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zhuklara [117]
3 years ago
12

A block slides along a frictionless surface with 220 J of kinetic energy. It then goes up a frictionless ramp. When the kinetic

energy drops to 55 J, how much gravitational potential energy does the block have?
Physics
1 answer:
aev [14]3 years ago
8 0

Hello!

For the explanation of this energy conservation exercise, where we'll use <u>energy conservation law</u>, let's see what this principle proposes.

How you should know, mechanical energy conserves in every point, that is to say mechanical energy is same in A point like B point. (Mechanical energy will be represented by "Me")

Once time we know that, let's take the 220 Joules momentum like A point, and when 55 Joules momentum like B point.

Then, let's use the <u>energy conservation principle:</u>

Me(A) = Me(B)

  • We know Mechanical energy in A point, so just lets replace according to our data:

220 J = Me(B)

  • In B point, we know kinetic energy, but <u>we dont know gravitational potential energy</u>, so lets descompose Mechanical energy, into kinetic energy and gravitational potential energy:

220 J = Ke + Gpe

  • We know kinetic energy value, so lets replace it:

220 J = 55 J + Gpe

  • Finally, just clean Gpe and resolve it:

Gpe = 220 J - 55 J = 165 J

Gravitational potential energy is of One hundred sixty five Joules <u>(165 J).</u>

                                                                                  ║Sincerely, ChizuruChan║

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The formula of kinetic energy is given as;

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   substitute the parameters in the equation;

                       600  = \frac{1}{2} x m x 15²

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Answer:

Explanation:

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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 )

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Expression to calculate the  A  is

A = π /4  d ²

Substitute the value in above expression

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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

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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

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Sveta_85 [38]

Answer:

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