_____ electrons are the most important electrons to an atom .
1 answer:
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Answer:
i) The motion of both particles are shown on the same speed-time curve included
ii) Approximately 19.5 seconds
Explanation:
We are given that;
Initial velocity of particle, P = 140 m/s
Start time of particle P = 6 s before start time of particle Q
Position of particle Q when velocity is 25 m/s = 125 m
Therefore, from the equation of motion, we have for particle Q;
v² = u² + 2·a·s
Where:
v = Final velocity = 25 m/s
u = Initial velocity = 0 m/s
a = Acceleration
s = Distance covered = 125 m
Therefore;
25² = 0² + 2×a×125
Which gives a = 25²/(2×125) = 2.5 m/s²
The time taken for particle Q to reach 125 m is found from the relation;
s = u·t + 1/2·a·t²
Where:
t = Time of journey
Therefore;
125 = 0×t + 1/2×2.5×t²
Which gives 125 = 1.25 × t²
Hence, t² = 125/1.25 = 100
t = √(100) = 10 s
The equation for particle Q is v = 0 + 2.5×t
Hence, since particle P starts deceleration 6 seconds before the commencement of motion of particle Q, the amount of seconds after the commencement of deceleration of the first particle P that it takes for particle P to come to rest is found as follows;
Hence, at t = 6 + 10 = 16 seconds particle P speed = 25 m/s
From the equation of motion, for particle P (decelerating) we have
v = u - a·t
Where:
v = 25 m/s
u = 140 m/s
t = 16 s
Hence, 25 = 140 - a×16
∴ 16·a = 140 - 25 = 115
a = 115/16 = 7.1875 m/s²
Therefore, the time it takes before particle P comes to rest is found from the same equation of motion, where v = 0 as follows;
v = u - a·t
0 = 140 - 7.1875 × t
∴7.1875·t = 140
t = 140/7.1875 = 19.48 s ≈ 19.5 seconds.
Answer:
t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number
Explanation:
To solve the problem/ we first write the differential equation governing the motion. So,
with m = 1 slug and k = 9 lb/ft, the equation becomes
The characteristic equation is
D² + 9 = 0
D = ±√-9 = ±3i
The general solution of the above equation is thus
x(t) = c₁cos3t + c₂sin3t
Now, our initial conditions are
x(0) = -1 ft and x'(0) = -√3 ft/s
differentiating x(t), we have
x'(t) = -3c₁sin3t + 3c₂cos3t
So,
x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)
x(0) = c₁cos(0) + c₂sin(0)
x(0) = c₁ × (1) + c₂ × 0
x(0) = c₁ + 0
x(0) = c₁ = -1
Also,
x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)
x'(0) = -3c₁sin(0) + 3c₂cos(0)
x'(0) = -3c₁ × 0 + 3c₂ × 1
x'(0) = 0 + 3c₂
x'(0) = 3c₂ = -√3
c₂ = -√3/3
So,
x(t) = -cos3t - (√3/3)sin3t
Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)
where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.
Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3
x(t) = (2√3/3)sin(3t + 4π/3)
So, the velocity v(t) = x'(t) = (2√3)cos(3t + 4π/3)
We now find the times when v(t) = 3 ft/s
So (2√3)cos(3t + 4π/3) = 3
cos(3t + 4π/3) = 3/2√3
cos(3t + 4π/3) = √3/2
(3t + 4π/3) = cos⁻¹(√3/2)
3t + 4π/3 = ±π/6 + 2kπ where k is an integer
3t = ±π/6 + 2kπ - 4π/3
t = ±π/18 + 2kπ/3 - 4π/9
t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9
t = π/18 - 4π/9 + 2kπ/3 or -π/18 - 4π/9 + 2kπ/3
t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3
Since t is not less than 0, the values of k ≤ 0 are not included
So when k = 1,
t = 5π/18 and π/6. So,
t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number