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Stells [14]
3 years ago
15

Sharks are generally negatively buoyant; the upward buoyant force is less than the weight force. This is one reason sharks tend

to swim continuously; water moving past their fins causes a lift force that keeps sharks from sinking. A 92 kg bull shark has a density of 1040 kg/m3. What lift force must the shark's fins provide if the shark is swimming in seawater? Bull sharks often swim into freshwater rivers. What lift force is required in a river?
Physics
1 answer:
Tresset [83]3 years ago
6 0

Answer:

8.67807 N

34.7123 N

Explanation:

m = Mass of shark = 92 kg

\rho_{se} = Density of seawater = 1030 kg/m³

\rho_{f} = Density of freshwater = 1000 kg/m³

\rho_{sh} = Density of shark = 1040 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Net force on the fin is (seawater)

F_n=mg-V_s\rho_{se}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{se}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1030\times 9.81\\\Rightarrow F_n=8.67807\ N

The lift force required in seawater is 8.67807 N

Net force on the fin is (freshwater)

F_n=mg-V_s\rho_{f}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{f}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1000\times 9.81\\\Rightarrow F_n=34.7123\ N

The lift force required in a river is 34.7123 N

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A sound wave has a frequency of 425 hz. what is the period of this wave? 0.00235 seconds 0.807 seconds 425 seconds 850 seconds
77julia77 [94]

The period of the sound wave at the given frequency is determined as 0.00235 second.

<h3>Period of the sound wave</h3>

The period of the sound wave at the given frequency is calculated as follows;

Period is reciprocal of frequency.

T = 1/f

T = 1/425

T = 0.00235 second

Thus, the period of the sound wave at the given frequency is determined as 0.00235 second.

Learn more about period here: brainly.com/question/10428039

#SPJ1

3 0
1 year ago
How to solve for the coefficient of friction
astraxan [27]
Their are two coefficients of friction, static and kinetic, regardless, they have basically the same formula:

u*N = F... aka
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Hope that helps!
3 0
3 years ago
What is the frequency, in units of kiloHertz, of an AC waveform that has a period of 12 microseconds?
Alik [6]

Answer:

83.3 kHz

Explanation:

The frequency of a waveform is equal to the reciprocal of its period:

f=\frac{1}{T}

where

f is the frequency

T is the period

In this problem, we have

T=12 \mu s=12\cdot 10^{-6} s

so, the frequency of the waveform is

f=\frac{1}{12 \cdot 10^{-6} s}=8.33\cdot 10^4 Hz

And by converting into kiloHertz,

f=8.33\cdot 10^4 Hz=83.3 kHz

3 0
3 years ago
A small glass bead has been charged to 20 nC. What is the magnitude of acceleration in m/s^2 of an electron that is 1.0 cm from
MAVERICK [17]

Answer:

The acceleration is 3.16x10¹⁷ m/s².

Explanation:

First, we need to find the magnitude of the Coulombs force (F):

|F| = \frac{Kq_{1}q_{2}}{d^{2}}

<u>Where</u>:

K is the Coulomb constant = 9x10⁹ Nm²/C²

q₁ is the charge = 20x10⁻⁹ C  

q₂ is the electron's charge = -1.6x10⁻¹⁹ C

d is the distance = 1.0 cm = 1.0x10⁻² m

|F| = \frac{Kq_{1}q_{2}}{d^{2}} = \frac{9\cdot 10^{9}Nm^{2}/C^{2}*20 \cdot 10^{-9} C*(-1.6\cdot 10^{-19} C)}{(0.01 m)^{2}} = 2.88 \cdot 10^{-13} N                                      

Now, we can find the acceleration:

a = \frac{F}{m} = \frac{2.88 \cdot 10^{-13} N}{9.1 \cdot 10^{-31} kg} = 3.16 \cdot 10^{17} m/s^{2}

Therefore, the acceleration is 3.16x10¹⁷ m/s².

I hope it helps you!    

7 0
3 years ago
A dramatic demonstration, called "singing rods," involves a long, slender aluminum rod held in the hand near the rod's midpoint.
Korvikt [17]
More vibration = higher frequency
8 0
3 years ago
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