;

q = 50 A.min = 50 × 60 = 3,000 C

V = 12 V

Thus;

qV = 3,000 × 12 = 36,000 J

E = 36,000 J

At a 100% conversion rate, the energy of the water totally powers the battery.

Thus;

(4900V) = (36,000)

4900V = 36,000

V = 36,000/4900

V = 7.35 m³

Answer:

a)  v₀ = - 164.62 m / s , c) y = 122.5 m

Explanation:

We can solve this exercise using the free fall kinematic relations.

We put our reference system on the floor, so the height of the skyscraper is y₀ = 180m and the floor level is y = 0 m

For the boy

         y = y₀ + v₀ t - ½ g t²

with free fall its initial speed is zero

        y = ½ g t2

For superman

        y = y₀ + v₀ (t-5) - ½ g (t-5)²

how superman grabs the lot just before hitting the ground

we look for the time it takes the boy down

         t = √ (2 y₀ / g)

         t = √ (2 180 / 9,8)

         t = 6.06 s

in the equation for superman, we clear the volume and calculate

         v₀ (t-5) = -y₀ + ½ g (t-5)²

         v₀ (6.06 -5) = -180 + ½ 9.8 (6.06 -5)²

         v₀ 1.06 = -174.49

         v₀ = - 174.49 / 1.06

         v₀ = - 164.62 m / s

the negative sign indicates that the initial speed is down

b) to graph the position of the two we use the table

  t (s)      Y_boy (m)   Y_superman (m)

    0             180                 180

   1              175.1               180

   5              57.5              180

   6                3.6                10.18

see attachment for the two curves

c) calculate the height that falls a lot in the 5 seconds (t = 5)

           y = -1/2 g t²

           y = ½ 9.8 5²

           y = 122.5 m

for this height superman has not yet left the skyscraper, so the boy hits the ground

Answer:

normal force = 10 N

Explanation:

Given data

frictional force = 0.400 N

coefficient of kinetic friction = 0.04

Solution

we get here normal force that is express as

normal force = [tex]\frac{Frictional\ force}{coefficient\ of\ friction}[/tex]        ............1

put here value and we will get value

normal force = [tex]\frac{0.400}{0.04}[/tex]  

solve it we get

normal force = 10 N

Answer: it's weight decreases

Explanation:

Answer: The angular velocity of the wheel at the beginning of the 4.00 s interval is 3.8 rad/s

Explanation: Please see the attachment below

The angular velocity of the wheel at the beginning of the 4.0 s time is 3.8 rad/s.

The given parameters:

The angular velocity of the wheel at the beginning of the 4.0 s time is calculated as follows;

[tex]\theta = \omega_i t + \frac{1}{2} \alpha t^2[/tex]

where;

[tex]\omega_i[/tex] is the initial angular velocity

[tex]30 = \omega_i (4) \ + \frac{1}{2}(1.85)(4)^2\\\\30 = 4\omega _i + 14.8\\\\4\omega _i = 30 - 14.8\\\\ 4\omega _i = 15.2\\\\\omega _i = \frac{15.2}{4} \\\\\omega _i = 3.8 \ rad/s[/tex]

Thus, the angular velocity of the wheel at the beginning of the 4.0 s time is 3.8 rad/s.

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Answer:

A. Ein = 8.05*10^-4 V/m

B. Clockwise sense

Explanation:

A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:

[tex]\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}[/tex]            (1)

Ein: induced electric field

ds: differential of a path of the ring

ФB: magnetic flux in the ring

The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):

[tex]\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}[/tex]   (2)

dB/dt = -0.280T/s     (it is decreasing)

A: area of the ring = π(r/2)^2= (π/4) r^2

r: radius of the ring = 4.60/2 = 2.30 cm

Then, you replace the values of all variables in the equation (2):

[tex]E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}[/tex]

hence, the induced electric field is 8.05*10^-4 V/m

B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.

