A student drops an object from rest above a force plate that records information about the force exerted on the object as a func
2 answers:
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Answer:
Explanation:
Light rays coming from moon is blocked by the pencil
so as per figure we know that angle subtended by pencil and angle subtended by moon must be same
so we have
so we have
so we have
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:
ΔK = 24 joules.
Explanation:
Δ Work done on the object
Work is equal to the dot product of force supplied and the displacement of the object.
* Δ
Δ can be found by subtracting the vectors (7.0, -8.0) and (11.0, -5.0), which is written as Δ
= (11.0 - 7.0, -5.0 - -8.0) which equals (4.0, 3.0).
This gives us
*
=
=
J