6 . . . . . a crest
7 . . . . . the amplitude
8 . . . . . the wavelength
9 . . . . . a trough
(a) 1200 rad/s
The angular acceleration of the rotor is given by:

where we have
is the angular acceleration (negative since the rotor is slowing down)
is the final angular speed
is the initial angular speed
t = 10.0 s is the time interval
Solving for
, we find the final angular speed after 10.0 s:

(b) 25 s
We can calculate the time needed for the rotor to come to rest, by using again the same formula:

If we re-arrange it for t, we get:

where here we have
is the initial angular speed
is the final angular speed
is the angular acceleration
Solving the equation,

The answer is; Irregular.
The height at time t is given by
h(t) = -4.91t² + 34.3t + 1
When the ball reaches maximum height, its derivative, h'(t) = 0.
That is,
-2(4.91)t+34.3 = 0
-9.82t + 34.3 = 0
t = 3.4929 s
Note that h''(t) = -9.82 (negative) which confirms that h will be maximum.
The maximum height is
hmax = -4.91(3.4929)² + 34.3(3.4929) + 1
= 60.903 m
Answer:
The ball attains maximum height in 3.5 s (nearest tenth).
The ball attains a maximum height of 60.9 m (nearest tenth)
Answer:
m = 375 [gram]
Explanation:
A triple Beam balance is an instrument very easy to use, since we only have to perform the arithmetic sum of each of the weights that are recorded in each beam
m = 300 + 70 + 5 = 375 [gram]
For a better understanding, the following image is attached, with values on each beam, which should be read.
The largest mass is in the indicator of 100 [gram], the second mass is in the indicator of 20 [gram] and the third is in the indicator of 5.8 [gram]. Thus the arithmetic sum corresponds to:
M= 100 + 20 + 5.8 = 125.8 [gram]
Note: it is important that when the instrument is in balance, the opposite end of the beam should indicate a position of zeros.