Answer:
37.8 m
Explanation:
At point 0, the ball is at height y₀.
At point 1, the ball is at height 30 m.
At point 2, the ball is at height 0 m.
Given:
y₁ = 30 m
y₂ = 0 m
v₀ = 0 m/s
a = -10 m/s²
t₂ − t₁ = 1.5 s
Find: y₀
Use constant acceleration equation.
y = y₀ + v₀ t + ½ at²
Evaluate at point 1.
y₁ = y₀ + v₀ t₁ + ½ at₁²
30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²
30 = y₀ − 5t₁²
Evaluate at point 2.
y₂ = y₀ + v₀ t₂ + ½ at₂²
0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²
0 = y₀ − 5t₂²
y₀ = 5t₂²
Substitute:
y₀ = 5 (1.5 + t₁)²
y₀ = 5 (2.25 + 3t₁ + t₁²)
y₀ = 11.25 + 15t₁ + 5t₁²
30 = 11.25 + 15t₁ + 5t₁² − 5t₁²
30 = 11.25 + 15t₁
t₁ = 1.25
30 = y₀ − 5t₁²
30 = y₀ − 5(1.25)²
y₀ ≈ 37.8
The rate of fuel burning in grams per hour if the DT reaction is used is 1.08 ×
J/g per hour
<h3>How is the rate of fuel burning in grams per hour calculated when the D-T reaction is used?</h3>
- D + T → He + n
- The D-T fusion reaction results in a Helium (He) and neutron (n)
E = 17.59 MeV
Mass = 2.014u + 3.016u
= 5.030u
Energy per Kg = (17.59×
×1.6×
) ÷ ( 5.030×1.66×
)
= 3.37×
J/Kg
= 3.0× 
J/g
Rate of fuel burning in grams per hour = 3.0× × 3600
= 3.6×3.0×
= 1.08 × J/g per hour
To learn more about fusion reactor and energy production, refer
brainly.com/question/13399644
#SPJ4
Low frequency, low amplitude, or both.
Answer: C
Explanation:
Production is expected that it will almost double by the end of the year is the answer because production in this context relates to a numerical quantity which is a function of time.
Therefore the probability there is on whether production will almost double to provide enough electricity and not if production will occur.
Start with saying “If...” and then say “the the....”
