12) If a man weighs 900 N-on the Earth, what would he weigh on Jupiter, where the acceleration due to gravity is 25.9 m/s?

Answer:

a) The initial total mechanical energy of the projectile is 498556.296 joules.

b) The work done on the projectile by air friction is 125960.4 joules.

c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

Explanation:

a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ([tex]E[/tex]) of the project is equal to the sum of gravitational potential energy ([tex]U_{g}[/tex]) and translational kinetic energy ([tex]K[/tex]), all measured in joules:

[tex]E = U_{g} + K[/tex] (Eq. 1)

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

[tex]E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}[/tex] (Eq. 1b)

Where:

[tex]m[/tex] - Mass of the projectile, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]y[/tex] - Initial height of the projectile above ground, measured in meters.

[tex]v[/tex] - Initial speed of the projectile, measured in meters per second.

If we know that [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y = 132\,m[/tex] and [tex]v = 126\,\frac{m}{s}[/tex], the initial mechanical energy of the earth-projectile system is:

[tex]E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}[/tex]

[tex]E = 498556.296\,J[/tex]

The initial total mechanical energy of the projectile is 498556.296 joules.

b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:

[tex]W_{loss} = E_{o}-E_{1}[/tex] (Eq. 2)

Where:

[tex]E_{o}[/tex] - Initial total mechanical energy, measured in joules.

[tex]E_{1}[/tex] - FInal total mechanical energy, measured in joules.

[tex]W_{loss}[/tex] - Work losses due to air friction, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

[tex]W_{loss} = E_{o}-K_{1}-U_{g,1}[/tex]

[tex]W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}[/tex] (Eq. 2b)

Where:

[tex]m[/tex] - Mass of the projectile, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]y_{1}[/tex] - Maximum height of the projectile above ground, measured in meters.

[tex]v_{1}[/tex] - Current speed of the projectile, measured in meters per second.

If we know that [tex]E_{o} = 498556.296\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 297\,m[/tex] and [tex]v_{1} = 89.3\,\frac{m}{s}[/tex], the work losses due to air friction are:

[tex]W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)[/tex]

[tex]W_{loss} = 125960.4\,J[/tex]

The work done on the projectile by air friction is 125960.4 joules.

c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:

[tex]E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}[/tex] (Eq. 3)

[tex]K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}[/tex]

Where:

[tex]E_{1}[/tex] - Total mechanical energy of the projectile at maximum height, measured in joules.

[tex]U_{g,2}[/tex] - Potential gravitational energy of the projectile, measured in joules.

[tex]K_{2}[/tex] - Kinetic energy of the projectile, measured in joules.

[tex]W_{loss}[/tex] - Work losses due to air friction during the upward movement, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}[/tex] (Eq. 3b)

[tex]m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}[/tex]

[tex]v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}[/tex]

[tex]v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }[/tex]

If we know that [tex]E_{1} = 372595.896\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{2} =0\,m[/tex] and [tex]W_{loss} = 125960.4\,J[/tex], the final speed of the projectile is:

[tex]v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }[/tex]

[tex]v_{2} \approx 82.475\,\frac{m}{s}[/tex]

The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

Answer:

1- the rate of flow of charge through a conductor is called current .whereas ,current density is the current per unit area of conductor

Answer:

 N = 136.77 N

Explanation:

This is an exercise in Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis. In the attachment we can see the applied forces.

Let's use trigonometry to decompose the force F1

           cos θ = F₁ₓ / F₁

           sin θ = F_{1y} / F₁

           F₁ₓ = F₁ cos θ

           F_{1y} = F₁ sin θ

now let's apply Newton's second law to each axis

X axis

        F₁ₓ - F4 = m a

Y axis

        N + F3 + F_{1y} -F₂ -W = 0

the acceleration can be calculated with kinematics

        v = v₀ + a t

since the object starts from rest, the initial velocity is zero v₀ = 0

        a = v / t

        a = 20/3

        a = 6.667 m / s²

we substitute in the equation

        F₁ₓ = F₄ + m a

        F₁ₓ = 42 + 15  6,667

        F₁ₓ = 142 N

        F₁ cos θ = 142

        cos θ = 142/206 = 0.6893

        θ = cos⁻¹ 0.6893

        θ = 46.42º

now let's work the y axis

        N = W + F₂ - F₃ - F_{1y}

        N  = 15 9.8 + 144 -5 - 206 sin 46.42

        N = 286 - 149.23

        N = 136.77 N

Answer:

