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schepotkina [342]
2 years ago
7

Which example best represents translational kinetic energy? A plucked string guitar. The moving blades of a ceiling fan. A spinn

ing top. A tuning fork struck on a rubber block. An apple falling from a tree
Physics
2 answers:
ValentinkaMS [17]2 years ago
6 0

Answer:

the answer is E

Explanation:

an apple falling from a tree

hope this helps u :)

MrRa [10]2 years ago
5 0

Answer:

The answer is e the apple one

Explanation:

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The product side of a chemical reaction is shown. → 7Ti2(SO4)3 Which number represents a coefficient?
solniwko [45]
7 is the answer.................................
3 0
3 years ago
Read 2 more answers
A pair of electrically charged objects attract each other with a force of 4 N when they are a distance of 3 m apart. If their ch
Rainbow [258]

Answer:

1 N

Explanation:

From coulomb's law,

The force of attraction between two charges is inversely proportional to the square of the distance between the charges.

From the question,

Assuming the charges are the same in both case,

F ∝ /r²....................... Equation 1

Fr² = k

F'r'² = Fr²........................... Equation 2

Where F' = First Force, r'² = First distance, F = second force, r² = second distance.

make F the subject of the equation,

F = F'r'²/r².................... Equation 3

Given: F' = 4 N, r' = 3 m, r = 6 m

Substitute into equation 3

F = 4(3²)/6²

F = 36/36

F = 1 N

7 0
3 years ago
An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 75
katen-ka-za [31]

Answer:

Part A) 3899 kPa  

Part B) 392.33 kJ/kg  

Part C) 0.523

Part D) 495 kPa

Explanation:

Part A

First from the temperature at state 1 the relative specific volume and the internal energy at that state are determined from:

u_{1} = 214.07 kJ/kg  

\alphar_{1} = 621.2  

The relative specific volume at state 2 is obtained from the compression ratio:  

\alphar_{2} = \frac{\alpha r_{1}  }{r}

     =621.2/ 8

    = 77.65  

From this the temperature and internal energy at state 2 can be determined using interpolation with data from A-17(table):  

T_{2} = 673 K

u_{2} = 491.2 kJ/kg  

The pressure at state 2 can be determined by manipulating the ideal gas relations at state 1 and 2:  

P_{2} =  P_{1} r\frac{T_{2} }{T_{1} }

       = 95*8*673/300

      = 1705 kPa  

Now from the energy balance for stage 2-3 the internal energy at state 3 can be obtained:  

deltau_{2-3} =q_{in}\\ u_{3} -u_{2} =q_{in}\\u_{3}=u_{2}+q_{in}

     = 1241.2 kJ/kg

From this the temperature and relative specific volume at state 3 can be determined by interpolation with data from A-17(table):  

T_{3} = 1539 K  

\alpha r_{3} = 6.588  

The pressure at state 3 can be obtained by manipulating the ideal gas relations for state 2 and 3:  

P_{3} =P_{2} \frac{T_{3} }{T_{2} }

     = 3899 kPa  

<u>Part B</u>

The relative specific volume at state 4 is obtained from the compression ratio:  

\alpha r_{4}= r\alpha r_{3}

      = 52.7

From this the temperature and internal energy at state 4 can be determined by interpolation with data from A-17:  

T_{4}=775 K

u_{4}= 571.74 kJ/kg  

The net work output is the difference of the heat input and heat rejection where the heat rejection is determined from the decrease in internal energy in stage 4-1:  

w=q_{in}-q_{out}\\q_{in}-(u_{4} -u_{1} )\\=392.33 kJ/kg

<u>Part C  </u>

The thermal efficiency is obtained from the work and the heat input:  

η=\frac{w}{q_{in} }

=0.523

<u>Part D  </u>

The mean effective pressure is determined from its standard relation:  

MEP=\frac{w}{\alpha_{1}- \alpha_{2} }

      =\frac{w}{\alpha_{1}(1- \frac{1}{r}  }

      =\frac{rwP_{1} }{RT_{1} (r-1) }

      =495 kPa

8 0
3 years ago
If the roller coaster car in the above problem were moving with twice the speed(20m/s), then what would be its new kinetic energ
Diano4ka-milaya [45]

Answer:

KE = 1.05 x105 Joules

Explanation:

KE = 4 * (1.04653 x 105 J) = 4.19 x 105 Joules.

4 0
2 years ago
A 125-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a
lbvjy [14]

Answer:

201.6 N

Explanation:

m = mass of disk shaped merry-go-round = 125 kg

r = radius of the disk = 1.50 m

w₀ = Initial angular speed = 0 rad/s

w = final angular speed = 0.700 rev/s = (0.700) (2π) rad/s = 4.296 rad/s

t = time interval = 2 s

α = Angular acceleration

Using the equation

w = w₀ + α t

4.296 = 0 + 2α

α = 2.15 rad/s²

I = moment of inertia of merry-go-round

Moment of inertia of merry-go-round is given as  

I = (0.5) m r² = (0.5) (125) (1.50)² = 140.625 kgm²

F = constant force applied

Torque equation for the merry-go-round is given as

r F = I α

(1.50) F = (140.625) (2.15)

F = 201.6 N

4 0
3 years ago
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