Concerned with citizen complaints of price gouging during past hurricanes, Florida's state government passes a law setting a price ceiling for a bottle of water equal to the market equilibrium price during normal times. After all, it seems unfair that sellers of water gain because of a hurricane.

Answer:

= 0.0532 cm^3

Explanation:

The computation of volume of the sphere is shown below:-

[tex]Density = \frac{Mass}{Volume}[/tex]

Where,

Density = 13.6 g/cm^3

Mass of sphere = 0.723 g

now we will put the values into the above formula to reach volume of the sphere which is here below:-

[tex]Volume = \frac{0.723}{13.6}[/tex]

= 0.0532 cm^3

Therefore for computing the volume of the sphere we simply applied the above formula.

Answer:

The power output of this engine is  [tex]P = 17.5 W[/tex]

The  the maximum (Carnot) efficiency is  [tex]\eta_c = 0.7424[/tex]

The  actual efficiency of this engine is  [tex]\eta _a = 0.46[/tex]

Explanation:

From the question we are told that

    The temperature of the hot reservoir is  [tex]T_h = 1250 \ K[/tex]

      The temperature of the cold reservoir  is  [tex]T_c = 322 \ K[/tex]

     The energy absorbed from the hot reservoir is [tex]E_h = 1.37 *10^{5} \ J[/tex]

       The energy exhausts into  cold reservoir is  [tex]E_c = 7.4 *10^{4} J[/tex]

The power output is mathematically represented as

      [tex]P = \frac{W}{t}[/tex]

Where t is the time taken which we will assume to be 1 hour =  3600 s  

W is the workdone which is mathematically represented as

      [tex]W = E_h -E_c[/tex]

substituting values

       [tex]W = 63000 J[/tex]

So

    [tex]P = \frac{63000}{3600}[/tex]

    [tex]P = 17.5 W[/tex]

The Carnot efficiency is mathematically represented as

          [tex]\eta_c = 1 - \frac{T_c}{T_h}[/tex]

         [tex]\eta_c = 1 - \frac{322}{1250}[/tex]

         [tex]\eta_c = 0.7424[/tex]

The actual efficiency is mathematically represented as

        [tex]\eta _a = \frac{W}{E_h}[/tex]

substituting values

         [tex]\eta _a = \frac{63000}{1.37*10^{5}}[/tex]

         [tex]\eta _a = 0.46[/tex]

the answer is 1.5 hope this helps

Answer:

1.5

Explanation:

0.5=σ/(15/0.5)

σ=3/2 or 1.5

Answer:

Explanation:

The velocity at the inlet and exit of the control volume are same [tex]V_i=V_e=V[/tex]

Calculate the inlet and exit velocity of water jet

[tex]V=V_j+V_e\\\\V=30+14\\\\V=44m/s[/tex]

The conservation of mass equation of steady flow

[tex]\sum ^e_i\bar V. \bar A=0\\\\(-V_iA_i+V_eA_e)=0[/tex]

[tex]A_i\ \texttt {is the inlet area of the jet}[/tex]

[tex]A_e\ \texttt {is the exit area of the jet}[/tex]

since inlet and exit velocity of water jet are equal so the inlet and exit cross section area of the jet is equal

The expression for thickness of the jet

[tex]A_i=A_e\\\\\frac{\pi}{4} D_j^2=2\pi Rt\\\\t=\frac{D^2_j}{8R}[/tex]

R is the radius

t is the thickness of the jet

D_j is the diameter of the inlet jet

[tex]t=\frac{(100\times10^{-3})^2}{8(230\times10^{-3}} \\\\=5.434mm[/tex]

(b)

[tex]R-x=\rho(AV_r)[-(V_i)+(V_c)\cos 60^o]\\\\=\rho(V_j+V_c)A[-(V_i+V_c)+(V_i+V_c)\cos 60^o]\\\\=\rho(V_j+V_c)(\frac{\pi}{4}D_j^2 )[V_i+V_c](\cos60^o-1)][/tex]

[tex]1000kg/m^3=\rho\\\\44m/s=(V_j+V+c)\\\\100\times10^{-3}m=D_j[/tex]

