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alekssr [168]
2 years ago
10

Oceanographers use submerged sonar systems, towed by a cable from a ship, to map the ocean floor. In addition to their downward

weight, there are buoyant forces and forces from the flowing water that allow them to travel in a horizontal path. One such submersible has a cross section area of 1.3m2 , a drag coeffecient of 1.2, and when towed at 4.3 m/s, the tow cable makes an angle of 30 degrees with the horizontal. What is the tension in the cable? Take the water density to be 1000 kg / m3
Physics
1 answer:
KATRIN_1 [288]2 years ago
7 0

Answer:

Tension in the cable is T = 16653.32 N

Explanation:

Give data:

Cross section Area A = 1.3 m^2

Drag coefficient CD = 1.2

Velocity V = 4.3 m/s

Angle made by cable with horizontal  =30 degree

Density \rho \ of\  water= 1000 kg/m3

 Drag force FD is given as

F_{D} = \fracP{1}{2} \rho v^{2} C_{D} A

        = 0.5\times 1000\times 4.32\times  1.2\times 1.3

Drag force = 14422.2 N acting opposite to the motion

As cable made angle  of 30 degree with horizontal  thus horizontal component is take into action to calculate drag force

TCos30 = F_D

T = \frac{F_D}{cos30}

T =\frac{ 14422.2}{cos 30}

T = 16653.32 N

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\Delta x=245\ mm

Explanation:

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Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approache
FromTheMoon [43]

Answer:

y = 54.9 m

Explanation:

For this exercise we can use the relationship between the work of the friction force and mechanical energy.

Let's look for work

      W = -fr d

The negative sign is because Lafourcade rubs always opposes the movement

On the inclined part, of Newton's second law

Y Axis  

      N - W cos θ  = 0

The equation for the force of friction is

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We replace at work

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The mechanical energy in the highest part, where it stopped

     Em_{f} = U = m g y

     W = ΔEm =  Em_{f} - Em₀

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Distance d and height (y) are related by trigonometry

     sin θ = y / d

     y = d sin θ

   

    - μ m g d cos θ = m g d sin θ - ½ m v²

We calculate the distance traveled

     d (g syn θ + μ g cos θ) = ½ v²

     d = v²/2 g (sintea + myy cos tee)

     d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)

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Let's use trigonometry to find the height

      sin 16 = y / d

      y = d sin 16

      y = 199.1 sin 16

      y = 54.9 m

8 0
2 years ago
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