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faltersainse [42]
3 years ago
15

Which considerations are used to calculate a windchill factor? Select two options.

Physics
2 answers:
vlada-n [284]3 years ago
7 0

Wind speed and air temperature are used to calculate a windchill factor.

<u>Explanation:</u>

<u></u>

Wind-chill factor is the reduction of body temperature due to the passing flow of lower-temperature air.

The air temperature  value is always higher than the wind chill numbers. the heat index will be used if the apparent temperature is higher than the air temperature.So, Wind speed and air temperature are mainly used to calculate a windchill factor.

There are many ways, the surface loses its heat through conduction, evaporation,radiation, and convection.The rate of convection depends on  the difference in temperature between the surface and the fluid surrounding the surface and the velocity of that fluid with respect to the surface. The air around the warm surface will be heated, an insulating  layer of warm air forms against the surface.The layer becomes a boundary between two. As the wind speed is high the surface cools down rapidly.

mamaluj [8]3 years ago
6 0

Answer:

wind speed and air temperature

Explanation:

\(0-0)/

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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o
matrenka [14]

Answers:

(a) 0.0073kg

(b) Volume gold: 3.79(10)^{-7}m^{3}, Volume cupper: 7.6(10)^{-8}m^{3}

(c) 17633.554kg/m^{3}

Explanation:

<h2>(a) Mass of gold </h2><h2 />

We are told the total mass M of the coin, which is an alloy  of gold and copper is:

M=m_{gold}+m_{copper}=7.988g=0.007988kg   (1)

Where  m_{gold} is the mass of gold and m_{copper} is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats K and mass is:

K=24\frac{m_{gold}}{M}   (2)

Finding {m_{gold}:

m_{gold}=\frac{22}{24}M   (3)

m_{gold}=\frac{22}{24}(0.007988kg)   (4)

m_{gold}=0.0073kg   (5)  This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density \rho of an object is given by:

\rho=\frac{mass}{volume}

If we want to find the volume, this expression changes to: volume=\frac{mass}{\rho}

For gold, its volume V_{gold} will be a relation between its mass m_{gold}  (found in (5)) and its density \rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}:

V_{gold}=\frac{m_{gold}}{\rho_{gold}}   (6)

V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}   (7)

V_{gold}=3.79(10)^{-7}m^{3}   (8)  Volume of gold in the coin

For copper, its volume V_{copper} will be a relation between its mass m_{copper}  and its density \rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}:

V_{copper}=\frac{m_{copper}}{\rho_{copper}}   (9)

The mass of copper can be found by isolating m_{copper} from (1):

M=m_{gold}+m_{copper}  

m_{copper}=M-m_{gold}  (10)

Knowing the mass of gold found in (5):

m_{copper}=0.007988kg-0.0073kg=0.000688kg  (11)

Now we can find the volume of copper:

V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}   (12)

V_{copper}=7.6(10)^{-8}m^{3}   (13)  Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density \rho_{coin will be a relation between its total mass M and its total volume V:

\rho_{coin}=\frac{M}{V} (14)

Knowing the total volume of the coin is:

V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3} (15)

\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}} (16)

Finally:

\rho_{coin}=17633.554kg/m^{3}} (17)  This is the total density of the British sovereign coin

6 0
3 years ago
What is a longitudinal wave?
GenaCL600 [577]

Answer:

Longitudinal waves are waves in which the displacement of the medium is in the same direction as, or the opposite direction to, the direction of propagation of the wave. ... In longitudinal waves, the displacement of the medium is parallel to the propagation of the waves.

4 0
3 years ago
Formed through longshore drift<br> a. sea stack<br> b. sandbar<br> c. spit<br> d. headland
ANTONII [103]
<h3>Answer;</h3>

<em><u>Sand Spit or Spit </u></em>

<h3><u>Explanation;</u></h3>
  • <em><u>Long shore drift is the process that occurs when a sheet of water moves on and off the beach, in other words the swash and back swash</u></em>, thus capturing and transporting sediment on the beach back out to the sea.
  • <em><u>Sandbar</u></em> is normally formed when the sandspit stretches across a bay and connects the two sides. <em><u>Headland</u></em> is a high piece of land that extends out onto the sea. <em><u>Sea stacks </u></em>on the other hand results from the collapsing of the roof of the arch.
7 0
4 years ago
Read 2 more answers
A rotating cylinder about 16 km long and 7.8km in diameter is designed to be used as a space colony.
Semmy [17]

Answer:

0.05 rad/s

Explanation:

7.8 km = 7800 m

For the residents inside the space cylinder to experience the same gravitation acceleration g = 9.81m/s2 on Earth, the centripetal acceleration must be the same as g

a = g

But centripetal acceleration is product of angular velocity squared and radius of rotation r

\omega^2r = 9.81

\omega^2 d/2 = 9.81

\omega^2 = \frac{2*9.81}{d} = \frac{19.62}{7800} = 0.0025

\omega = \sqrt{0.0025} = 0.05 rad/s

3 0
3 years ago
Answers to question 2,3,4
Sati [7]

Answer:

OMG

these are sooooooo hard

Explanation:

5 0
3 years ago
Read 2 more answers
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