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Andrej [43]
3 years ago
7

A window in a house has a rectangular shape of 2.0 m by 1.0 m. The glass in the window is 0.5 cm thick, with a thermal conductiv

ity of 0.84 W/(m oC). The temperature in the house is 20.5 °C, and the temperature outside is 4 oC. How much heat is lost through this window in one minute? Don't use scientific notations
Physics
1 answer:
KonstantinChe [14]3 years ago
7 0

Answer:

 P = 5280 W

Explanation:

The conductivity of the materials determines that heat flows from the hot part to the cold part, the equation for thermal conductivity transfer is

    P = Q / t = k A (T_{h} -T_{c}) / L

Where k is the thermal conductivity of the glass 0.8 W / ºC, A the area of ​​the window, T the temperature and L is glass thickness

Let's calculate the window area

    A = l * a

    A = 2.0 1.0

    A = 2.0 m²

Let's replace

    L = 0.5 cm (1 m / 100 cm) = 0.005 m

    P = 0.8 2 (20.5 - 4) / 0.005

    P = 5280 W

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Answer:

Correct answer: Ek = 2,028.6 J

Explanation:

Ek = m v²/2 = 0.023 · 220² = 0.023 · 176,400/2 = 2,028.6 J

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8 0
3 years ago
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What are the components of the "Earth Radiation Budget?"
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4 0
3 years ago
An object is placed perpendicular to the the principal axis of convex lens of focal length 8,the distance of the object from the
Digiron [165]

Answer:

v = 24 cm and inverted image

Explanation:

Given that,

The focal length of the object, f = +8 cm

Object distance, u = -12 cm

We need to find the position &nature of the image​. Let v be the image distance. Using lens formula to find it :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Put all the values,

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{8}+\dfrac{1}{(-12)}\\\\v=24\ cm

So, the image distance from the lens is 24 cm.

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{24}{-12}\\\\m=-2

The negative sign of magnification shows that the formed image is inverted.

7 0
2 years ago
Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30 second time interval. We take th
Natalija [7]

Let say for every 5 s of time interval the speed will remain constant

so it is given as

v(mi/h)   16    21    23    26    33    30     28

now we have to convert the speed into ft/s as it is given that 1 mi/h = 5280/3600 ft/s

so here we will have

v(ft/s)      23.5    30.8     33.73     38.13     48.4     44     41.1

now for each interval of 5 s we will have to find the distance cover for above interval of time

d = v \times t

d = (23.5 + 30.8 + 33.73 + 38.13 + 48.4 + 44 + 41.1) \times 5

d = 1298.1 ft

so here it will cover 1298.1 ft distance in 30 s interval of time

4 0
3 years ago
Read 2 more answers
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