5m

Explanation:

Answer:

5

Explanation:

because that's the average of how far it when the most

Answer:

Considering first question

    Generally the coefficient of performance of the air condition  is mathematically represented as

   [tex]COP  =  \frac{T_i}{T_o - T_i}[/tex]

Here [tex]T_i[/tex] is the inside temperature

while  [tex]T_o[/tex] is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes   [tex] \frac{T_i}{T_o - T_i}[/tex] heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

         [tex]Q \ \alpha \ (T_o - T_i)[/tex]

=>        [tex]Q= k (T_o - T_i)[/tex]

Here k is the constant of proportionality

So  

    since  1 unit of electricity  removes   [tex] \frac{T_i}{T_o - T_i}[/tex]  amount of heat

   E  unit of electricity will remove  [tex]Q= k (T_o - T_i)[/tex]

So

      [tex]E =  \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }[/tex]

=>   [tex]E = \frac{k}{T_i} (T_o - T_i)^2[/tex]

given that  [tex]\frac{k}{T_i}[/tex] is constant

    =>  [tex]E \  \alpha  \  (T_o - T_i)^2[/tex]

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

 Considering the  second question

Assuming that  [tex]T_i   =  30 ^oC[/tex]

 and      [tex]T_o  =  40 ^oC[/tex]

Hence  

     [tex]E = K (T_o - T_i)^2[/tex]

Here K stand for a constant

So  

        [tex]E = K (40 -  30)^2[/tex]

=>      [tex]E = 100K [/tex]

Now if  the  [tex]T_i   =  20 ^oC[/tex]

Then

       [tex]E = K (40 -  20)^2[/tex]

=>      [tex]E = 400 \ K[/tex]

So  from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low  is  much higher than the electricity required when the inside temperature is higher

Considering the  third question

Now in the case where the  heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

       [tex]Q = k (T_o - T_i )^{\frac{1}{2} }[/tex]

So

       [tex]E =  \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }[/tex]

=>   [tex]E =  \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }[/tex]

Assuming [tex]\frac{k}{T_i}[/tex] is a constant

Then  

     [tex]E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }[/tex]

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root  of the cube of the  temperature difference.

Answer:

It would take 5 seconds

Explanation:

I can't find the 'delta' sign nor the vector sign so just pretend that displacement, time and velocity has them.

V = d / t

34 = 170 / t

34 x t = 170

34t = 170

t = 170 / 34

t = 5

Answer:

The tension in the web is 0.017738 N

Explanation:

Net Force

The net force exerted on an object is the sum of the vectors of each individual force applied to an object.

If the net force equals 0, then the object is at rest or moving at a constant speed.

The spider described in the question is hanging at rest. It means the sum of the forces it's receiving is 0.

A hanging object has only two forces: The tension of the supporting string (in our case, the web) and its weight. If the object is in equilibrium, the tension is numerically equal to the weight:

T=W=m.g

The mass of the spider is m=1.81 gr = 0.00181 Kg, thus the tension is:

[tex]T = 0.00181\ Kg\cdot 9.8\ m/s^2[/tex]

[tex]T=0.017738\ N[/tex]

The tension in the web is 0.017738 N

Answer:

Explanation:

To find the momentum of an object given it's mass and velocity we use the formula

From the question

mass = 20 kg

velocity = 5 m/s

We have

momentum = 20 × 5

We have the final answer as

Hope this helps you

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Answer:

40 m/s^2

Explanation:

Mass= 0.40 kg

Force= 16 N

Therefore the acceleration can be calculated as follows

F = ma

16= 0.40 × a

16= 0.40 a

a= 16/0.40

a= 40 m/s^2

Hence the acceleration is 40 m/s^2

Answer:

Vy = 80.5 [m/s]

Explanation:

In order to solve this problem we must use the Pythagorean theorem.

