Answer:
The correct answer is b, x = 9 cos (pi / 2 t)
Explanation:
The equation that describes a simple pendulum is
θ = θ₀ cos (wt + φ)
The angle is measured is radians
θ = x / L
We replace
d / L = x₀ / L cos (wt + φ)
x₀ = 9 in
We replace
d = 9 cos (wt + φ)
Angular velocity is related to frequency and period.
w = 2π f = 2π / T
The period is the time of a complete oscillation T = 4 s
w =2π / 4
w = π / 2
Let's replace
x = 9 cos (π/2 t + φ)
As the system is released from the root x = x₀ for t = 0 s
x₀ = x₀ cos φ
Cos φ = 1
φ = 0°
The final equation is
x = 9 cos (pi / 2 t)
The correct answer is b
Answer:
A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes
Explanation:
Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
Vf = Final Velocity of Car = 0 mi/h
Vi = Initial Velocity of Car = 50 mi/h
a = deceleration of car
s = distance covered
Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.
So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:
s = vt
FOR SOBER DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 0.33 s
s = s₁
Therefore,
s₁ = (73.33 ft/s)(0.33 s)
s₁ = 24.2 ft
FOR DRUNK DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 1 s
s = s₂
Therefore,
s₂ = (73.33 ft/s)(1 s)
s₂ = 73.33 ft
Now, the distance traveled by drunk driver's car further than sober driver's car is given by:
ΔS = s₂ - s₁
ΔS = 73.33 ft - 24.2 ft
<u>ΔS = 49.13 ft</u>
Allotropes are different forms of the same element. Different bonding arrangements between atoms result in different structures with different chemical and physical properties. Allotropes occur only with certain elements, in Groups 13 through 16 in the Periodic Table.
Therefore, if the block moves from its position of maximum spring stretch to maximum spring compression in 0.25 s, the time required for a full cycle is twice as much; T = 0.5 s.
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