(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
<span>Amplitude is the correct answer. I hope this helps.</span>
Answer:
(a) 40.6 degree
Explanation:
When refraction takes place from slab to water, the critical angle is 60 degree.
Use Snell's law
refractive index of water with respect to slab
μs = 1.536
Now for slab air interface, the critical angle is C.
1 / 1.536 = Sin C
C = 40.6 degree
Answer:
Explanation:
The classification will be made into 3 categories, which are
Ones that shortens wavelengths
Ones that lengthens wavelengths
Ones that has no effect on wavelengths
Shortens wavelengths -> Increase frequency
Lengthens wavelengths -> Decrease frequency
No effect -> Increase amplitude, decrease amplitude, increase damping, decrease damping.