V = 90 [m/s]

The components are Vx and Vy:

Therefore:

[tex]v=\sqrt{v_{x}^{2} + v_{y}^{2} }[/tex]

where:

Vy = 2*Vx ; because one is twice of the other.

[tex]90 = \sqrt{v_{x}^{2} +(2*v_{x})^{2} }\\ 90 =\sqrt{v_{x}^{2}+4*v_{x}^{2}} \\90 =\sqrt{5v_{x}^{2}} \\90=2.23*v_{x} \\v_{x}=40.25[m/s][/tex]

and the bigger vector is:

Vy = 40.25*2

Vy = 80.5 [m/s]

Answer:

The thermal conductivity  [tex]k  = 1.4094   W/ m\cdot K[/tex]

Explanation:

From the question we are told that

  The  depth of the thermocouple from the surface is  x =  10 mm  = 0.01 m

   The  temperature is  [tex]T_f  =  100 ^o C[/tex]

   The  initial temperature is   [tex]T_i  =  30 ^o C[/tex]

    The  temperature of the thermocouple after t =  2 minutes( 2 * 60 =  120 \ seconds)  is   [tex]T_t  =  65 ^o C[/tex]

    The  density of the material  is  [tex]\rho =  2200 kg/m^3[/tex]

     The specific heat of the solid [tex]c_s  =  700 J/kg \cdot K[/tex]

Generally  the  equation for  semi -infinite medium  is mathematically as  

    [tex]\frac{T_s - T }{T_i - T} =  erf [\frac{x}{2 \sqrt{\alpha  * t} } ][/tex]

     [tex]\frac{65 - 100 }{30 - 100} =  erf [\frac{x}{2 \sqrt{\alpha  * t} } ][/tex]

        [tex]0.5 =  erf [\frac{0.01}{2 \sqrt{\alpha  * 120} } ][/tex]

Here [tex]\alpha[/tex] is a constant with unit [tex]m^2 /s[/tex]

  [tex]\frac{0.01}{ 2 (\sqrt{\alpha  *  120 } )}[/tex]   this is from the Gaussian function table  

    [tex]0.0 1 =  0.954 * (\sqrt{\alpha * 120  } )[/tex]

=>   [tex]\sqrt{\alpha  * 120  } =  \frac{0.01 }{0.954 }[/tex]

=>   [tex]\alpha =  9.1525 *10^{-7} \  m^2 /s[/tex]

Generally the thermal  conductivity is mathematically represented as

     [tex]k  =  \alpha  *  \rho * c_s[/tex]

      [tex]k  = 9.1525 *10^{-7}   *  2200 * 700[/tex]

     [tex]k  = 1.4094   W/ m\cdot K[/tex]

Answer:

 E ’= E / 8

therefore the correct answer is A

Explanation:

Let's calculate the electric field in an insulating sphere with a radius r <R, let's use Gauus's law, with a spherical Gaussian surface

       Фi = ∫ E. dA = [tex]q_{int}[/tex] /ε₀

       E (4πr²) = q_{int} / ε₀

density is

      ρ = q_{int} / V

      q_{int} = ρ V = ρ 4/3 π r³

we substitute        

        E (4π r²) = ρ 4/3 π r³ /ε₀

        E = 1 /3ε₀   ρ r

let's change the density by

       ρ = Q / V = ​​Q / (4/3 π R³)

         E = 1 / 4πε₀ Q r / R³

if we now distribute the same charge on a sphere of radius R' = 2R

        E ’= 1 / 4pieo Q r / (2R)³

        E ’= 1 / 4ft Qr / R³ ⅛

        E ’= E / 8

therefore the correct answer is A

Answer:Surface 1 is blacktop, Surface 2 is gravel, and Surface 3 is ice.

