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Airida
6 days ago
5

What is the thickness required of a masonry wall having thermal conductivity 0.75 W/m K if the heat rate is to be 80% of the hea

t rate through a composite structural wall having a thermal conductivity of 0.25 W/m K and a thickness of 100 mm? Both walls are subjected to the same surface temperature difference.

Physics
1 answer:
Maru [1K]6 days ago
5 0

Answer:

The required thickness for a masonry wall is L = 375mm

Explanation:

The methodology with clear steps and proper application of Fourier's law for heat conduction is detailed in the attached file.

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Which trailer has more downward pressure where it attaches to the car?
serg [1198]

The trailer that is loaded the most. The total weight does not matter; it is about how the load is distributed. For example, our 12,000 lb snow cat trailer has weight distribution that results in less than 100 lbs of tongue weight. Heavy tongue weight can create issues, as it can shift the weight off the front wheels of the towing vehicle, causing instability.

4 0
4 days ago
An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how mu
Yuliya22 [1153]

Answer:

Wnet, in, = 133.33J

Explanation:

Provided that

Pump heat QH = 1000J

Hot temperature TH= 300K

Cold temperature TL= 260K

Given the heat pump is entirely reversible, the performance coefficient expression is formulated as follows:

According to the first law of thermodynamics,

COP(HP, rev) = 1/(1-TL/TH)

COP(HP, rev) = 1/(1-260/300)

COP(HP, rev) = 1/(1-0.867)

COP(HP, rev) = 1/0.133

COP(HP, rev) = 7.5

The power necessary to operate the heat pump is given by

Wnet, in = QH/COP(HP, rev)

Wnet, in = 1000/7.5

Wnet, in = 133.333J. QED

Thus, the 133.33J represents the initial work input during the heat transfer process.

<padditionally...><pbased on="" the="" first="" law="" rate="" at="" which="" heat="" is="" extracted="" from="" lower="" temperature="" reservoir="" calculated="" as="">

QL=QH-Wnet, in

QL=1000-133.333

QL=866.67J

</pbased></padditionally...>
5 0
1 day ago
Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller
Softa [913]

Answer:

σ₁ = 3.167 * 10^{-6} C/m²

σ₂ = 7.6 * 10 ^{-6}  C/m²

Explanation:

Provided Information:

i) Smaller sphere's radius ( r ) = 5 cm.

ii) Larger sphere's radius ( R ) = 12 cm.

iii) Electric field at the larger sphere's surface  ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m

E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )

358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }

Charge (Q₁) = 572.8 * 10^{-9} C

Since the electric field inside a conductor is zero, the electric potential ( V ) remains constant.

V = constant

∴\frac{Q_{1} }{R} = \frac{Q_{2} }{r}

Q_{2} = \frac{r}{R} *Q_{1}

Q_{2} = \frac{5}{12} *572.8*10^{-9}   = 238.666 *10^{-9} C

Surface charge density ( σ₁ ) for the larger sphere.

Calculated Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = \frac{Q_{1} }{A_{1} } = \frac{572.8 *10^{-9} }{0.180864} = 3.167 * 10^{-6}  C/m².

Surface charge density ( σ₂ ) for the smaller sphere.

Calculated Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  = 0.0314 m².

σ₂ =\frac{Q_{2} }{A_{2} } = \frac{238.66 *10^{-9} }{0.0314} = 7.6 * 10 ^{-6} C/m²

8 0
4 days ago
While looking at bromine (Br) on the periodic table, a student needs to find another element with very similar chemical properti
Sav [1105]

Response: Numerous elements can be found, all situated within the same vertical column as bromine.

Explanation:

Elements are organized by their atomic numbers on the periodic table. Those in the same vertical column (known as groups) exhibit the same valence electron configurations, resulting in similar chemical characteristics. Consequently, there are numerous elements sharing analogous chemical properties grouped with Bromine.

7 0
12 days ago
Read 2 more answers
A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e
kicyunya [1025]

Answer:

Maximum emf = 5.32 V

Explanation:

Provided data includes:

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions each second

Magnetic field, B = 0.5 T

We are tasked to determine the maximum emf produced in the loop, which is founded on Faraday's law. The induced emf can be calculated by:

\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t

For the maximum emf, \sin\omega t=1

Therefore,

\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V

Hence, the maximum emf generated in the loop is 5.32 V.

3 0
8 days ago
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