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Rufina [12.5K]
3 years ago
10

Imagine we cook an egg by immersing it into water which is boiled by an electric heater. The heater utilizes a current I =10 A a

t a voltage V = 120 volts for a time t = 120 s lf the change in energy of the newly-cooked egg over its raw torm is given by ΔEngg = 28.8 kJ, what is the amount of energy wasted in the process?
Physics
1 answer:
tatuchka [14]3 years ago
8 0

Answer:

Q' = 115.2 KJ

Explanation:

Given that

Current I = 10 A

Voltage = 120 V

Time t= 120 s

The energy supplied Q= V I t

Q= 10 x 120 x 120 J

Q= 144 x 1000 J

Q= 144 KJ

The change in the energy ΔEngg = 28.8 kJ

By using energy conservation

Q= Q' +  ΔEngg

Q'=Wasted energy

Now by putting the values in the above equation

144 = Q' + 28.8 KJ

Q' = 115.2 KJ

Therefore the waste energy will be 115 kJ.

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DerKrebs [107]

Answer:

they rise in temperature

Explanation:

when there being compressed theres more pressure causing heat

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If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2
kaheart [24]

Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

\Delta T is the temperature of difference that we want to find

\Delta x=2.5 cm =0.025 m represent the thickness of the material

If we solve \Delta T in absolute value from the equation (1) we got:

\Delta T =\frac{Q \Delta x}{Ak}

First we convert 3KW to W and we got:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

And we have everything to replace and we got:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

5 0
3 years ago
A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness
Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

\sigma_h = \frac{Pd}{2t}

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

\sigma_l = \frac{Pd}{4t}

The letters have the same meaning as before.

Then he hoop stress would be,

\sigma_h = \frac{Pd}{2t}

\sigma_h = \frac{75 \times 8}{2\times 0.25}

\sigma_h = 1200psi

And the longitudinal stress would be

\sigma_l = \frac{Pd}{4t}

\sigma_l = \frac{75\times 8}{4\times 0.25}

\sigma_l = 600Psi

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

\tau_{max} = \frac{\sigma_h}{2}

\tau_{max} = \frac{1200}{2}

\tau_{max} = 600Psi

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

6 0
3 years ago
The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the dist
Sidana [21]

Complete Question

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2

Answer:

The acceleration due to gravity is  g = 167.2 \ m/s^2  

Explanation:

From the question we are told that

     The length of the simple pendulum is L = 1.081.08 \ m

      The number of cycles is  N =  101

       The time take is  t =  2.00 *10^{2 \ }s

Generally the period of this oscillation is mathematically evaluated as

         T = \frac{N}{t }

substituting values

         T = \frac{101}{2.0*10^2 }

        T = 0.505 \  s

The period of this oscillation is mathematically represented  as

               T = 2 \pi \sqrt{\frac{l}{g} }

making g the subject of the formula we have

              g = \frac{L}{[\frac{T}{2 \pi } ]^2 }

              g = \frac{4 \pi ^2 L }{T^2 }

Substituting values

               g = \frac{4 * 3.142 ^2  * 1.08 }{505.505^2 }

               g = \frac{4 * 3.142 ^2  * 1.08 }{0.505^2 }  

              g = 167.2 \ m/s^2  

7 0
3 years ago
The combustion of propane (C3H8) in the presence of excess oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g) W
zimovet [89]

Answer:

1.2

Explanation:

2.0 mol O₂ × (3 mol CO₂ / 5 mol O₂) = 1.2 mol CO₂

4 0
3 years ago
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