Explanation:

Answer: its C

Explanation:

Answer:

The final speed of the second package is twice as much as the final speed of the first package.

Explanation:

Free Fall Motion

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

[tex]v=gt[/tex]

And the distance traveled downwards is:

[tex]\displaystyle y=\frac{gt^2}{2}[/tex]

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

[tex]\displaystyle t=\sqrt{\frac{2y}{g}}[/tex]

Replacing into the first equation:

[tex]\displaystyle v=g\sqrt{\frac{2y}{g}}[/tex]

Rationalizing:

[tex]\displaystyle v=\sqrt{2gy}[/tex]

Let's call v1 the final speed of the package dropped from a height H. Thus:

[tex]\displaystyle v_1=\sqrt{2gH}[/tex]

Let v2 be the final speed of the package dropped from a height 4H. Thus:

[tex]\displaystyle v_2=\sqrt{2g(4H)}[/tex]

Taking out the square root of 4:

[tex]\displaystyle v_2=2\sqrt{2gH}[/tex]

Dividing v2/v1 we can compare the final speeds:

[tex]\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}[/tex]

Simplifying:

[tex]\displaystyle v_2/v_1=2[/tex]

The final speed of the second package is twice as much as the final speed of the first package.

Answer:

478.75 J

Explanation:

W=force* displacement

constant speed= (a=0) net F=0

Horizontal component of tension

Tcosx

125Ncos40= 95.76 N

W= (95.76 N)(5 m)

=478.75 J

The work done in moving the crate across the given distance is 478.75 J.

The given parameters;

The work done in moving the crate is the product of the horizontal component of the tension and the distance through which the crate is moved.

The work-done in moving the crate is calculated as;

W = Tcos(Ф) x d

W = 125cos(40) x 5

W = 478.75 J.

Thus, the work done in moving the crate across the given distance is 478.75 J.

Learn more here: https://brainly.com/question/19498865

Answer:

A soil

Explanation:

Sand is a granular material composed of finely divided rock and mineral particles. It is defined by size, being finer than gravel and coarser than silt. Sand can also refer to a textural class of soil or soil type; i.e., a soil containing more than 85 percent sand-sized particles by mass.

Answer:

mixture

sand is a mixture

Answer:

When the mouse broke free from the hawk, its vertical velocity is zero since the mouse just fell from the grips of the hawk

We are given:

initial vertical velocity (v) = 0 m/s

initial horizontal velocity (u) = 10.1 m/s

height from the ground (h) = 15 m

acceleration due to gravity (a) = 10 m/s/s

final vertical velocity (vf) = v m/s

Solving for Final vertical velocity:

from the third equation of motion

(vf)² - (v)² = 2ah

replacing the variables

v² - 0² = 2(10)(15)

v² = 300

v = √300

v = 10√3     OR    17.3 m/s

Answer:

We are given:

mass of the truck (m) = 3000 kg

velocity of the truck (v) = 25 m/s

Calculating the Momentum:

We know that the momentum of an object is equal to its mass times it's velocity

P = mv           (where P is the momentum of the truck)

Momentum of the truck:

replacing with the values of the truck

P = mv

P = 3000 * 25

P = 75000 kg m/s

Therefore, the momentum of the truck is 75000 kg m/s

Answer:

Making an observation

Explanation:

The use of senses as a tool to gather information is described as making an observation.

While making an observation, the senses must be at alert.

Therefore, the act of using senses or tools to gather information is called making an observation.

The question is missing parts. The complete question is here.

The human circulatory system is closed, that is, the blood pumped out of the left ventricle of the heart into the arteries is constrained to a series of continuoous, branching vessels as it passes through the capillaries and then into the veins as it returns to the heart. The blood in each of the heart's four chambers comes briefly to rest before it is ejected by cotraction of the heart muscle. If the aorta (diameter [tex]d_{a}[/tex]) branches into two equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches?

