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3 years ago

Which method is not a technique currently used by ocean scientists to map the topography of the ocean floor?

Physics

1 answer:

raketka [301]3 years ago

4 0

Answer:

Magnetometer

Explanation:

Magnetometer technique is not using by scientists for studying the ocean floor.The scientists currently is using SONAR ( sound navigation and ragging) technique for studying the ocean floor.SONAR is used sound waves sound waves for studying the ocean floor or we can say that SONAR is based on sound propagation.

Therefore answer is Magnetometer

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What valves in this diagram need to be opened to allow liquid to flow?

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I'd guess at valve B. more information about the interesting question would help.

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4 years ago

What part of a circuit has the main job of conducting the electricity?

Vesnalui [34]

The main job of conducting electricity is the power source.

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3 years ago

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The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an

9966 [12]

Answer:

$-4.941*10^8J.$

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

$\Delta U = \Delta_{perogee}-\Delta_{Apogee}$

Where,

$U= \frac{-GmM_e}{r}$

Replacing

$\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}$

$\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})$

Our values are given by,

$m = 85.5Kg$

$M_e = 5.97*10^{24}Kg$

$r_A = 7330Km$

$r_p = 6610Km$

$G = 6.67*10^{-11}Nm^2/Kg^2$

Replacing at the equation,

$\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})$

$\Delta U = -4.941*10^8J$

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was $-4.941*10^8J.$

4 0

3 years ago

Is it true or false Objects with different masses can’t have the same momentum.

xenn [34]

Answer:

true

Explanation:

3 0

2 years ago

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S GP A projectile of mass m moves to the right with a speed vi (Fig. P11.51a). The projectile strikes and sticks to the end of a

Andrews [41]

What is the kinetic energy of the system after the collision?

$K_f=\frac{3}{2} \frac{m^{2}v_i^{2} }{(M+3m)}$

How this is calculated?

Given:

Initial speed= $v_i$

mass of rod=M

Let, Initial kinetic energy = $K_i$

Final kinetic energy= $K_f$

Moment of inertia =I

What is the moment of inertia?

$I=(I_p)_0+(I_{rod})_0\\I=m(\frac{d}{2})^{2} +\frac{Md^{2} }{12} \\I=\frac{(M+3m)d^{2} }{12}$

What is the angular momentum?

By conservation of angular momentum,

$L_i=L_f$

$mv_i\frac{d}{2}=\frac{(M+3m)d^{2}\omega }{12} \\\omega=\frac{6mv_i^{2} }{d(M+3m)}$

We know that, the final kinetic energy is given by,

$K_f=I\omega^{2}\\K_f=\frac{1}{2} *\frac{(M+3m)d^{2} }{12} *\frac{36m^{2}v_i^{2}}{d^{2}(M+3m)^{2}}\\ K_f=\frac{3}{2} \frac{m^{2}v_i^{2} }{(M+3m)}$

What is the kinetic energy?

In physics, the kinetic energy of an object is the energy that it possesses due to its motion.
It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.

To know more about kinetic energy, refer:

brainly.com/question/114210

#SPJ4

8 0

1 year ago