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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

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A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 375 N

spayn [35]

Answer:

The conservation of energy should be used to answer this question.

At the position where the spring is unstretched, the elastic potential energy of the spring is zero.

$K_1 + U_1 - W_f = K_2 +U_2\\K_1 - W_f = U_2$

since $U_1$ and $K_2$ is equal to zero.

$W_f = F_fx\\\\U_2 = \frac{1}{2}kx^2\\\\19 - (10)x = \frac{1}{2}(375)x^2\\\\375x^2 + 20x - 38 = 0$

The roots of this quadratic equation can be solved by using discriminant.

$\Delta = b^2 - 4ac\\x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a}$

$x_1 = -0.346\\x_2 = 0.292$

We should use the positive root, so

x = 0.292 m.

We should use energy conservation between the point where the spring is momentarily at rest, and the point where the spring is unstretched.

$K_2 + U_2 - W_f = K_3 + U_3\\U_2 - W_f = K_3$

since the kinetic energy at point 2 and the potential energy at point 3 is equal to zero.

$\frac{1}{2}kx^2 - F_fx = K_3\\K_3 = 15.987 - 2.92 = 13.067 J$

Explanation:

In questions with springs, the important thing is to figure out the points where kinetic or potential energy terms would be zero. When the spring is unstretched, the elastic potential energy is zero. And when the spring is at rest, naturally the kinetic energy is equal to zero.

In part b) the cookie slides back to its original position, so the distance traveled, x, is equal to the distance in part a). The frictional force is constant in the system, so it is quite simple to solve part b) after solving part a).

8 0

2 years ago

"A student bikes to school by traveling first dN = 0.900" and "Similarly, let d⃗ W be the displacement vector corresponding to t

raketka [301]

Complete Question

A student bikes to school by traveling first dN = 0.900 miles north, then dW = 0.300 miles west, and finally dS = 0.100 miles south.

Similarly, let d⃗ W be the displacement vector corresponding to the second leg of the student's trip. Express d⃗ W in component form.

Express your answer as two numbers separated by a comma. Be careful with your signs.

Answer:

The value is $dT = ( -0.3, 0.8)$