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bija089 [108]
3 years ago
11

An 81.5-kg man stands on a horizontal surface.

Physics
1 answer:
OLga [1]3 years ago
7 0

Answer:

a)V= 0.0827 m³

b)P=181.11 x 10²  N/m²

Explanation:

Given that

m = 81.5 kg

Density ,ρ = 985 kg/m³

As we know that

Mass = Volume x Density

81.5 = V x 985

V= 0.0827 m³

The force exerted by weight = m g

 F= m g= 81.5 x 10 = 815 N      ( Take ,g= 10 m/s²)

Area ,A= 4.5 x 10⁻² m²

The Pressure P

P=\dfrac{F}{A}

P = \dfrac{815}{4.5\times 10^{-2}}\ N/m^2

P=181.11 x 10²  N/m²

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Large wind turbines with a power capacity of 8 MW and blade span diameters of over 160 m are available for electric power genera
IrinaK [193]

Answer:

The electric power generated by the wind turbine is 1105.84 kWh

The amount of electric energy generated is 26540.17 kWh

The revenue generated per day is $2388.62

Explanation:

Consider a wind turbine with a blade

Span diameter of 100 m installed at a site

subjected to steady winds at 8 m/s

l.e wind speed v = 8 m/s

Span diameter d = 100 m

A, sweap area = πd² / 4

                        = π x 100² / 4

                        = 7853.98 m²

Lets solve for wind speed v = 8 m/s

Density of Area ρ = 1.25 kg/m³

η = 44%

P = 1/2 ρAv³η

  = 1/2 x 1.25 x 785.98 x 8³ x 44/100

  = 1/2 x 1.25 x 7853.98 x 512 x 0.44

  = 1105840.38

  = 1.10584038 mw

  = 1105.84038 kWh

  = 1105.84 kWh

Energy generated by wind turbine per day

⇒ P x H

      = 1105.84038 x 24

      = 26540.16912 kwh

      = 26540.17 kwh      

Revenue generated per day = Energy x 0.09 kwh

                                                = 26540.16912 x 0.09

                                                = $2388.615

                                                = $2388.62

4 0
3 years ago
A 480N sphere 40.0cm in radius rolls without slipping 1200cm down a ramp that is inclined at 53 0 with the horizontal. What is t
IgorC [24]

As we know that sphere roll without slipping so there is no loss of energy in this case

so here we can say that total energy is conserved

Initial Kinetic energy + initial potential energy = final kinetic energy + final potential energy

\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2+ mgh'

as we know that ball start from rest

v_i = 0

height of the ball initially is given as

h = Lsin\theta

h = 1200sin53 = 960 cm

also we know that

I = \frac{2}{5}mR^2

also for pure rolling

v = r\omega

also we know that

480 = m*9.8

m = 49 kg

now plug in all data in above equation

480*9.60 + 0 = \frac{1}{2}*49*(0.40*\omega)^2 + \frac{1}{2}*\frac{2}{5}*49*(0.40)^2\omega^2 + 0

4608 = 3.92\omega^2 + 1.568\omega^2

\omega^2 = 839.65

\omega = 29 rad/s

So speed at the bottom of the inclined plane will be 29 rad/s

7 0
3 years ago
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