Answer:

A) 4D

Explanation:

The distance traveled by the cars before coming to rest can be determined by 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,

s = distance traveled

Vf = Final Speed = 0 m/s

Vi = Initial Speed

a = deceleration rate

First, we consider Car B and we assign a subscript 2 for it:

Vf₂ = 0 m/s  (As, car finally stops)

s₂ = D

a₂ = - a  (due to deceleration)

D = (0² - Vi₂²) /(-2a)

D = Vi₂²/2a    -------- equation (1)

Now, we consider Car A and we assign a subscript 1 for it:

Vf₁ = 0 m/s  (As, car finally stops)

s₁ = ?

a₁ = - a  (due to deceleration)

Vi₁ = 2 Vi₂  (Since, car A was initially traveling at twice speed of car B)

s₁ = (0² - Vi₁²) /(-2a)

s₁ = (2Vi₂)²/2a

s₁ = 4 (Vi₂²/2a)

using equation (1), we get:

s₁ = 4D

Therefore, the correct option is:

A) 4D

Answer:

It's true

Explanation:

Because the ship is mafe up of aluminium, which is a light metal.

Answer:

False

Explanation:

Took The Quiz

ANSWER: num 1 is black hole

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  [tex]Q =2.094 C[/tex]

Explanation:

From the question we are told that

    The diameter of the wire is  [tex]d = 0.205cm = 0.00205 \ m[/tex]

     The radius of  the wire is  [tex]r = \frac{0.00205}{2} = 0.001025 \ m[/tex]

     The resistivity of aluminum is [tex]2.75*10^{-8} \ ohm-meters.[/tex]

       The electric field change is mathematically defied as

         [tex]E (t) = 0.0004t^2 - 0.0001 +0.0004[/tex]

Generally the charge is  mathematically represented as

       [tex]Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt[/tex]

Where A is the area which is mathematically represented as

       [tex]A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2[/tex]

 So

       [tex]\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega[/tex]

Therefore

      [tex]Q = 120 \int\limits^{t}_{0} { E(t) } \, dt[/tex]

substituting values

      [tex]Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt[/tex]

     [tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.[/tex]

From the question we are told that t =  5 sec

           [tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.[/tex]

          [tex]Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }[/tex]

         [tex]Q =2.094 C[/tex]

The charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds is 2.094 Coulomb.

Given the following data:

Conversion:

Diameter of wire = 0.205 cm to m = 0.00205 meter.

Radius = [tex]\frac{Diameter}{2} =\frac{0.00205}{2} =0.001025\;meter[/tex]

To determine the charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds, we would apply Gauss's law in an electric field for a surface charge:

First of all, we would find the area of the wire.

[tex]Area = \pi r^2\\\\Area = 3.142 \times 0.001025^2\\\\Area = 3.3 \times 10^{-6}\;m^2[/tex]

Mathematically, Gauss's law in an electric field for a surface charge is given by the formula:

[tex]Q = \int\limits^t_0 {\frac{A}{\rho } E(t)} \, dt[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]Q= \int\limits^t_0 {\frac{3.3 \times 10^{-6}}{2.75\times 10^{-8} } (0.0004t^2-0.0001t+0.0004)} \, dt\\\\Q=120\int\limits^t_0 1{ (0.0004t^2-0.0001t+0.0004)} \, dt[/tex]

[tex]Q=120(\frac{0.0004t^3}{3} -\frac{0.0001t^2}{2} +0.0004t |\left{5} \atop {0} \right[/tex]

When t = 5 seconds:

[tex]Q=120(\frac{0.0004[5]^3}{3} -\frac{0.0001[5]^2}{2} +0.0004[5])\\\\Q=120(\frac{0.03}{3} -\frac{0.0025}{2} +0.002)\\\\Q=120(0.0167-0.00125+0.002)\\\\Q=120(0.01745)[/tex]

Q = 2.094 Coulomb.

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Answer:

☟

Explanation:

Fuels made from oil mixtures containing large hydrocarbon molecules are not efficient as they do not flow easily and are difficult to ignite. Crude oil often contains too many large hydrocarbon molecules and not enough small hydrocarbon molecules to meet demand. This is where cracking comes in.

Cracking allows large hydrocarbon molecules to be broken down into smaller, more useful hydrocarbon molecules. Fractions containing large hydrocarbon molecules are heated to vaporise them. They are then either:

heated to 600-700°C

passed over a catalyst of silica or alumina

These processes break covalent bonds in the molecules, causing thermal decompositionreactions. Cracking produces smaller alkanesand alkenes (hydrocarbons that contain carbon-carbon double bonds). For example:

hexane → butane + ethene

C6H14 → C4H10 + C2H4

Some of the smaller hydrocarbons formed by cracking are used as fuels, and the alkenes are used to make polymers in plastics manufacture. Sometimes, hydrogen is also produced during cracking.

