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alekssr [168]
3 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]3 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:

2nd and 4th

Explanation:

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3 years ago
*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude
kvv77 [185]

Answer:

Approximately 3.86\; {\rm N} (given that the magnitude of this charge is -7.35\; {\rm \mu C}.)

Explanation:

If a charge of magnitude q is placed in an electric field of magnitude E, the magnitude of the electrostatic force on that charge would be F = E\, q.

The magnitude of this charge is q = 7.35\; {\rm \mu C}. Apply the unit conversion 1\; {\rm \mu C} = 10^{-6}\; {\rm C}:

\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}.

An electric field of magnitude E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}} would exert on this charge a force with a magnitude of:

\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}.

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0
2 years ago
How are the electric field lines around a positive charge affected when a second positive charge is near it?
Lesechka [4]

Answer: The field lines bend away from the second positive charge

Explanation: opposite attracts, same repulse

4 0
3 years ago
Read 2 more answers
A simple pendulum consists of a mass m (the "bob") hanging on the end of a thin string of length l and negligible mass. The bob
svetlana [45]

Answer:

The pendulum frequency  is (c) the same, or very close to it

Explanation:

The simple pendulum corresponds to a simple harmonic movement, to reach this approximation in the expression of the force the sine of the angle (θ) approaches an angle value, this is only true for small angles, generally less than 15º

Sine (15th) = 0.2588

The angle in radians is 15º π / 180º = 0.26180.2588 / 0.2618

The difference between these two values ​​is less than 1.2%

for smaller angle the difference is reduced more

Therefore, the period for both the 5º and 10º angles is almost the same

5 0
3 years ago
Can you please tell me what this is
Anna [14]

Answer:

200000 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) of roller coaster = 1000 Kg

Velocity (v) of roller coaster = 20 m/s

Kinetic energy (KE) =?

Kinetic energy is simply defined as the energy possess by an object in motion. Mathematically, it can be expressed as:

KE = ½mv²

Where

KE => is the kinetic energy.

m =>is the mass of the object

V => it the velocity of the object.

With the above formula, we can obtain the kinetic energy of the roller coaster as follow:

Mass (m) of roller coaster = 1000 Kg

Velocity (v) of roller coaster = 20 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 1000 × 20²

KE = 500 × 400

KE = 200000 J

Therefore, the kinetic energy of the roller coaster is 200000 J.

4 0
3 years ago
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