Answer:
The tangential speed of the tack is 8.19 m/s.
Explanation:
The wheel rotates 3.37 times a second that means wheel complete 3.37 revolutions in a second. Therefore, the angular speed ω of the wheel is given as follows:

Use the relation of angular speed with tangential speed to find the tangential speed of the tack.
The tangential speed v of the tack is given by following expression
v = ω r
Here, r is the distance to the tack from axis of rotation.
Substitute 21.174 rad/s for ω, and 0.387 m for r in the above equation to solve for v.
v = 21.174 × 0.387
v = 8.19m/s
Thus, The tangential speed of the tack is 8.19 m/s.
Answer:
3.6 seconds
Explanation:
Given:
y₀ = y = 0 m
v₀ = 31 sin 35° m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
0 = 0 + (31 sin 35°) t + ½ (-9.8 m/s²) t²
0 = 17.78t − 4.9t²
0 = t (17.78 − 4.9t)
t = 0 or 3.63
Rounded to the nearest tenth, the ball lands after 3.6 seconds.
Answer:

Explanation:
From the concept of fluids mechanics we know that if a tank has a hole at the bottom, the equation that we need to use is:

Since we know gravity and its hight

20/40=0.5 g/cm^3 becuase, mass/volume=density.