Fractional distillation of crude oil

Fractional distillation separates a mixture into a number of different parts, called fractions.

A tall fractionating column is fitted above the mixture, with several condensers coming off at different heights. The column is hot at the bottom and cool at the top. Substances with high boiling points condense at the bottom and substances with lower boiling points condense on the way to the top.

Crude oil is a mixture of hydrocarbons. The crude oil is evaporated and its vapours condense at different temperatures in the fractionating column. Each fraction contains hydrocarbon molecules with a similar number of carbon atoms and a similar range of boiling points.

Oil fractions

The diagram below summarises the main fractions from crude oil and their uses, and the trends in properties. Note that the gases leave at the top of the column, the liquids condense in the middle and the solids stay at the bottom.

As you go up the fractionating column, the hydrocarbons have:

lower boiling points

lower viscosity (they flow more easily)

higher flammability (they ignite more easily).

Other fossil fuels

Crude oil is not the only fossil fuel.

Natural gas mainly consists of methane. It is used in domestic boilers, cookers and Bunsen burners, as well as in some power stations.

Coal was formed from the remains of ancient forests. It can be burned in power stations. Coal is mainly carbon but it may also contain sulfur compounds, which produce sulfur dioxide when the coal is burned. This gas is a cause of acid rain. Also, as all fossil fuels contain carbon, the burning of any fossil fuel will contribute to global warming due to the production of carbon dioxide.

In fractional distillation, the crude oil is added to the chamber and heated. The components with the highest boiling point will condense in the lower part of the column and the components with the lower boiling point will condense at the top of the column. Petrol with a low boiling point is collected from the top of the column.

Fractional distillation can be described as the separation of a mixture into its component fractions. The chemical compound is separated by heating them to a temperature at which fractions of the mixture will vaporize.

Generally, the components have boiling points that differ by less than 25 °C  from each other under one atmosphere. When the mixture is heated, the component with the lower boiling point boils and changes to vapours.

The more volatile component remains in a vapour state and repeated distillations are used in the process, and the mixture is separated into component parts.

Therefore, the petrol from the crude oil can easily be separated as it has a boiling point of about 25-60°C.

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Answer:

Its called PSY

Explanation: I so do not know why they named it this way but, hope i helped.

Answer:

Explanation:

 Time taken by sound waves to cover distance between politician and reporter = time taken by em waves to travel distance between politician and the television viewer.

5.5 / 343 = d / 3 x 10⁸ + 2.7 / 343

d is distance between politician and television set + time taken by sound to travel distance between television and its viewer.

.0160349 = d / 3 x 10⁸  + .0078717

d / 3 x 10⁸  = .0081632

d = 2448960 m

= 2448.96 km

= 2449 km .

Answer: A chemical change results from a chemical reaction, while a physical change is when matter changes forms but not chemical identity. Examples of chemical changes are burning, cooking, rusting, and rotting. Examples of physical changes are boiling, melting, freezing, and shredding. Often, physical changes can be undone, if energy is input.

Explanation: hope this helps have a good day

Answer:

If the reaction is a chemical change, new substances with different properties and identities are formed. This may be indicated by the production of an odor, a change in color or energy, or the formation of a solid.

Answer:

a) 463.29 K

b) 8065.65 Pa

c) 0 J

Explanation:

The parameters given are;

Volume of the tank, V = 3.72 m³

Number of moles of gas present in the tank, n = 22.1 moles

Temperature of the gas before heating, T₁ = 300 k

Heat added to the gas, ΔQ = 4.5 × 10⁴ J

Specific heat capacity at constant volume, [tex]c_v[/tex], for monatomic gas = 12.47 J/K/mole

Avogadro's number = 6.022 × 10²³ particles per mole

a) ΔQ = n × [tex]c_v[/tex] × ΔT

Where:

ΔT = T₂ - T₁

T₂ = Final temperature of the gas

Hence, by plugging in the values, we have;

4.5 × 10⁴ = 22.1 × 12.47 × (T₂ - 300)

[tex]T_{2} - 300 = \frac{4.5\times 10^{4}}{22.1\times 12.47}[/tex]

T₂ = 300 + 163.29 = 463.29 K

b) The pressure of the gas is found from the relation;

P×V = n×R×T

[tex]P = \dfrac{n \times R \times T}{V}[/tex]

Where:

P = Pressure of the gas

R = Universal gas constant = 8.3145 J/(mol·K)

