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Wittaler [7]
3 years ago
14

A house is advertised as having 1460 square feet under roof. What is the area of this house in square meters?

Physics
2 answers:
Vladimir [108]3 years ago
8 0
1400 is the story of the president Barack Obama
Wewaii [24]3 years ago
6 0

Answer: 131.92 m

Explanation: The problem give above is a conversion problem. To convert any measurement from one unit to another, we need a relation for the original units to that of the desired unit. This relation is called the conversion factor. It is used by multiplying or dividing the factor depending on what is the unit of the original measurement. For this case, we need to convert from ft^2 to m^2. The conversion factor would be 1 foot is equal to 0.3048 m. So, we would be getting the square of the factor and then multiply to the original value.

1420 ft^2 ( 0.3048 m / 1 ft)^2 = 131.92 m

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A ball bearing of radius of 1.5 mm made of iron of density
Serjik [45]

Answer:

\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine (\sf \eta)

Explanation:

\boxed{ \bold{v =  \frac{2}{9}  \frac{( {r}^{2} ( \rho -  \sigma)g)}{ \eta} }}

\sf \implies \eta =  \frac{2}{9}  \frac{( {r}^{2}( \rho -  \sigma)g )}{v}

Substituting values of r, ρ, σ, v & g in the equation:

\sf \implies \eta =  \frac{2}{9}  \frac{( {(0.15)}^{2}  \times  (7.85 - 1.25) \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times  \frac{145.6191}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times 64.7196

\sf \implies \eta =  2 \times 7.191

\sf \implies \eta =  14.382 \: poise

6 0
3 years ago
After using soap to wash dishes by hand, if is sometimes difficult to keep your hands from remaining stick. Explain why rinshing
gulaghasi [49]

<h2>HOPE THIS HELPS YOU ....</h2><h3>PLEASE MARK ME AS BRAINILIST...</h3>

6 0
3 years ago
At what wavelength would a star radiate the greatest amount of energy if the star has a surface temperature of 60,000 K?
kompoz [17]

Answer:

\lambda=4.81\times 10^{-8}\ m

Explanation:

We have,

The surface temperature of the star is 60,000 K

It is required to find the wavelength of a star that radiated greatest amount of energy. Wein's displacement law gives the relation between wavelength and temperature such that :

\lambda T=2.89\times 10^{-3}

Here,

\lambda = wavelength

\lambda=\dfrac{2.89\times 10^{-3}}{60000}\\\\\lambda=4.81\times 10^{-8}\ m

So, the wavelength of the star is 4.81\times 10^{-8}\ m.

7 0
3 years ago
Determine the magnitude of the resultant force FR=F1+F2FR=F1+F2. Assume that F1F1F_1 = 235 lblb and F2F2F_2 = 350 lb
Svetlanka [38]

Answer:

<em>585lb</em>

Explanation:

Given the formula for calculating the magnitude of the resultant force as;

F_R = F_1 + F_2

<em>Given </em>

<em></em>F_1 = 235 lb \ and \ F_2 = 350 lb\\F_R = 235lb + 350lb\\F_R = 585lb\\<em></em>

<em>Hence the magnitude of the resultant force is 585lb</em>

4 0
3 years ago
Is this statement true or false concerning squall line thunderstorm development? These often form ahead of the advancing front b
Pavlova-9 [17]

Answer: The following statement is true about squall line thunderstorm development: <em><u>These often form ahead of the advancing front but rarely behind it because lifting of warm, humid air and the generation of a squall line usually occur in the warm sector ahead of an advancing cold front. Behind a cold front, the air motions are usually downward, and the air is cooler and drier.</u></em>

<em>An upper-level wave, accountable for the fabrication of a squall line, extend in front of and backside a cold front, the air backside the front is cold, steady and settling while the air ahead of the front is hot and co-seismic.</em>

3 0
3 years ago
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