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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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Answer:

Explanation:

Acceleration is the time rate of change of velocity.

Acceleration and velocity are vectors

If east and north are the positive directions, the east moving vector is reduced to zero and the north moving vector increases from zero to 4 m/s.

There are 3 hours or 10800 seconds between 10 AM and 1 PM

a1 = √((-4)² + 4²) / 10800 = (√32) / 10800 m/s² ≈ 4.2 x 10⁻⁴ m/s²

There are 14400 seconds between 10 AM and 2 PM

The velocity changes are still the same

a2 = √((-4)² + 4²) / 10800 = (√32) / 14400 m/s² ≈ 3.9 x 10⁻⁴ m/s²

7 0
2 years ago
Please help! will give brainlest!!!!!!!!!!!!
eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

a=0

Therefore the net force is zero as well:

\sum F = 0

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

\sum F = F cos \theta - F_f = 0

where

F cos \theta is the horizontal component of the applied force, with

F = 22.5 N

\theta=32.0^{\circ}

F_f is the force of friction

And solving for F_f, we find

F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N

Learn more about friction:

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2 years ago
_____ discovered that light also showed properties commonly found in waves.
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Through the work of Max Planck<span>, Einstein, </span>Louis de Broglie<span>, </span>Arthur Compton<span>, </span>Niels Bohr<span>, current scientific theory holds that all particles also have a wave nature (and vice versa).</span>
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2 years ago
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A swimming pool contains x (less than 0.02) grams of chlorine per cubic meter. the pool measures 5 meters by 50 meters and is 2
zubka84 [21]
The solution for this problem would be:(10 - 500x) / (5 - x) 
so start by doing: 
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3 0
2 years ago
29. 2072 Set C Q.No. 10c
Annette [7]

Answer:

90.2^{\circ}C

Explanation:

Considering the thermal conductivity of aluminium and brass as k_{al}=205 W/mK and k_{br}=109 W/mk respectively  

The temperature at the end of aluminium and brass are given as T_{al}=150^{\circ}C and T_{br}=20^{\circ}C respectively with length of rod L=1.3 m , Length of aluminium L_{al}=0.8 m, length of brass L_{br}=0.5 m and letting temperature at steady state be T

At steady state, thermal conductivity of both aluminium and brass are same hence

H_{br}=H_{al}

k_{al}A\frac {T_H-T}{L_{al}}= k_{br}A\frac {T-T_H}{L_{br}}

Upon re-arranging

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(205)\frac {150-T}{0.8}=109\frac {T-20}{0.5}

T=\frac {(109*0.8*20)+(205*0.5*150)}{(109*0.8)+(205*0.5)}

T=90.2^{\circ}C

Therefore, the temperatures at which the metals are joined is 90.2^{\circ}C

6 0
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