Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
Complete question:
(b) How much energy must be supplied to boil 2kg of water? providing that the specific latent heat of vaporization of water is 330 kJ/kg. The initial temperature of the water is 20 ⁰C
Answer:
The energy that must be supplied to boil the given mass of the water is 672,000 J
Explanation:
Given;
mass of water, m = 2 kg
heat of vaporization of water, L = 330 kJ/kg
initial temperature of water, t = 20 ⁰C
specific heat capacity of water, c = 4200 J/kg⁰C
Assuming no mass of the water is lost through vaporization, the energy needed to boil the given water is calculated as;
Q = mc(100 - 20)
Q = 2 x 4200 x (80)
Q = 672,000 J
Q = 672,000 J
Q = 672,000 J
Therefore, the energy that must be supplied to boil the given mass of the water is 672,000 J
Answer:
A place where organic and non organic materials interact to make a living space
Quit cheating and figure it out your self, the app wasn’t made to cheat it was made to help.
Answer:
potential
Explanation:
Potential energy is stored
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