The speed of the roller coater at the bottom of the hill is 31 m/s.
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Speed of the roller coater at the bottom of the hill</h3>
Apply the principle of conservation of mechanical energy as follows;
K.E(bottom) = P.E(top)
¹/₂mv² = mgh
v² = 2gh
v = √2gh
where;
- v is the speed of the coater at bottom hill
- h is the height of the hill
- g is acceleration due to gravity
v = √(2 x 9.8 x 49)
v = 31 m/s
Thus, the speed of the roller coater at the bottom of the hill is 31 m/s.
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Answer:
A. Endothermic reaction.
B. +150KJ.
C. 250KJ.
Explanation:
A. The graph represents endothermic reaction because the heat of the product is higher than the heat of the reactant.
B. Determination of the enthalpy change, ΔH for the reaction. This can be obtained as follow:
Heat of reactant (Hr) = 50KJ
Heat of product (Hp) = 200KJ
Enthalphy change (ΔH) =..?
Enthalphy change = Heat of product – Heat of reactant.
ΔH = Hp – Hr
ΔH = 200 – 50
ΔH = +150KJ
Therefore, the enthalphy change for the reaction is +150KJ
C. The activation energy for the reaction is the energy at the peak of the diagram.
From the diagram, the activation energy is 250KJ.
It is the subtance betwen the person or the object
Answer: m∠P ≈ 46,42°
because using the law of sines in ΔPQR
=> sin 75°/ 4 = sin P/3
so ur friend is wrong due to confusion between edges
+) we have: sin 75°/4 = sin P/3
=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16
=> m∠P ≈ 46,42°
Explanation: