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inna [77]
3 years ago
13

What are the dimensions of a square that has an area of 4cm^2

Physics
2 answers:
Nikolay [14]3 years ago
8 0

Answer:

Explanation:

A square has 4 sides of equal length  

(

l

)

Area of square  

=

l

2

here  

l

=

4

⇒

Area  

=

4

2

=

16

square cm

nevsk [136]3 years ago
6 0

Answer:

square has equal length and width of dimension 2 cm^2

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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
4 years ago
A stretched string is observed to have four equal segments in a standing wave driven at a frequency of 480 hz. what driving freq
Korvikt [17]

600Hz is the driving frequency needed to create a standing wave with five equal segments.

To find the answer, we have to know about the fundamental frequency.

<h3>How to find the driving frequency?</h3>
  • The following expression can be used to relate the fundamental frequency to the driving frequency;

                                        f(n) = n * f (1)

where, f(1) denotes the fundamental frequency and the driving frequency f(n).

  • The standing wave has four equal segments, hence with n=4 and f(n)=4, we may calculate the fundamental frequency.

                                          f(4) = 4× f (1)

                                          480 = 4× f(1)

                                         f(1) = 480/4 =120Hz.

So, 120Hz is the fundamental frequency.

  • To determine the driving frequency necessary to create a standing wave with five equally spaced peaks?
  • For, n = 5,

                      f(n) = n 120Hz,

                      f(5) = 5×120Hz=600Hz.

Consequently, 600Hz is the driving frequency needed to create a standing wave with five equal segments.

Learn more about the fundamental frequency here:

brainly.com/question/2288944

#SPJ4

8 0
1 year ago
A car with a mass of 1.50x10^3 kg starts from rest and accelerates to a speed of 18.0m/s in 12.0 s. assume that the force of res
Luden [163]
The first thing you should know for this case is the definition of distance.
 d = v * t
 Where,
 v = speed
 t = time
 We have then:
 d = v * t
 d = 9 * 12 = 108 m
 The kinetic energy is:
 K = ½mv²
 Where,
 m: mass
 v: speed
 K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
 The work due to friction is
 w = F * d
 Where,
 F = Force
 d = distance:
 w = 400 * 108 = 4.32 * 10 ^ 4
 The power will be:
 P = (K + work) / t
 Where,
 t: time
 P = 2.86 * 10 ^ 5/12 = 23.9 kW
 answer:
 the average power developed by the engine is 23.9 kW
8 0
3 years ago
A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

I = MR^2

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

\Omega = \frac{Mgd}{I\omega}

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

\omega= Angular velocity

Replacing the value for moment of inertia

\Omega= \frac{MgR}{MR^2 \omega}

\Omega = \frac{g}{R\omega}

The value for our angular velocity is not in SI, then

\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})

\omega = 104.7rad/s

Replacing our values we have that

\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}

\Omega = 1.17rad/s

The precession frequency is

\Omega = \frac{2\pi rad}{T}

T = \frac{2\pi rad}{\Omega}

T = \frac{2\pi}{1.17}

T = 5.4 s

Therefore the precession period is 5.4s

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3 years ago
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3 years ago
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