(a) [tex]\sqrt{d_{a}}[/tex]

(b) [tex]\frac{d_{a}}{\sqrt{2} }[/tex]

(c) [tex]2d_{a}[/tex]

(d) [tex]\frac{d_{a}}{2}[/tex]

Answer: (b) [tex]\frac{d_{a}}{\sqrt{2} }[/tex]

Explanation: The cross-sectional area of a vessel is a circle. So area is:

[tex]A=\pi.r^{2}[/tex]

Radius is half of a diameter, i.e.:

[tex]r=\frac{d}{2}[/tex]

Suppose [tex]d_{b}[/tex] is diameter of the branches, radius of one the branches is:

[tex]r_{b}=\frac{d_{b}}{2}[/tex]

Both branches are equal sized, which means, both have the same radius. So, combined area of both branches is:

[tex]A_{b}=\pi\frac{d_{b}^{2}}{4} +\pi\frac{d_{b}^{2}}{4}[/tex]

[tex]A_{b}=2\frac{d_{b}^{2}}{4}[/tex]

Area of aorta is

[tex]A_{a}=\pi.(\frac{d_{a}}{2})^{2}[/tex]

[tex]A_{a}=\pi.\frac{d_{a}^{2}}{4}[/tex]

Area of aorta is equal the combined area of the branches, then:

[tex]2\pi\frac{d_{b}^{2}}{4}= \pi\frac{d_{a}^{2}}{4}[/tex]

Rearraging:

[tex]d_{b}^{2}=\frac{d_{a}^{2}}{2}[/tex]

[tex]d_{b}=\frac{d_{a}}{\sqrt{2} }[/tex]

The diameter of one of the branches is [tex]\frac{d_{a}}{\sqrt{2}}[/tex].

Answer:

Explanation: someone help me please you want free points right

Answer:

C

Explanation:

Static electricity is what makes your hair stand up when you rub a balloon against it or gives you a shock from your doorknob. In static electricity, electrons are moved around mechanically (i.e. by someone rubbing two things together).

HOPE THIS HELPED AND LETTER D MADE ME LAUGH

Given parameters:

Mass of puck = 162g  = 0.162kg (1000g = 1kg)

Force exerted on puck = 171N

Final velocity  = 42.3m/s

Unknown

A.  time of the acceleration

B. radius of the curve?

Solution:

A. time of the acceleration

the initial velocity of the puck = 0m/s

    We know that;

              Force  = mass x acceleration

      Acceleration  = [tex]\frac{Final velocity - Initial velocity}{time taken}[/tex]

       Acceleration  = [tex]\frac{42.3 - 0}{t}[/tex]

  So force  = mass x [tex]\frac{42.3 }{t}[/tex]

 Input the parameters and solve for time;

             171 = 0.162 x [tex]\frac{42.3 }{t}[/tex]

             171 = [tex]\frac{6.85}{t}[/tex]  

                 t = [tex]\frac{6.85}{171}[/tex]   = 0.04s

The time of acceleration is 0.04s

B. radius of the curve;

      to solve this, we apply the centripetal force formula;

               F  = [tex]\frac{mv^{2} }{r}[/tex]

  where;

      F is the centripetal force

     m is the mass

      v is the velocity

      r is the radius

               Since the force exerted on the puck is 151;

      input the parameters and solve for r²;

             151 = [tex]\frac{0.162 x 42.3^{2} }{r}[/tex]

               151r  = 0.162 x 42.3²

                r  = 1.92m

The radius of the circular curve is 1.92m

Answer:

D. The liquid is a mixture because it has a gas dissolved in a liquid.

Answer:

The time for the change in the angular velocity to occur is 14.08 secs

Explanation:

From the question,

the angular acceleration is - 4.46 rad/s²

Angular acceleration is given by the formula below

[tex]\alpha =\frac{\omega -\omega _{o} }{t - t_{o} }[/tex]

Where [tex]\alpha[/tex] is the angular acceleration

[tex]\omega[/tex] is the final angular velocity

[tex]\omega _{o}[/tex] is the initial angular velocity

[tex]t[/tex] is the final time

[tex]t_{o}[/tex] is the initial time

From the question

[tex]\alpha[/tex] = - 4.46 rad/s²

[tex]\omega _{o}[/tex] = 0 rad/s (starting from rest)

