Hooke's Law states that the extension is directly proportional to the force applied so:
F/x = constant
F₁/x₁ = F₂/x₂
2 / 0.02 = 1600 / x₂
x₂ = 16 m
Elastic work = 1/2 Fx
= 1/2 * 1600 * 16
= 12.8 kJ
Answer:
it will take 36.12 ms to reduce the capacitor's charge to 10 μC
Explanation:
Qi= C×V
then:
Vi = Q/C = 30μ/20μ = 1.5 volts
and:
Vf = Q/C = 10μ/20μ = 0.5 volts
then:
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
τ = 20µ x 1.5k = 30 ms
v = v₀e^(t/τ)
0.5 = 1.5e^(t/30ms)
e^(t/30ms) = 10/3
t/30ms = 1.20397
t = (30ms)(1.20397) = 36.12 ms
Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.
Answer:
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Answer:
Work done = 422.45 kJ
Explanation:
given,
weight of equipment = 6 kN
coefficient of kinetic friction = 0.05
distance up to which it is pulled = 1000 m
constant acceleration = 0.2 m/s²
Work done by the camper = ?
actual acceleration acting a'
m a = m a' - μ mg
a' = a + μ g
a' = 0.2 + 0.05 x 9.8
a' = 0.69 m/s²
Work done = Force x distance
F = m a'
F = 422.44897 N
Work done = F x d
Work done = 422.44897 x 1000
Work done = 422449 J
Work done = 422.45 kJ
