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timurjin [86]
3 years ago
15

HELP PLEASEEEE How might the concept of momentum be applied to seafloor spreading?

Physics
2 answers:
Artemon [7]3 years ago
7 0

Answer:

please mark me as brainlest

11/28/20      2:14 pm   name tom

hello i hope this is what you needed

What does seafloor spreading result in?

Explanation:

Seafloor spreading occurs at divergent plate boundaries. As tectonic plates slowly move away from each other, heat from the mantle's convection currents makes the crust more plastic and less dense. The less-dense material rises, often forming a mountain or elevated area of the seafloor. Eventually, the crust cracks.

svet-max [94.6K]3 years ago
6 0

Answer:

<h3>the momentum  of the earth</h3>

Explanation:

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At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 fi
Amanda [17]

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

(5+0.18, 2-0.1)

= ( 5.18, 1.9)

3 0
3 years ago
Three 20.0 ohm resistors are
V125BC [204]

Answer:

6.67 ohm

Explanation:

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

Resistor 1 (R₁) =20 ohm

Resistor 2 (R₂) = 20 ohm

Resistor 3 (R₃) = 20 ohm

Equivalent Resistance (R) =?

Since the resistors are arranged in parallel connection, the equivalent resistance can be obtained as follow:

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/20 + 1/20 + 1/20

1/R = (1 + 1 + 1) / 20

1/R = 3/20

Invert

R = 20/3

R = 6.67 ohm

Therefore, the equivalent resistance is 6.67 ohm.

5 0
2 years ago
If a crow flies west for 60 km and then south for 45 km, what is the direction of its displacement?
son4ous [18]
That's 105 km that he flew, or 65.2 miles !  I'm absolutely positive
that the crow must have landed and gotten some rest when you
weren't looking.  But that had no effect on his displacement when
he got where he was going, so we can continue to solve the problem:


The displacement is the distance and direction from the place
where the crow took off to the place where he landed.

-- It's distance is the hypotenuse of the right triangle whose legs
are 60 km and 45 km.

        D²  =  (60 km)²  +  (45 km)²

              =    3,600 km²  +  2,025 km²  =  5,625 km²

         D  =  √(5625 km²)  =  75 km .    
 
-- It's direction is the angle whose tangent is  (45 S / 60 W).

         tan⁻¹ (45/60)  =  tan⁻¹ (0.75)  =  36.9° south of west

                                                         =  53.1° west of south.

                                                         =  not exactly southwest but close.
7 0
3 years ago
Which instrument is used to measure depth of ocean ?
skelet666 [1.2K]
Echo sounding is a type of SONAR used to determine the depth of water by transmitting sound pulses into water. The time interval between emission and return of a pulse is recorded, which is used to determine the depth of water along with the speed of sound in water at the time.
5 0
2 years ago
A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal
Gemiola [76]

Answer:

Time taken, T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

T\ cos\theta-mg=0

T=\dfrac{mg}{cos\theta}

Sum of forces in x direction,

T\ sin\theta=\dfrac{mv^2}{r}

mg\ tan\theta=\dfrac{mv^2}{r}.............(1)

Also, r=l\ sin\theta

Equation (1) becomes :

mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}

v=\sqrt{gl\ tan\theta.sin\theta}...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

T=\dfrac{2\pi r}{v}

Put the value of T from equation (2) to the above expression:

T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}

T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}

On solving above equation :

T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Hence, this is the required solution.

4 0
3 years ago
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