[tex]\omega[/tex] = -31.4 rad/s

[tex]t_{o}[/tex] = 0 s

Now, we will determine t

From [tex]\alpha =\frac{\omega -\omega _{o} }{t - t_{o} }[/tex], then

[tex]-4.46 = \frac{-31.4 - 0}{t - 0}[/tex]

[tex]-4.46 = \frac{-31.4}{t}[/tex]

[tex]t = \frac{-31.4}{-4.46}[/tex]

t = 7.04 secs

This is the time spent in one direction,

Since the angular displacement of the wheel is zero ( it returned to its initial position), then the time required for the change in the angular velocity will be twice this time, that is 2t

Hence,

The time is 2×7.04 secs = 14.08 secs

This is the time for the change in the angular velocity to occur.

Answer:

Explanation:

The density of a substance can be found by using the formula

[tex]density = \frac{mass}{volume} \\[/tex]

From the question

mass = 100 g

volume = 50 mL

We have

[tex] density = \frac{100}{50} = \frac{10}{5} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

transition metal, and inner transition metals groups are numbered 1-18 from left to right

Answer:

a) Time it will taken for the column surface temperature to rise to 27°C is  

17.1 hours

b) Amount of heat transfer is 5320 kJ  

c) Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ

Explanation:

Given that;

Diameter D = 30 cm

Height H = 4m

heat transfer coeff h = 14 W/m².°C

thermal conductivity k = 0.79 W/m.°C

thermal diffusivity α  = 5.94 × 10⁻⁷ m²/s

Density p = 1600 kh/m³

specific heat Cp = 0.84 Kj/kg.°C

a)

the Biot number is

Bi = hr₀ / k

we substitute

Bi = (14 W/m².°C × 0.15m) / 0.79 W/m.°C

Bi = 2.658

From the coefficient for one term approximate of transient one dimensional heat conduction The constants λ₁ and A₁ corresponding to this Biot number are,  

λ₁ = 1.7240

A₁ = 1.3915

Once the constant J₀ = 0.3841 is determined from corresponding to the constant λ₁

the Fourier number is determined to be  

[ T(r₀, t) -T∞ ] / [ Ti - T∞]  = A₁e^(-λ₁²t') J₀ (λ₁r₀ / r₀)

(27 - 28) / (14 - 28)   = (1.3915)e^-(17240)²t (0.3841)  

t' = 0.6771

Which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used. Then the time it will take for the column surface temperature to rise to 27°C becomes  

t =  t'r₀² / ₐ

= (0.6771 × 0.15 m)² /  (5.94 x 10⁻⁷ m²/s)

= 23,650 s

= 7.1 hours

Time it will taken for the column surface temperature to rise to 27°C is  

17.1 hours

b)

The heat transfer to the column will stop when the center temperature of column reaches to the ambient temperature, which is 28°C.  

Maximum heat transfer between the ambient air and the column is

m = pV

= pπr₀²L

= (1600 kg/m³ × π × (0.15 m)² × (4 m)

= 452.389 kg

Qin = mCp [T∞ - Ti ]

= (452.389 kg) (0.84 kJ/kg.°C) (28 - 14)°C

= 5320 kJ  

Amount of heat transfer is 5320 kJ  

(c)

the amount of heat transfer until the surface temperature reaches to 27°C is

(T(0,t) - T∞) / Ti - T∞  = A₁e^(-λ₁²t')

= (1.3915)e^-(1.7240)² (0.6771)

= 0.1860

Once the constant J₁ = 0.5787 is determined from Table corresponding to the constant λ₁, the amount of heat transfer becomes  

(Q/Qmax)cyl = 1 - 2((T₀ - T∞) / ( Ti - T∞)) ((J₁(λ₁)) / λ₁)

= 1 - 2 × 0.1860 × (0.5787  / 1.7240)  

= 0.875

Q = 0.875Qmax

Q = 0.875(5320 kJ)  

Q = 4660 kJ

Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