[tex]R_x=[1000\times(44)\frac{\pi}{4} (10\times10^{-3})^2[(44)(\cos60^o-1)]]\\\\=-7603N[/tex]

The negative sign indicate that the direction of the force will be in opposite direction of our assumption

Therefore, the horizontal force is -7603N

Answer:

3.2 m/s

Explanation:

Given:

Δx = 1000 m

v₀ = 23 m/s

a = -0.26 m/s²

t = 76 s

Find: v

This problem is over-defined.  We only need 3 pieces of information, and we're given 4.  There are several equations we can use.  For example:

v = at + v₀

v = (-0.26 m/s²) (76 s) + (23 m/s)

v = 3.2 m/s

Or:

Δx = ½ (v + v₀) t

(1000 m) = ½ (v + 23 m/s) (76 s)

v = 3.3 m/s

Or:

v² = v₀² + 2aΔx

v² = (23 m/s)² + 2(-0.26 m/s²)(1000 m)

v = 3.0 m/s

Or:

Δx = vt − ½ at²

(1000 m) = v (76 s) − ½ (-0.26 m/s²) (76 s)²

v = 3.3 m/s

As you can see, you get slightly different answers depending on which variables you use.  Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.

Answer:

the correct answer is B

Explanation:

We analyze this exercise a little, the block goes into the air and is under the acceleration of gravity. The ball is fired by the hand and is describing a parabolic movement, subjected to the acceleration of gravity.

For the ball to hit the block we must have the distance the ball goes up equal to the distance the block moves, therefore we must shoot the ball at the block at its highest point.

Let's write the kinematic equation for the two bodies

The block. At the highest point of the path

      y = - ½ g t2

The ball, in its vertical movement

     y = vo t - ½ g t2

therefore the correct answer is B

Answer:

The angular momentum is  [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]

Explanation:

From the question we are told that

    The mass of the woman is  [tex]m = 50 \ kg[/tex]

     The angular  speed of the rim is  [tex]w = 0.80 \ rev/s = 0.8 * [\frac{2 \pi}{1} ] = 5.024 \ rad \cdot s^{-1}[/tex]

      The mass of the disk is  [tex]m_d = 110 \ kg[/tex]

       The radius of the disk is [tex]r_d = 4.0 \ m[/tex]

The moment of inertia of the disk is mathematically represented as

        [tex]I_D = \frac{1}{2} m_d r^2_d[/tex]

substituting values

          [tex]I_D = \frac{1}{2} * 110 * 4^2[/tex]

          [tex]I_D = 880 \ kg \cdot m^2[/tex]

The moment of inertia of the woman is  

          [tex]I_w = m * r_d^2[/tex]

substituting values

        [tex]I_w = 50 * 4^2[/tex]

       [tex]I_w =800\ kg[/tex]

The moment of inertia of the system (the woman + the large disk ) is  

        [tex]I_t = I_w + I_D[/tex]

substituting values  

      [tex]I_t = 880 +800[/tex]

     [tex]I_t =1680 \ kg \cdot m^2[/tex]

The angular momentum of the system is

      [tex]L = I_t w[/tex]

substituting values  

      [tex]L = 1680 * 5.024[/tex]

      [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]

The dragster's velocity v at time t with constant acceleration a is

[tex]v=at[/tex]

since it starts at rest.

After 2.1 s, it will attain a velocity of

[tex]v=\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)[/tex]

or 92.4 m/s.

Doubling the time would double the final velocity,

[tex]v=a(2t)=2at[/tex]

so the velocity would be twice the previous one, 184.8 m/s.

The dragster undergoes a displacement x after time t with acceleration a of

[tex]x=\dfrac12at^2[/tex]

if we take the starting line to be the origin.

After 2.1 s, it will have moved

[tex]x=\dfrac12\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)^2[/tex]

or 88 m.

Doubling the time has the effect of quadrupling the displacement, since

[tex]x=\dfrac12a(2t)^2=4\left(\dfrac12at^2\right)[/tex]

so after 4.2 s it will have moved 352 m.

Answer:

The speed of the freight car decreases.

