Answer:
#2
Step-by-step explanation:
N/A
Yes, triangle ZHY and JHI are similar triangles because if you multiply each side on Triangle ZHY by 2, they equal the sides of the Triangle from IHJ.
For example, ZH x 2= JH , YZ x 2= IJ
This is the missing equation that models the hieght and is misssing in the question:
<span>h= 7cos(π/3 t)
</span>
Answers:
<span>a. Solve the equation for t.
</span>
<span>1) Start: h= 7cos(π/3 t)
</span>
2) Divide by 7: (h/7) = <span>cos(π/3 t)
</span>
3) Inverse function: arc cos (h/7) = π/3 t
4) t = 3 arccos(h/7) / π ← answer of part (a)
b. Find the times at which the weight is first at a height of 1 cm, of 3
cm, and of 5 cm above the rest position. Round your answers to the
nearest hundredth.
<span>1) h = 1 cm ⇒ t = 3 arccos(1/7) / π</span>
t = 1.36 s← answer
2) h = 3 cm ⇒ t = 3arccos (3/7) / π = 1.08s← answer
3) h = 5 cm ⇒ 3arccos (5/7) / π = 0.74 s← answer
c. Find the times at which the weight is at a height of 1 cm, of 3 cm, and of 5 cm below the rest position for the second time.
Use the periodicity property of the function.
The periodicity of <span>cos(π/3 t) is 6.
</span><span>
</span><span>
</span><span>So, the second times are:
</span><span>
</span><span>
</span><span>1) h = 1 cm, t = 6 + 0.45 s = 6.45 s ← answer
</span>
2) h = 3 cm ⇒ 6 + 1.08 s = 7.08 s← answer
3) h = 5 cm ⇒ t = 6 + 0.74 s = 6.74 s ← answer
Answer:
The acceleration of the particle at any time t
Step-by-step explanation:
Given:
The position of a particle moving in a straight line at any time t is.
The velocity of the particle 
So the velocity of the particle 
The acceleration of the particle 
In this condition the acceleration does not depending upon the time, so the acceleration is constant
Therefore the acceleration of the particle at any time t
Answer:
i think option 2 maybe i dont know
Step-by-step explanation:
sorry if I'm wrong I haven't done this in a while
