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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.

Explanation:

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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jack peddles a stationary bicycle. the pedals of the bicycle are attached to a 1000 turn coil of area 0.10 m2. the coil rotates

tatuchka [14]

The maximum voltage generated in the coil is 3.14 V.

According to the question, we are provided that a bicycle turns 1000 of the area of 0.10 m2, where the coil rotates half a revolution per second, which is placed in a uniform magnetic field of 0.01 T.

Therefore, no .of turns the coil took, N = 1000

Area of the coil, A= 0.10 m²

Frequency of the system =0.5 revolutions per second

= 0.5 Hz

The intensity of the Magnetic field , B =0.01 T

So the Equation we refer to for the calculating the desired value

E(ind) = $-db/dt$

= -d(NBAcosωt) / dt

= − NBA d(cosωt) /dt

= NBAωsinωt

if sinωt=1,

maximum E(ind)= NBAω

Since we know, ω=2πf

Therefore, (Eind)max = NBA(2πf)

= (1000)(0.01)(0.1)(2π)(0.5)

(E(ind))max= 3.14v.

the value for the voltage we got is 3.14 V

To know more about Magnetic Field brainly.com/question/23096032?referrer=searchResults.

#SPJ4

7 0

2 years ago

Astronauts on a distant planet set up a simple pendulum of length 1.2 m. The pendulum executes simple harmonic motion and makes

Iteru [2.4K]

Answer:

If you use P = 2 * pi * (L / g)^1/2 for the period of the simple pendulum

g = 4 * pi^2 * 1.2 / 2.8^2 = 6.04 m/s2

Note: omega = 2 pi * f = 2 pi / P and omega = (g / L)^1/2

7 0

3 years ago

A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c

viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = $71\times 10^{- 9} C$

Q' = + 42 nC = $42\times 10^{- 9} C$

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

$\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E} = 708.03 N/C$

Now,

Electric field at the mid-point due to charge Q' is given by:

$\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E'} = 418.84 N/C$

Now,

The net Electric field is given by:

$\vec{E_{net}} = \vec{E} - \vec{E'}$

$\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C$

5 0

3 years ago

A current of 3.60A flows for 15.3s through a conductor. Calculate the number of electrons that pass through a point in the condu

Artyom0805 [142]

3.60 A = 3.60 coulombs of charge per second

(3.60 Coul/sec) x (15.3 sec) = 55.08 coulombs of charge

1 coulomb of charge is carried by 6.25 x 10^18 electrons

Number of electrons =

(55.08 Coul) x (6.25 x 10^18 e/coul) = 3.4425 x 10^20 electrons

7 0

4 years ago

Please Help Physics!

Anton [14]

Answer:

The answer is below!!

Explanation:

The ratio of the wavelength of the waves that travel on the string to the wavelength of the generated sound wave is 0.585

Hope I Helped!!

6 0

2 years ago