Answer:
7.54 m/s²
Explanation:
uₓ(t) = (0.980 m/s³) t²
Acceleration is the derivative of velocity with respect to time.
aₓ(t) = 2 (0.980 m/s³) t
aₓ(t) = (1.96 m/s³) t
When uₓ = 14.5 m/s, the time is:
14.5 m/s = (0.980 m/s³) t²
t = 3.85 s
Plugging into acceleration equation:
aₓ = (1.96 m/s³) (3.85 s)
aₓ = 7.54 m/s²
Work = (force) x (distance)
The work he did: Work = (700 N) x (4m) = 2,800 joules
The rate at which
he did it (power): Work/time = 2,800 joules/2 sec
= 1,400 joules/sec
= 1,400 watts
= 1.877... horsepower (rounded)
Answer:
0.247 J = 247 mJ
Explanation:
From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.
So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁) where m = mass of particle = 5.4 × 10⁻² kg, q = charge of particle = 5.10 × 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.
So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁)
= 1/2 × 5.4 × 10⁻²kg × ((3m/s)² - (2 m/s)²) + 5.10 × 10⁻⁵ C(7850 - 5650)
= 0.135 J + 0.11220 J
= 0.2472 J
≅ 0.247 J = 247 mJ
A. Bohr
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