Ask question

Ask question

All categories

2 years ago

6

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 kgbull 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?

1 answer:

e-lub [12.9K]2 years ago

4 0

Answer:

866.92 N

Explanation:

mass of shark, m = 92 kg

density, d = 1040 kg/m^3

Volume, V = mass / density

V = 92 / 1040 = 0.08846 kg/m^3

The lift force is the buoyant force acting on the shark is

Lift force = Volume x density of water x g

Lift force = 0.08846 x 1000 x 9.8

Lift force = 866.92 N

Thus, the lift force is 866.92 N.

You might be interested in

If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g

EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408 × 10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M_1, M_2$ are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

$\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}$

$\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2$

Since $M_2 = 2m_2$ and $r = R/3$

$\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5$

So gravity would have been decreased by a factor of 4.5

8 0

2 years ago

You've always wondered about the acceleration of the elevators in the 101 story-tall Empire State Building. One day, while visit

love history [14]

To develop this problem we will proceed to convert all units previously given to the international system for which we have to:

$140 lb = 63.5 kg \rightarrow 63.5kg (9.8m/s) =622.3 N$

$120 lb = 54.4 kg \rightarrow 54.4kg (9.8m/s)= 533 N$

$170 lb = 77.1 kg \rightarrow 77.1 kg (9.8m/s) =756 N$

PART A ) From the given values the minimum acceleration will be given for 120Lb and maximum acceleration when 170Lb is reached therefore:

$F = 756 - 622.3$

$F = 133.7N$

Through the Newtonian relationship of the Force we have to:

$F= ma$

$a = \frac{F}{m}$

$a = \frac{133.7}{63.5}$

$a = 2.1m/s^2$

PART B) For the maximum magnitude of the acceleration downward we have that:

$F = 622.3 - 533$

$F = 89.3N$

Through the Newtonian relationship of the Force we have to:

$F= ma$

$a = \frac{F}{m}$

$a = \frac{89.3}{63.5}$

$a = 2.1m/s^2$

$a = 1.04 m/s^2$

7 0

2 years ago

A water-skier with weight Fg = mg moves to the right with acceleration a. A horizontal tension force T is exerted on the skier b

Degger [83]

Answer:

The correct relationships are T-fg=ma and L-fg=0.

(A) and (C) is correct option.

Explanation:

Given that,

Weight Fg = mg

Acceleration = a

Tension = T

Drag force = Fa

Vertical force = L

We need to find the correct relationships

Using balance equation

In horizontally,

The acceleration is a

$T-Fd=ma$ ...(I)

In vertically,

No acceleration

$w=L$

$mg-L=0$

Put the value of mg

$L-fg=0$ ....(II)

Hence, The correct relationships are T-fg=ma and L-fg=0.

(A) and (C) is correct option.

3 0

2 years ago

A small mass m is tied to a string of length L and is whirled in vertical circular motion. The speed of the mass v is such that

adell [148]

Answer:

(mv^2/R)/(mg)=1/2

v^2=R/2g

7 0

2 years ago

Consider a steel guitar string of initial length l=1.00m and cross-sectional area a=0.500mm2. the young's modulus of the steel i

laiz [17]

L = 1.00 m, the original length
A = 0.5 mm² = 0.5 x 10⁻⁶ m², the cross sectional area
E = 2.0 x 10¹¹ n/m², Young's modulus
P = 1500 N, the applied tension

Calculate the stress.
σ = P/A = (1500 N)/(0.5 x 10⁻⁶ m²) = 3 x 10⁹ N/m²

Let δ = the stretch of the string.
Then the strain is
ε = δ/L

By definition, the strain is
ε = σ/E = (3 x 10⁹ N/m²)/(2 x 10¹¹ N/m²) = 0.015
Therefore
δ/(1 m) = 0.015
δ = 0.015 m = 15 mm

Answer: 15 mm

4 0

2 years ago