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DerKrebs [107]
3 years ago
6

The arrows on the diagram show the ocean floor spreading from the Redge what three kinds of evidence scientists have found suppo

rt this idea?
Physics
1 answer:
Blizzard [7]3 years ago
5 0
Edit: You do mean Ridge?

Rocks near Mid-Ocean Ridge are younger than rocks near the trenches.

Seismic data shows oceanic crust is sinking into the mantle at the trenches.

Matching bands of magnetic rock are found on either side of the Ridge. Earth's magnetic fields change these bands over time.
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What is the kinetic energy of a 1,350kg car traveling at a speed of 12m/s?
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Kinetic energy= 1/2 m v^2

so... 1/2(1350)(12^2)

kinetic energy = 97200Joules
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3 years ago
Sound waves with frequency higher that 20,000 Hz are referred to as
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Hecto the Mornar a fursa explain i magnituse of the force acting right angle to the moment arm​
mixer [17]
<h2>~<u>Solution</u> :-</h2>
  • Here, the <u>moment arm</u> is defined as follows;

The magnitude of two forces, which when acting at right angle produce resultant force of VlOkg and when acting at 60° produce resultant of Vl3 kg. These forces are D. gravitational force of attraction towards the centre of the earth. A sample of metal weighs 219 gms in air, 180 gms in water, 120 gms in an <em>unknown fluid</em>.

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7 0
3 years ago
A teapot with a surface area of 700 cm2 is to be plated with silver. It is attached to the negative electrode of an electrolytic
Brums [2.3K]
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5 0
4 years ago
Need help solving this question.
MatroZZZ [7]

Answer:

See the answers below.

Explanation:

to solve this problem we must make a free body diagram, with the forces acting on the metal rod.

i)

The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.

We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.

For the summation of forces we will take the forces upwards as positive and the negative forces downwards.

ΣF = 0

-15+T-W=0\\T-W=15

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.

ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.

ΣM = 0

(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]

7 0
3 years ago
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