Explanation:

According to the law of conservation of momentum indicates that for colliding in an isolated system, the total momentum pre and post collision is same for the two objects this is done because the momentum that one item has lost is same for the momentum that the other received

In the given situation, the freight car travels at constant speed along a frictionless railroad line. The top floor freight car is open. Then a huge load of coal is dumped inside the car.

Therefore the speed of the freight car decreased by applying the law of conservation of momentum i

Answer:

a. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

[tex]W = F\times d[/tex]

[tex]KE = 0.5\times m\times v^2[/tex]

As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.

F = 3.70 cos 45 = 2.62 N

[tex]W = F \times d = 2.62 N \times 100 m[/tex]

[tex]W = 261.6 N\times m[/tex]

We know that

KE1 = Initial kinetic energy

KE2 = kinetic energy following 100 m

The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.

So, the equation is

KE2 = KE1 - W

[tex]0.5 m\times v2^2 = 0.5 m\ v1^2 - W[/tex]

Now solve for v2

[tex]v2 = \sqrt{v1^2 - {\frac{2W}{M}}}[/tex]

[tex]= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}[/tex]

= 2.668 m/s

b. Now the minimum value of Ug is

As we know that

Ff = force of friction

Us = coefficient of static friction

N = Normal force = weight of skater

So,

[tex]Ff = Us\times N[/tex]

Now solve for Us

[tex]= \frac{Ff}{N}[/tex]

[tex]= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}[/tex]

= 0.00494

Answer:

(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v

Explanation:

Solution

Given that:

A refrigerator magnet about = 2 mm

The estimated magnetic field strength of the magnet is = 5 m T

The Area = 8 cm²

Now,

(A) The magnetic flux ΦB = BA

Thus,

ΦB  = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²

So,

ΦB =  4* 6 ^ ⁻6 T m²

(B)By applying Faraday's Law we have the following formula given below:

Ε = Bℓυ

Here,

ℓ = 2 cm the same as 2 * 10 ^⁻2 m

B = 5 m T = 5 * 10 ^ ⁻3 T

υ = 2 cm/s  = 2 * 10 ^ ⁻2 m/s

Thus,

Ε = (5 * 10 ^ ⁻3 T) *  (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v

E =2 * 10 ^ ⁻6 v

A) The magnetic flux ΦB into the refrigerator door behind the magnet :

B) The estimated emf induced under the rectangle on the door's surface ;

Given data :

magnetic field strength of magnet ( B )  = 5 mT

size of refrigerator magnet = 2 mm

Area of magnet ( A )  = 4 * 2 = 8 cm²

where ; ΦB  = BA

ΦB = ( 5 * 10⁻³ ) * ( 4 * 10⁻² ) * ( 2 * 10⁻² ) Tm²

      =  4 * 6⁻⁶ Tm²

To determine the estimated emf we will apply Faraday's law

Ε = Bℓυ ---- ( 2 )

where : B = 5 * 10⁻³ T,  ℓ = 2 * 10⁻² m,  υ = 2 * 10⁻² m/s

insert values into equation 2

E = ( 5 * 10⁻³ ) * ( 2 * 10⁻² ) * ( 2 * 10⁻² )

  = 2 * 10⁻⁶ v

Hence we can conclude that The magnetic flux ΦB is 4 * 6⁻⁶ Tm² and The estimated emf induced is  2 * 10⁻⁶ v

Learn more about magnet flux : https://brainly.com/question/4721624

Answer:

Explanation:

At height h , potential energy of coaster car  having mass m = mgh .

The car will lose potential energy and gain kinetic energy.

height lost by car when it is at the top of loop of radius R

= h - 2R

potential energy lost = mg ( h - 2R )

kinetic energy gained = mg ( h - 2R )

kinetic energy = 0 + mg ( h - 2R )

= mg ( h - 2R )

b )

For the car to remain in contact with the track , if v be the minimum velocity

centripetal force at top = mg

m v² / R = mg

v² = gR

kinetic energy = 1/2 mv²

= 1/2 m x gR

= mgR /

If h be the minimum height that can give this kinetic energy

mg ( h - 2R ) = mgR / 2

h - 2R = R / 2

h = 2.5 R .

