Squid use jet propulsion for rapid escapes. A squid pulls water into its body and then rapidly ejects the water backward to prop
el itself forward. A 1.5 kg squid (not including water mass) can accelerate at 20 m/s2 by ejecting 0.15 kg of water.?
1 answer:
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Answer:
magnetic flux ΦB = 0.450324 × weber
current I = 1.02484 A
Explanation:
Given data
length a = 2.2 cm = 0.022 m
width b = 0.80 cm = 0.008 m
Resistance R = 0.40 ohms
current I = 4.7 A
speed v = 3.2 mm/s = 0.0032 m/s
distance r = 1.5 b = 1.5 (0.008) = 0.012
to find out
magnitude of magnetic flux and the current induced
solution
we will find magnitude of magnetic flux thorough this formula that is
ΦB = ( μ I(a) /2 π ) ln [(r + b/2 ) /( r -b/2)]
here μ is 4π × put all value
ΦB = (4π × 4.7 (0.022) /2 π ) ln [(0.012+ 0.008/2 ) /( 0.012 -0.008/2)]
ΦB = 0.450324 × weber
and
current induced is
current = ε / R
current = μ I(a) bv / 2πR [(r² ) - (b/2 )² ]
put all value
current = μ I(a) bv / 2πR [(r² ) - (b/2 )² ]
current = 4π × (4.7) (0.022) (0.008) (0.0032) / 2π(0.40) [(0.012² ) - (0.008/2 )² ]
current = 1.02484 A
Answer:
v = √(Lsinθ tanθ)
Explanation:
From the diagram attached,
v = the tangential speed.
r = the radius of the horizontal circle.
T = tension in the string.
θ = the angle that the string makes with the vertical
m = Bob's mass (mg = the weight)
F = centripetal force
L = the length of the string
From geometry,
r = Lsin θ
Thus, the centripetal acceleration is given as;
a = v²/r = v²/Lsin θ
Force = mass x acceleration
Thus,
Centripetal force = mv²/Lsin θ
Let's balance the vertical forces to obtain,
T cosθ - mg = 0
Thus, T cosθ = mg - - - - (eq1)
Similarly, let's balance the horizontal forces to obtain;
T sinθ = F = mv²/Lsin θ
So, T sinθ = mv²/Lsin θ - - - - (eq2)
Let's divide eq 2 by eq 1 to get;
Tsinθ/Tcosθ = (mv²/Lsin θ)/mg
tanθ = gv²/Lsinθ
Thus, v² = Lsinθ tanθ
v = √(Lsinθ tanθ)
