This problem, a squid at rest suddenly sees a predator coming toward it and needs to escape. Assume the following:______.
1 answer:
Answer:
6.79 m/s
Explanation:
By applying the principle of conservation of momentum.
The total momentum = MV - mv = 0 (since the squid is beginning at rest)
the mass of the squid (M) in absence of water in its cavity = (6.5 - 1.75) kg
= 4.75 kg
speed of the squid (V) = 2.5 m/s
mass of the water expelled (m) = 1.75 kg
speed of the water (v) = ???
∴
4.75 × 2.5 = 1.75 × v
v = 6.79 m/s
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Answer:
a)Amplitude ,A = 2 mm
b)f=95.49 Hz
c)V= 30 m/s ( + x direction )
d) λ = 0.31 m
e)Umax= 1.2 m/s
Explanation:
Given that
As we know that standard form of wave equation given as
A= Amplitude
ω=Frequency (rad /s)
t=Time
Φ = Phase difference
So from above equation we can say that
Amplitude ,A = 2 mm
Frequency ,ω= 600 rad/s (2πf=ω)
ω= 2πf
f= ω /2π
f= 300/π = 95.49 Hz
K= 20 rad/m
So velocity,V
V= ω /K
V= 600 /20 = 30 m/s ( + x direction )
V = f λ
30 = 95.49 x λ
λ = 0.31 m
We know that speed is the rate of displacement
The maximum velocity
Umax = 1200 mm/s
Umax= 1.2 m/s
Answer:
Explanation:
As we know that the speed of the sound is given as
now at t = 273 k = 0 degree
so we have
now when temperature is changed to 313 K we have
now we have
now from two equations we have
so we have