Answer:

It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4

Answer:

the magnitude is 7 and sign of the point charge on the surface shell is -13

Explanation:

Answer:

St Lawrence Sea way

Explanation:

The great lake connects the middle of North America which is at the Canada-United states border connecting to the Atlantic Ocean through the St Lawrence River.

Answer: B

Explanation:

I would say the shrinking of the polar ice caps because in order for ice caps to shrink, they would have to obviously melt. This will cause the sea level and total volume of sea water to rise and cover up the land bridges

Answer:B :)

Explanation:

Answer:

10 Sig Figs

Explanation:

Just start counting at the first non zero after the decimal so in this case the nine, and count all of the numbers including zeros after that.

Answer:

25 m/s

Explanation:

Data provided in the question

20 m/s for 10 minutes

And, the 30 m/s for another 10 minutes

Based on the above information, the average speed is

As we know that

[tex]Average\ speed = \frac{Total\ distance}{Total\ time}[/tex]

[tex]= \frac{20\times10\times60 + 30\times10\times60 }{20\times60}[/tex]

= 25 m/s

1 hour = 60 minutes

1 minute = 60 seconds

Hence, the average speed is 25 m/s

In the question,  there are miles is given but instead of this we use the minutes as we have to find out the average speed and time should not be in miles it should be in minutes, hour or seconds

Therefore we considered the same

Answer:

P1 = 0 gage

P2 = 87.9 lb/ft³

Explanation:

Given data

Airplane flying = 200 mph = 293.33 ft/s

altitude height = 5000-ft

air velocity relative to the airplane = 273 mph = 400.4 ft/s

Solution

we know density at height 5000-ft is 2.04 × [tex]10^{-3}[/tex] slug/ft³

so here P1 + [tex]\frac{\rho v1^2}{2}[/tex]  = P2 + [tex]\frac{\rho v2^2}{2}[/tex]

and here

P1 = 0 gage

because P1 = atmospheric pressure

and so here put here value and we get

P1 + [tex]\frac{\rho v1^2}{2}[/tex]  = P2 + [tex]\frac{\rho v2^2}{2}[/tex]

0 + [tex]\frac{2.048 \times 10^{-3} \times 293.33^2}{2}[/tex]  [tex]= P2 + \frac{2.048 \times 10^{-3} \times 400.4^2}{2}[/tex]  

solve it we get

P2 = 87.9 lb/ft³

Answer:

107 N, option d

Explanation:

Given that

mass of the ball, m = 0.2 kg

initial velocity of the ball, u = 20 m/s

final velocity of the ball, v = -12 m/s

time taken, Δt = 60 ms

Solving this question makes us remember "Impulse Theorem"

It states that, "that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object"

Mathematically, it is represented as

FΔt = m(v - u), where

F = the average force

Δt = time taken

m = mass of the ball

v = final velocity of the ball

u = initial velocity of the ball

From the question we were given, if we substitute the values in it, we have

F = ?

Δt = 60 ms = 0.06s

m = 0.2 kg

v = -12 m/s

u = 20 m/s

F = 0.2(-12 - 20) / 0.06

F = (0.2 * -32) / 0.06

F = -6.4 / 0.06

F = -106.7 N

Thus, the magnitude is 107 N

Answer:

0.1575 m/s^2

Explanation:

Solution:-

- Acceleration ( a ) is expressed as the rate of change of velocity ( v ).

- We are given that the trains starts from rest i.e the initial velocity ( vo ) is equal to 0. Then the train travels from reference point ( so = 0 ) to ( sf = 5.6 km ) from the reference.

- During the travel the train accelerated uniformly to a speed of ( vf =42 m/s ).

- We will employ the use of 3rd kinematic equation of motion valid for constant acceleration ( a ) as follows:

                         [tex]v_f^2 = v_i^2 + 2*a*( s_f - s_o )[/tex]

- We will plug in the given parameters in the equation of motion given above:

                         [tex]42^2 = 0^2 + 2*a* ( 5600 - 0 )\\\\1764 = 11,200*a\\\\a = \frac{1,764}{11,200} \\\\a = 0.1575 \frac{m}{s^2}[/tex]

Answer: the acceleration during the first 5.6 km of travel is 0.1575 m / s^2

Answer:

2C

Explanation:

The equivalent capacitance of a parallel combination of capacitors is the sum of their capacitance.

