At what speed does a 1800 kg compact car have the same kinetic energy as a 15000 kg truck going 21 km/h ? Express your answer in
1 answer:
Answer:
v = 60.62 km/h
Explanation:
Kinetic energy
Kinetic energy is that which is due to the movement of bodies and is calculted like this:
K = (1/2)mv² Formula (1)
Where:
K : Kinetic energy (J)
m: mass (kg)
v : speed (m/s)
Data for the truck
m = 15000 kg
v= 21 km/h = 21 *(1000m)/ (3600s)
v= 5.83 m/s
Calculation of the kinetic energy of the truck
We replace data in the formula (1)
K = (1/2)mv²
K = (1/2)(15000)(5.83)²
K = 255208.3 J
Data for the compact car
K = 255208.3 J = 255208.3 (N*m)
m = 1800 kg
Calculation of the speed of the compact car
We replace data in the formula (1)
K = (1/2)mv²
255208.3 = (1/2)( 1800)(v)²
255208.3 = ( 900)(v)²
v² = 255208.3 / ( 900)
v = 16.84 m/s
1 km = 1000 m
1 h = 3600 s
v = 16.84*(3600/1000)km/h
v = 60.62 km/h
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Answer:
More energy is transferred in situation A
Explanation:
Each of the situations are analyzed as follows;
Situation A
The temperature of the cup of hot chocolate = 40 °C
The temperature of the interior of the freezer in which the chocolate is placed = -20 °C
We note that at 0°C, the water in the chocolate freezes
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Where;
m = The mass of the chocolate
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ΔT₁ = The change in temperature from 40 °C to 0°C
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The heat the coffee gives to turn to ice is given as follows;
E₂ = m·
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= The latent heat of fusion = 334 kJ/kg
∴ E₂ = m × 334 kJ/kg = 334·m kJ
The heat required to cool the frozen ice to -20 °C is given as follows;
E₃ = m·c₂·ΔT₂
Where;
c₂ = The specific heat capacity of ice = 2.108 kJ/(kg·K)
Therefore, we have;
E₃ = m × 2.108 ×(0 - (-20)) = 42.16
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Situation B
The temperature of the cup of hot chocolate = 90 °C
The temperature of the room in which the chocolate is placed = 25 °C
The heat transferred by the hot cup of coffee, E, is given as follows;
E = m×4.184×(90 - 25) = 271.96
∴ E = 271.96 kJ/(kg·K)
Therefore, the total heat transferred in situation A is approximately twice the heat transferred in situation B and is therefore more than the heat transferred in situation B
Energy transferred in situation A = 543.52 kJ/(kg·K)
Energy transferred in situation B = 271.96 kJ/(kg·K)
Energy transferred in situation A ≈ 2 × Energy transferred in situation B
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