Answer: a. a leaping frog
Explanation: In a projectile motion the object must have a certain path or trajectory with respect to a certain angle. For the leaping frog its velocity can be resolve into components having the final velocity at the highest point to be zero.
Heat lost or gained, H = mc(θ₂ - θ₁)
Where m = mass, c = Specific heat capacity, θ₂= final temperature, θ₁ = initial temperature
m = 200g, c = 0.444 J/g°C, θ₁ = 22 °C (Since it was cooled).
H = 6.9 kj = 6.9 *1000J = 6900 J
6900 = 200*0.444* (θ₂ - 22)
6900/(200*0.444) = θ₂ - 22
77.70 = θ₂ - 22
θ₂ - 22 = 77.7
θ₂ = 77.7 + 22 = 99.7
So initial temperature before cooling ≈ 100°C . Option C.
The action or process of magnifying something or being magnified, especially visually. Hope this helped