So, if the capacitance of each capacitor is half the previous one, we have a geometric series with first term = C and rate = 0.5.

Using the formula for the sum of the infinite terms of a geometric series, we have:

Sum = First term / (1 - rate)

Sum = C / (1 - 0.5)

Sum = C / 0.5 = 2C

So the equivalent capacitance of this parallel connection is 2C.

Answer:

m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg

Explanation:

First, we find the the surface area of the cell wall. Since, the cell is spherical in shape. Therefore, surface area of cell wall will be:

A = 4πr²

where,

A = Surface Area = ?

r = Radius of Cell = Diameter/2 = 2.2 μm/2 = 1.1 μm = 1.1 x 10⁻⁶ m

Therefore,

A = 4π(1.1 x 10⁻⁶ m)²

A = 15.2 x 10⁻¹² m²

Now, we find the volume of the cell wall. For that purpose, we use formula:

V = At

where,

V = Volume of the Cell Wall = ?

t = Thickness of Wall = 40 nm = 4 x 10⁻⁸ m

Therefore,

V = (15.2 x 10⁻¹² m²)(4 x 10⁻⁸ m)

V = 60.82 x 10⁻²⁰ m³

Now, to find mass of cell wall, we use formula:

ρ = m/V

m = ρV

where,

ρ = density of water = 1000 kg/m³

m = Mass of Wall = ?

Therefore,

m = (1000 kg/m³)(60.82 x 10⁻²⁰ m³)

m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg

The mass of the cell wall in mg is 6.082 × 10⁻¹⁰ mg

Since we assume the cell to be spherical and the wall to be a thin spherical shell, the volume of the cell wall V = At where

So, V = 4πR²t

Substituting the values of the variables into the equation, we have

V = 4πR²t

V = 4π(1.1 × 10⁻⁶ m)² × 40 × 10⁻⁹ m.

V = 4π(1.21 × 10⁻¹² m²) × 40 × 10⁻⁹ m.

V = 193.6π × 10⁻²¹ m³

V = 608.21 × 10⁻²¹ m³

V = 6.0821 × 10⁻¹⁹ m³

V ≅ 6.082 × 10⁻¹⁹ m³

We know that density of cell wall, ρ = m/v where m = mass of cell wall and V = volume of cell wall.

Making m subject of the formula, we have

m = ρV

Since we assume the density of the cell wall to be equal to that of pure water, ρ = 1000 kg/m³

So, m = ρV

m = 1000 kg/m³ × 6.082 × 10⁻¹⁹ m³

m = 6.082 × 10⁻¹⁶ kg

Converting to mg, we have

m = 6.082 × 10⁻¹⁶ kg × 10⁶ mg/kg

m = 6.082 × 10⁻¹⁰ mg

So, the mass of the cell wall in mg is 6.082 × 10⁻¹⁰ mg

Learn more about mass of cell wall here:

https://brainly.com/question/13173768

Answer: in this question, the only charge in the cavity is Q. Inside the conducting spherical shell, the electric field is zero.

While outside the shell, the electric field is given by: k(q + Q)/r²

Where;

K= is a constant which is given as, 8.99 x 10^9 N m² / C².

Q= source charge which creates the electric field

q= is the test charge which is used to measure the strength of the electric field at a given location.

r= is the radius

Explanation: Inside the conducting spherical shell, the electric field is zero since the Electric field vanishes everywhere inside the volume of a good conductor.

Answer:

The difference between AC and DC lies in the direction in which the electrons flow. In DC, the electrons flow steadily in a single direction, or "forward." In AC, electrons keep switching directions, sometimes going "forward" and then going "backward."

Answer:

2000 m

Explanation:

since the boat is 1000 m from the river bank, the beam must be focused at this point. This indicates that the focal length is 1000 m

for a spherical mirror, the focal length is given by

f = R/2

where R is the radius of curvature

1000 = R/2

R = 2000 m

R = 2000 m

this means that the radius of curvature must be 2000 m

Answer:

The thermal conductivity of the insulated metal rod is [tex]202.92\,\frac{W}{m\cdot K}[/tex].

