Ilang ektarya ang lawak na itinakda ni llda na maaaring gawin palaisdaan
1 answer:
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Answer:
The work done by gravity during the roll is 490.6 J
Explanation:
The work (W) is:
<em>Where</em>:
F: is the force
d: is the displacement = 20 m
The force is equal to the weight (W) in the x component:
<em>Where:</em>
m: is the mass of the bowling ball = 5 kg
g: is the gravity = 9.81 m/s²
θ: is the degree angle to the horizontal = 30°
Now, we can find the work:
Therefore, the work done by gravity during the roll is 490.6 J.
I hope it helps you!
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
The energy transferred by the chocolate to the freezer before freezing is given approximately as follows;
E₁ = m×c₁×ΔT₁
Where;
m = The mass of the chocolate
c₁ = The specific heat capacity of water = 4.184 kJ/(kg·K)
ΔT₁ = The change in temperature from 40 °C to 0°C
Therefore, we have;
E₁ = m×4.184×(40 - 0) = 167.360·m kJ
The heat the coffee gives to turn to ice is given as follows;
E₂ = m·
Where;
= 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
E₃ = 42.16·m kJ/(kg·K)
The total heat transferred = (167.360 + 334 + 42.16)·m kJ/(kg·K) = 543.52·m kJ/(kg·K)
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
∴ Energy transferred in situation A > Energy transferred in situation B.
Answer:
The pressure after passing the valve is 23,8 [Kpa] ( 0,234 atm) and the pressure drop is about 1,53 [Kpa]
Explanation:
We need to use the formula of bernoulli, in the attached image we can see the fluid throw the pipe, we also can calculate the velocity inside the pipe using the flow rate and the cross sectional area.
For this case, we don't use the elevation difference and therefore those terms can be cancelled.
When the area has reduced the velocity of the fluid is increased but there is a drop pressure through the valve.
