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3 years ago

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Physics

1 answer:

wolverine [178]3 years ago

8 0

The layers that make up the upper mantle are the following.

The lithosphere and the asthenosphere.
The lithosphere and the asthenosphere are the layers that make up the upper mantle.
The rocky outer part of the Earth is called the lithosphere.
It is very rigid, solid, and it is not as hot as the other parts.
According to scientists, its thickness is about 90 to 100 kilometers from the surface of the earth.
The layer of the Earth that moves the lithospheric plates is called the asthenosphere.
Lithospheric plates are 60 miles long on average and are composed of continental crust.

We conclude that the lithosphere and the asthenosphere are the layers that make up the upper mantle. The new lithosphere is being created at oceanic ridges due to plate tectonics. Yes, planet Earth is getting larger constantly.

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Allison wants to determine the density of a bouncing ball. which metric measurements must she use?

lubasha [3.4K]

Density depends on mass and volume so option D is correct answer. Hope this helps!

3 0

3 years ago

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0

3 years ago

Hey guys.. ans me

Stels [109]

The time taken by the swimmer was 1 hour.

Why?

Since the swimmer is maintaining an angle of 150° while he was swimming, there were two components of the speed (horizontal and vertical). If we want to calculate the time taken by him to cross the river, we need to calculate the vertical speed and consider that the flow's speed is compensated by his horizontal speed.

We can calculate both components of the speed using the following formula:

$HorizontalSpeed=Speed*Cos(150\°)\\\\VerticalSpeed=Speed*Sin(150\°)$

Now, calculating we have:

$HorizontalSpeed=2\frac{Km}{h} *Cos(150\°)=-\sqrt{3} \frac{Km}{h} \\\\VerticalSpeed=2\frac{Km}{h} *Sin(150\°)=1\frac{Km}{h}$

Therefore, we have that the horizontal speed is compesating the flow's speed while his vertical speed is used to cross the river which is 1 Km wide.

Hence, we have that the tame taken is:

$Time=\frac{RiverWidth}{VerticalSpeed}=\frac{1Km}{1\frac{Km}{h} } =1h$

Have a nice day!

7 0

3 years ago

8. What is the rate of change in an object’s position?

adelina 88 [10]

A similar but separate notion is that of velocity, which the rate of change of position. Example . If p(t) is the position of an object moving on a number line at time t (measured in minutes, say), then the average rate of change of p(t) is the average velocity of the object, measured in units per minute.

4 0

3 years ago

Read 2 more answers

A ball player catches a ball 3.24 s after throwing it vertically upward.With what speed did he throw it?

Firdavs [7]

Doesn't seem like we know much here, but we can answer it. Let's talk about what we know. We know it takes 3.24 s for the ball to go up and drop back down again. We know that gravity is the only force acting after the ball leaves the hand, so a = 9.8 m/s^2 (we'll say it's negative in our equations because down being negative is intuitive). We also know that it stops moving for a brief moment at the top of the arc, where v = 0 m/s. Because gravity is the only force, and it slows it down on the way up at the same rate it speeds it up on the way down and the distance covered in upward and downward motion is the same, we can confidently say that it will reach the top of its arc (where v = 0 and it turns around) in half the total time it is in the air, so it takes 1.62 s to reach the peak. Now we can use a kinematics equation, let's use vf = vi + a*t, where vf is final velocity and is 0, vi is initial velocity and is some unknown v we need to solve for, a is acceleration and is -9.8 m/s^2 and t is time and since this is just to the top of the arc, we'll use half the time so 1.62 s. We can solve for vi and plug stuff in like so: v = -a*t = -(-9.8m/s^2)*(1.62s) = 15.876 m/s.

8 0

3 years ago