Explanation:

This is a situation of one-dimensional thermal conduction of a metal rod in a temperature gradient. The heat transfer rate through the metal rod is calculated by this expression:

[tex]\dot Q = \frac{k_{rod}\cdot A_{c, rod}}{L_{rod}}\cdot \Delta T[/tex]

Where:

[tex]\dot Q[/tex] - Heat transfer due to conduction, measured in watts.

[tex]L_{rod}[/tex] - Length of the metal rod, measured in meters.

[tex]A_{c,rod}[/tex] - Cross section area of the metal rod, measured in meters.

[tex]k_{rod}[/tex] - Thermal conductivity, measured in [tex]\frac{W}{m\cdot K}[/tex].

Let assume that heat conducted to melt some ice was transfered at constant rate, so that definition of power can be translated as:

[tex]\dot Q = \frac{Q}{\Delta t}[/tex]

Where Q is the latent heat required to melt the ice, whose formula is:

[tex]Q = m_{ice}\cdot L_{f}[/tex]

Where:

[tex]m_{ice}[/tex] - Mass of ice, measured in kilograms.

[tex]L_{f}[/tex] - Latent heat of fussion, measured in joules per gram.

The latent heat of fussion of water is equal to [tex]330000\,\frac{J}{g}[/tex]. Hence, the total heat received by the ice is:

[tex]Q = (6.15\,g)\cdot \left(330\,\frac{J}{g} \right)[/tex]

[tex]Q = 2029.5\,J[/tex]

Now, the heat transfer rate is:

[tex]\dot Q = \frac{2029.5\,J}{(10\,min)\cdot \left(60\,\frac{s}{min} \right)}[/tex]

[tex]\dot Q = 3.382\,W[/tex]

Turning to the thermal conduction equation, thermal conductivity is cleared and computed after replacing remaining variables: ([tex]L_{rod} = 0.75\,m[/tex], [tex]A_{c,rod} = 1.25\times 10^{-4}\,m^{2}[/tex], [tex]\Delta T = 100\,K[/tex], [tex]\dot Q = 3.382\,W[/tex])

[tex]\dot Q = \frac{k_{rod}\cdot A_{c, rod}}{L_{rod}}\cdot \Delta T[/tex]

[tex]k_{rod} = \frac{\dot Q \cdot L_{rod}}{A_{c,rod}\cdot \Delta T}[/tex]

[tex]k_{rod} = \frac{(3.382\,W)\cdot (0.75\,m)}{(1.25\times 10^{-4}\,m^{2})\cdot (100\,K)}[/tex]

[tex]k_{rod} = 202.92\,\frac{W}{m\cdot K}[/tex]

The thermal conductivity of the insulated metal rod is [tex]202.92\,\frac{W}{m\cdot K}[/tex].

Answer:

The location are [tex]x_1 = 2 \ and \ x_2 = 3[/tex]

Explanation:

From the question we are told that

    The potential energy is  [tex]U(x) = (2.0 \ J/m^3) * x^3 - (15 \ J/m) * x^2 + (36 \ J/m) * x - 23 \ J[/tex]

The force on the mass can be mathematically evaluated as  

      [tex]F = - \frac{d U(x)}{d x } = -( 6 x^2 - 30x +36)[/tex]

The negative sign shows that the force is moving in the opposite  direction of the potential energy

       [tex]F = - 6 x^2 + 30x - 36[/tex]

At critical point

      [tex]\frac{d U(x)}{dx} = 0[/tex]

So  

     [tex]- 6 x^2 + 30x - 36 = 0[/tex]

     [tex]- x^2 + 5x - 6 = 0[/tex]

Using quadratic equation formula to solve this we have that

       [tex]x_1 = 2 \ and \ x_2 = 3[/tex]

Answer:

the arrow symbol ⇒ in irreversible reactions and doble arrow symbol in reversible reactios⇔

Explanation:

i hope this will help you