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3 years ago

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>

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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:

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a 1.5 kg ball is thrown vertically upward with an initial speed of 15 m/s. if the initial potential energy is taken as zero, fin

trapecia [35]

Answer:

a) $E_{p} = 0$

$E_{k} = 168.7 J$

$E_{m} = 168.7 J$

b) $E_{p} = 73.6 J$

$E_{k} = 95.8 J$

$E_{m} = 169.4 J$

c) $E_{p} = 169.2 J$

$E_{k} = 0$

$E_{m} = 169.2 J$

Explanation:

We have:

m: is the ball's mass = 1.5 kg

v₀: is the initial speed = 15 m/s

g: is the gravity acceleration = 9.81 m/s²

a) In the initial position we have:

h: is the height = 0

The potential energy is given by:

$E_{p} = mgh = 0$

The kinetic energy is:

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(15)^{2} = 168.7 J$

And the mechanical energies:

$E_{m} = E_{p} + E_{k} = 0 + 168.7 J = 168.7 J$

b) At 5 m above the initial position we have:

h = 5 m

The potential energy is:

$E_{p} = mgh = 1.5*9.81*5 = 73.6 J$

Now, to find the kinetic energy we need to calculate the speed at 5 m:

$v_{f}^{2} = v_{0}^{2} - 2gh = (15)^{2} - 2*9.81*5 = 126.9$

$v_{f} = \sqrt{126.9} = 11.3 m/s$

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(11.3)^{2} = 95.8 J$

And the mechanical energies:

$E_{m} = E_{p} + E_{k} = 73.6 + 95.8 J = 169.4 J$

c) At its maximum height:

$v_{f}$ : is the final speed = 0

$h = \frac{v_{0}^{2}}{2g} = \frac{(15)^{2}}{2*9.81} = 11.5 m$

Now, the potential, kinetic and mechanical energies are:

$E_{p} = mgh = 1.5*9.81*11.5 = 169.2 J$

$E_{k} = \frac{1}{2}mv^{2} = 0$

$E_{m} = 169.2 J + 0 = 169.2 J$

I hope it helps you!

7 0

2 years ago

Two campers dock a canoe. One camper has a mass of 100.0 kg and moves forward at 3.0 m/s as he leaves the canoe to step onto the

Solnce55 [7]

Answer:

The combined speed of camper and canoe is 1.71 m/s.

Explanation:

Given that,

Mass of camper 1, m = 100 kg

Speed of camper 1, v = 3 m/s

The combined mass of another camper and canoe is, M = 175 kg

We need to find the combined speed of camper and canoe. According to the conservation of linear momentum, the momentum of first camper is equal to linear momentum of the canoe and the second camper.

$mv=MV\\\\V=\dfrac{mv}{M}\\\\V=\dfrac{100\times 3}{175}\\\\V=1.71\ m/s$