A rocket is fired at an initial speed v0 = 151.0 m/s from ground level, at an angle θ = 46 degrees above the horizontal. A wall
is located at d = 41.0 m. Its heigh is h = 30.0 m. Ignore air resistance. The magnitude of the gravitational acceleration is 9.8 m/s2. Choose the RIGHT as positive x-direction. Choose UPWARD as psotitive y-direction Keep 2 decimal places in all answers Find v0x, the x component of the initial velocity (in m/s)
1 answer:
Answer:
v0x = 104.89 m/s
The vector v0x will be:
v0x = (104.89 m/s, 0)
Explanation:
Please, see the attached figure for a better understanding of the problem.
To obtain v0x, we have to use this trigonometric rule:
cos θ = adjacent / hypotenuse
Seeing the figure, notice that the side adjacent to the angle θ is the x-component of the initial velocity v0. The hypotenuse is the magnitude of the vector v0. Then:
cos θ = magnitude v0x / magnitude v0
magnitude v0 * cos θ = magnitude v0x
151.0 m/s * cos 46 = magnitude v0x
magnitude v0x = 104.89 m/s
The vector v0x will be:
v0x = (104.89 m/s, 0)
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Answer:
The specific latent heat of ice is approximately 341 J/g
The correct option is;
b) 341 J/g
Explanation:
The given parameters are;
The mass of ice the pupil adds to the water, m₁ = 37 g
The initial temperature of the ice, T₁ = 0°C
The mass of the water to which the ice is added, m₂ = 100 g
The initial temperature of the water, T₂₁ = 30°C
The final temperature of the water and the melted ice, T₂₂ = 0°C
The specific heat capacity of the water, c₂ = 4.2 J/(g·°C)
By the principle of conservation of energy, we have;
The heat gained by the ice = The heat lost by the water = ΔQ₂
Given that the ice is only melted with no change in temperature, we have;
The heat gained by the ice = The latent heat needed for melting the ice
ΔQ₂ = m₂ × c₂ × (T₂₂ - T₂₁) = 100 × 4.2 × (0 - 30) = -12,600 J
The heat gained by the ice = m₁ ×
Where;
represents the specific latent heat of fusion of ice;
We have;
12,600 = 37 ×
= 12,600/37 = 340.54 J/g ≈ 341 J/g
The specific latent heat of fusion of ice = ≈ 341 J/g.