In this graph, what is the displacement of the particle in the last two seconds?of the particle in the last two seconds?
<span>0.2 meters
2 meters
4 meters
6 meters</span>
In this graph, the displacement of the particle in the last two seconds is 2 meters.
Answer
given,
v = 128 ft/s
angle made with horizontal = 30°
now,
horizontal component of velocity
vx = v cos θ = 128 x cos 30° = 110.85 ft/s
vertical component of velocity
vy = v sin θ = 128 x sin 30° = 64 m/s
time taken to strike the ground
using equation of motion
v = u + at
0 =-64 -32 x t
t = 2 s
total time of flight is equal to
T = 2 t = 2 x 2 = 4 s
b) maximum height
using equation of motion
v² = u² + 2 a h
0 = 64² - 2 x 32 x h
64 h = 64²
h = 64 ft
c) range
R = v_x × time of flight
R = 110.85 × 4
R = 443.4 ft
You would flip forward or to the side
Answer:
f_D = =3.24 N/m
Explanation:
data given
properties of air
k = 0.0288 W/m.K
WE KNOW THAT
Reynold's number is given as
= 1.941 *10^4
drag coffecient is given as
solving for f_D
Drag coffecient for smooth circular cylinder is 1.1
therefore Drag force is
f_D = =3.24 N/m
