Answer:
The final angular velocity is 20rad/s
Explanation:
We are given;
mass, m = 12 kg
radius, r = 0.25 m
Work done;W = 75 J
Moment of inertia of cylinder, I = (1/2) mr²
Thus,
I = (1/2) x 12 x 0.25² = 0.375 kg.m²
Now, from work energy theorem,
Work done = Change in kinetic energy
So, W = KE_f - KE_i
Now, Initial Kinetic Energy (KE_i) = 0
Final Kinetic Energy; KE_f = (1/2)Iω²
So, KE_f = (1/2) x 0.375 x ω²
KE_f = 0.1875 ω²
Now, W = 75 J
Thus,
From, W = KE_f - KE_i, we have;
75 = 0.1875 ω² - 0
75 = 0.1875 ω²
ω² = 75/0.1875
ω² = 400
ω = √400
ω = 20 rad/s
Answer: Colby we both dumb if we need brainly lol
Explanation:
Answer:
a) # lap = 301.59 rad
, b) L = 90.48 m
Explanation:
a) Let's use a direct proportions rule (rule of three). If one turn of the wire covers 0.05 cm, how many turns do you need to cover 24 cm
# turns = 1 turn (24 cm / 0.5 cm)
# laps = 48 laps
Let's reduce to radians
# laps = 48 laps (2 round / 1 round)
# lap = 301.59 rad
b) Each lap gives a length equal to the length of the circle
L₀ = 2π R
L = # turns L₀
L = # turns 2π R
L = 48 2π 30
L = 9047.79 cm
L = 90.48 m
Answer:
H_w = 2.129 m
Explanation:
given,
Width of the weir, B = 1.2 m
Depth of the upstream weir, y = 2.5 m
Discharge, Q = 0.5 m³/s
Weir coefficient, C_w = 1.84 m
Now, calculating the water head over the weir
now, level of weir on the channel
H_w = y - H
H_w = 2.5 - 0.371
H_w = 2.129 m
Height at which weir should place is equal to 2.129 m.
