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Montano1993 [528]
3 years ago
7

A variable is defined as anything that

Physics
1 answer:
LenKa [72]3 years ago
3 0
Answer : Quantity or quality that varies like manipulating a variable to see what will happen to another variable
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g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located
Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is:  P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P  /RT ⋯ ⋯( I )

Here  

m  is the mass of the air,  

v  is the volume of the air,  

P  is the atmospheric pressure,  

T  is the standard temperature at the atmospheric pressure and  

R  is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101  kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here  A  is the area of the windmill

Expression to calculate the  A  is

A = π /4  d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94  k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94  k g / s ( (6m/s)²/2 )

P w = 160144.92 W  × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

n=13.11

8 0
3 years ago
A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 18.5 kg, and the
meriva

Answer:

a.2.86 m/s^2

b.1058 N

Explanation:

We are given that

Mass of each dog,M=18.5 kg

Mass of sled with rider,m=250 kg

a.Average force,F=185 N

\mu_s=0.14

g=9.8 m/s^2

By Newton's second law

8F-f=(8M+m)a

a=\frac{8F-f}{8M+m}=\frac{8(185)-(0.14)(9.8)(250)}{8(18.5)+250}

a=2.86 m/s^2

b.By Newton's second law

T=ma+\mu_s mg

Substitute the values

T=250\times 2.86+0.14(250)(9.8)=1058 N

Hence, the force in the coupling between the dogs and the sled=1058 N

8 0
3 years ago
The unit for measuring electric power is the
bazaltina [42]
C.) The measuring unit of "Electrical Power" is "Watt"

Hope this helps!
4 0
3 years ago
Un objeto de 1400 g de masa se mueve bajo la acción de una fuerza constante con una aceleración de 0,5 m/s2 , sobre una superfic
igomit [66]

Answer:

if you spoke this in english i can help you out

Explanation:

3 0
3 years ago
Describe a situation in which an electron will be affected by an external electric field but will not be affected by an external
irina [24]

Answer:

Explanation:

Situations in which an electron will be affected by an external electric field but will not be affected by an external magnetic field

a ) When an electron is stationary in the electric field and magnetic field , he will be affected by electric field but not by magnetic field. Magnetic field can exert force only on mobile charges.

b ) When the electron is moving parallel to electric field and magnetic field . In this case also electric field will  exert force on electron but magnetic field field will not exert force on electrons . Magnetic field can exert force only on the perpendicular component of the velocity of charged particles.

Situations when electron is affected by an external magnetic field but not by an external electric field

There is no such situation in which electric field will not affect an electron . It will always affect an electron .

6 0
2 years ago
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