One game at the amusement park has you push a puck up a long, frictionless ramp. You win a stuffed animal if the puck, at its hi
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Answer:
50 J
Explanation:
The net force acting on the box is given by the algebraic sum of the two forces, so:
The net work done on the box is equal to (assuming the net force is parallel to the displacement of the object)
where
F = 5 N is the net force on the object
d = 10 m is the displacement of the object
Substituting,
Explanation:
It is given that, Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m/s².
The second equation of kinematics gives the relationship between the height reached and time taken by it.
Here, the ball is droped under the action of gravity. The value of acceleration due to gravity on Mars is positive.
We want to know how many seconds the basketball is in the air before it hits the ground. So, the formula is :
t is time taken by the ball to hit the ground
is initial speed of the ball
So, the correct option is (A).
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.