T = Temperature of the gas

V = Volume of the gas = 3.72 ³ (constant)

n = Number of moles of gas present = 22.1 moles (constant)

Hence the change in pressure is given by the relation;

[tex]\Delta P = \dfrac{n \times R \times (T_2 - T_1)}{V} = \dfrac{n \times R \times \Delta T}{V}[/tex]

Plugging in the values, we have;

[tex]\Delta P = \dfrac{22.1 \times 8.3145 \times 163.29}{3.72} = 8065.65 \, Pa[/tex]

c) Work done, W, by the gas is given by the area under the pressure to volume graph which gives;

W = f(P) × ΔV

The volume given in the question is constant

∴ ΔV = 0

Hence, W =  f(P) × 0 = 0 J

No work done by the gas during the process.

Answer:

To determine wavelength of a wave on a string.

Explanation:

The Standing Waves lab study the parameters that affect standing waves in various strings. The effects of string tension and density on wavelength and frequency will be studied.

Answer:

Exothermic

Explanation:

Depending on the unit you are in, the answer may vary.

This is an exothermic reaction because it produces heat (the beaker gets warm).

Answer:

FInal speed (v) = 0.509 m/s (Approx)

Explanation:

Given:

Mass of ant (m) = 12 mg

Force (f) = 47 N

Time taken (t) = 0.13 ms

Find:

FInal speed (v) = ?

Computation:

Initial velocity (u) = 0

Impulse = change in momentum

Force × TIme = change in momentum

47 × 0.13 = mv - mu

6.11 = 12 (V)

FInal speed (v) = 0.509 m/s (Approx)

Answer:

The length of organ pipe A is [tex]L = 0.3611 \ m[/tex]

The length of organ pipe B is  [tex]L_b = 0.2708 \ m[/tex]

Explanation:

From the question we are told that

    The fundamental frequency is  [tex]f = 475 Hz[/tex]

     The speed of sound is  [tex]v_s = 343 \ m/s[/tex]

The fundamental frequency of the organ pipe A  is mathematically represented as

        [tex]f= \frac{v_s}{2 L}[/tex]

Where L is the length of  organ pipe

   Now  making L the subject

        [tex]L = \frac{v_s}{2f}[/tex]

substituting values

        [tex]L = \frac{343}{2 *475}[/tex]

        [tex]L = 0.3611 \ m[/tex]

The second harmonic frequency of the  organ pipe A is mathematically represented as

       [tex]f_2 = \frac{v_2}{L}[/tex]

The third harmonic frequency of the  organ pipe B is mathematically represented as      

      [tex]f_3 = \frac{3 v_s}{4 L_b }[/tex]

So from the question

       [tex]f_2 = f_3[/tex]

So

    [tex]\frac{v_2}{L} = \frac{3 v_s}{4 L_b }[/tex]

Making  [tex]L_b[/tex] the subject

     [tex]L_b = \frac{3}{4} L[/tex]

substituting values

    [tex]L_b = \frac{3}{4} (0.3611)[/tex]

    [tex]L_b = 0.2708 \ m[/tex]

Answer:

a)   E = 1.58 10²¹ J , b) Oil = 4,236 107 liter ,  e)   T = 54.3 C

Explanation:

a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface

     I = P / A

     A = 4π r²

in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m

     I = P / A

     I = P / 4π r²

let's calculate

     I = 3,828 10²⁵/4 pi (1.5 10¹¹)²

     I = 1.3539 10²W / m² = 135.4 W / m2

the energy that reaches the disk of the Earth is

    E = I A

the area of ​​a disc

    A = π r²

    E = I π r²

where r is the radius of the Earth 6.37 10⁶ m

     E = 135.4 π(6.37 10⁶)

     E = 1,726 10¹⁶ W

This is the energy per unit of time that reaches Earth

    t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s

    E = 1,826 10¹⁶ 86400

     E = 1.58 10²¹ J

b) for this part we can use a direct proportions rule

      Oil = 1.58 10²¹ (1 / 37.3 10⁶)

      Oil = 4,236 10⁷ liter

c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law

       P = σ A e T⁴

       T = [tex]\sqrt[4]{P/Ae}[/tex]

nos indicate the refect, therefore the amount of absorbencies

       P_absorbed = 0.7 P

let's calculate

       T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)

       T = RER (8,676 106)

       T = 54.3 C

b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere