All categories

3 years ago

What is the heat-loss rate through the slab if the ground temperature is 5 ∘C while the interior of the house is 25 ∘C?

Physics

1 answer:

Gala2k [10]3 years ago

3 0

Complete question :

A 12 m x 15 m house is built on a 12-cm-thick concrete slab.

What is the heat-loss rate through the slab if the ground temperature is 5°C while the interior of the house is 25°C

Answer:

3kW

Explanation:

Given the following :

Dimension of house :

Length = 12m

Width = 15m

Thickness of concrete slab (t) = 12cm

t in metres :

100cm = 1m

12cm = (12/100)m

= 0.12m

Ground temperature (Tg) = 5°C

Interior temperature = (Th) = 25°C

Thermal conductivity of concrete (K) is approximately 1 Wm/k

Using the relation:

Q = KA * [ (Th - Tg) / d]

A = Length * width = (12 *15) = 180

Q = (1 * 180) * [(25°C - 5°C) / 0.12]

Q = 180 * (20/0.12)

Q = 180 * 16.6666

Q = 3,000W = 3kW

You might be interested in

Please help!! This is a final and I need a good grade

Ket [755]

Diagram 4 is the correct answer.

6 0

3 years ago

A bus contains a 1440 kg flywheel (a disk that has a 0.63 m radius) and has a total mass of 10200 kg. Calculate the angular velo

CaHeK987 [17]

Answer: $\omega =93.51 rad/s$

Explanation:

Given

mass of Flywheel $m_1=1440 kg$

mass of bus $m_b=10200 kg$

radius of Flywheel $r=0.63 m$

final speed of bus $v=21 m/s$

Conserving Energy i.e.

0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus

Let $\omega$ be the angular velocity of Flywheel

$0.9\cdot \frac{I\omega ^2}{2}=\frac{m_bv^2}{2}$

$I=moment\ of\ Inertia =mr^2=1440\cdot 0.63^2=571.536 kg-m^2$

$0.9\cdot \frac{571.536\cdot \omega ^2}{2}=\frac{10200\cdot 21^2}{2}$

$\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}$

$\omega =21\times 4.45=93.51 rad/s$

8 0

2 years ago

A "sound" is different than a "sound wave"

e-lub [12.9K]

Answer:

sound is how an animal or human perceives a sound wave

5 0

1 year ago

What determines if a star turns into a red giant or a red supergiant?

mixer [17]

Answer:

Explanation:

As stars age, they evolve away from the main sequence to become red giants or supergiants. The core of a red giant is contracting, but the outer layers are expanding as a result of hydrogen fusion in a shell outside the core. The star gets larger, redder, and more luminous as it expands and cools.

3 0

2 years ago

A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the

Aleksandr-060686 [28]

Answer:

(a): The magnitude of the electric force on the small sphere = $\dfrac{q\sigma}{2\epsilon_o}.$

(b): Shown below.

Explanation:

<u>Given:</u>

m = mass of the small sphere.
q = charge on the small sphere.
L = length of the silk fiber.
$\sigma$ = surface charge density of the large vertical insulating sheet.

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

$\rm E = \dfrac{\sigma}{2\epsilon_o}.$

<em>where,</em>

$\epsilon_o$ is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

$\rm F_e=qE.$

Thus, the magnitude of the electric force on the small sphere is given by

$\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.$

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

$\rm \tan \theta = \dfrac{F_e}{W}.$

<em>where,</em>

$\rm F_e$ = electric force on the sphere, acting along left.
$\rm W$ = weight of the sphere, acting vertically downwards.

<em />

$\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .$

g is the acceleration due to gravity.

6 0

3 years ago