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Stella [2.4K]
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?
Physics
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.

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A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the
Andrej [43]

Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

Mass of the truck = 7100 kg

Angle = 15°

velocity = 15m/s

Acceleration = 1.5 m/s²

Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinθ

⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes  m1*gsinθ .

Resultant force on truck is F = T – m1*gsinθ  

F causes the acceleration of the truck: F= m*a

This gives the equation:

T – m1*gsinθ = m1*a  

T = m1(a + gsinθ)

Combining both equations gives:

(m1+m2)*g*sinθ = m1*(a + gsinθ)

m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ

m2*g*sinθ = m1*a

Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:

m2*g*sinθ = (7100 – m2)*a

m2*g*sinθ = 7100a – m2a

m2*gsinθ + m2*a = 7100a

m2* (gsinθ + a) = 7100a

m2 = 7100a/(gsinθ  + a)

m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)

m2 = 2636.8 kg

The load has a mass of 2636.8 kg

6 0
2 years ago
Question 1
Vesna [10]

Answer:

1)  g = 4π² / m, 3) xaxis the  length of the pendulums and the y axis the period squared

Explanation:

a) students can approximate this system to a simple pendulum, in this case the angular velocity is

         w = √ g / l

angular velocity, frequency and period are related

         w = 2π f = 2π / T

we substitute

         T = 2π√ l / g

with this equation they can determine the value of the acceleration of gravity, for this they measure the period for various lengths of the pendulum and graph

        T² = 4π²  l / g

We graph T² vs l

where this is the equation of a line if the independent variable is y = T² and x = l

        y = (4π² / g)  l

so the slope is

         m = 4π² / g

clearing

         g = 4π² / m

where the slope can be found with the values ​​of the line not the experimental values.

2) to carry out the experiment, or the thread is attached to the sphere, the length of the pendulum that goes from the pivot point to the center of the sphere is measured with a tape measure and a small finished angle is turned or less than 10th is released, it is good to wait for the first oscillation to walk, the time of a determined number of oscillations is generally measured 10 or 20, the period is calculated

    T = t / n

a table of T² against the length is made and it is plotted with the length in the ax ax, we look for the slope and hence the acceleration of gravity

3) on the independent x-axis, the controlled variable must be plotted, which is the length of the pendulums, and on the y-axis, the dependent variable is the period squared

4) of the equation of the line

            m = 4pi2 / g

                 where it ends up reaching the floor

            g = 4pi2 / m

5) when the spring is cut, the sphere remains under the effect of gravity acceleration, the harmonic movement disappears and the sphere is in a vertical movement

5 0
2 years ago
A soccer ball kicked with a force of 13.5 n accelerates at 6.5 m/s^2 to the right. what is the mass of the ball?
natita [175]

Answer:

2.08 kg

Explanation:

Newton's second law states that the acceleration of an object is proportional to the force applied to the object, according to the equation:

F=ma

where F is the force applied, m is the mass of the object and a its acceleration.

In this situation, the soccer ball is kicked with a force F=13.5 N and its acceleration is a=6.5 m/s^2, therefore its mass is

m=\frac{F}{a}=\frac{13.5 N}{6.5 m/s^2}=2.08 kg

6 0
2 years ago
A light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N
Mnenie [13.5K]

Answer:

8N and 32N

Explanation:

Given that a  light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N weight is placed between the two sawhorses, 3.0 m from the edge and 2.0 m from the center.

To calculate the forces that are exerted by the sawhorses on the board, we must consider the equilibrium of forces acting on the board.

Let the two upward forces produce by the saw horses be P1 and P2

Assuming that the weight is negligible

Sum of the upward forces = sum of the downward forces.

P1 + P2 = 40 ....... (1)

Also, the sum of the clockwise moment = sum of the anticlockwise moments.

Let's assume that the board is uniform. The weight will act at the centre.

Taking moment at the centre:

P1 × 5 + 40 × 2 = 0

P1 = 40 / 5

P1 = 8N

Substitute P1 into equation 1

8 + P2 = 40

P2 = 40 - 8

P2 = 32N

3 0
2 years ago
Read 2 more answers
A ball on a string travels once around a circle with a circumference of 2.0 m. The tension in the string is 5.0 N. how much work
SpyIntel [72]

Answer:0

Explanation:

Given

circumference of circle is 2 m

Tension in the string T=5 N

2\pi r=2

r=\frac{2}{2\pi }=\frac{1}{\pi }=0.318 m

In this case Force applied i.e. Tension is Perpendicular to the Displacement therefore angle between Tension and displacement is 90^{\circ}

W=\int\vec{F}\cdot \vec{r}

W=\int Fdr\cos 90

W=0

4 0
2 years ago
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