Answer:
638 m.
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 94 m/s
Final velocity (v) = 22 m/s
Time (t) = 11 s
Distance (s) =?
We can obtain the distance travelled by using the following formula:
s = (u + v) t /2
s = (94 + 22) × 11 /2
s = 116 × 11 /2
s = 1276 /2
s = 638 m
Thus, the distance travelled is 638 m.
Answer: It should be A or the very left red circle that you can click on
Explanation: Because when the wind is moving downward and the earth is spinning the spot the wind ends up will never be directly down from where it was to begin with
Answer:
26 m/s
69 m
Explanation:
Given:
v₀ = 20 m/s
a = 2 m/s²
t = 3 s
Find: v and Δx
v = at + v₀
v = (2 m/s²) (3 s) + 20 m/s
v = 26 m/s
Δx = v₀ t + ½ at²
Δx = (20 m/s) (3 s) + ½ (2 m/s²) (3 s)²
Δx = 69 m
Answer:
(a) f= 622.79 Hz
(b) f= 578.82 Hz
Explanation:
Given Data
Frequency= 600 Hz
Distance=1.0 m
n=120 rpm
Temperature =20 degree
Before solve this problem we need to find The sound generator moves on a circular with tangential velocity
So
Speed of sound is given by
c = √(γ·R·T/M)
............in an ideal gas
where γ heat capacity ratio
R universal gas constant
T absolute temperature
M molar mass
The speed of sound at 20°C is
c = √(1.40 ×8.314472J/molK ×293.15K / 0.0289645kg/mol)
c= 343.24m/s
The sound moves on a circular with tangential velocity
vt = ω·r.................where
ω=2·π·n
vt= 2·π·n·r
vt= 2·π · 120min⁻¹ · 1m
vt= 753.6 m/min
convert m/min to m/sec
vt= 12.56 m/s
Part A
For maximum frequency is observed
v = vt
f = f₀/(1 - vt/c )
f= 600Hz / (1 - (12.56m/s / 343.24m/s) )
f= 622.789 Hz
Part B
For minimum frequency is observed
v = -vt
f = f₀/(1 + vt/c )
f= 600Hz / (1 + (12.56m/s / 343.24m/s) )
f= 578.82 Hz
