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4 years ago

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5. Describe the properties of each of these parts of the Earth’s interior: lithosphere, mantle, and core. What are they made of?

How hot are they? What are their physical properties?

1 answer:

Arisa [49]4 years ago

8 0

Hey there!

Okay so, the Lithosphere is made out of Earth’s outermost layer, which is composed of rocks in the crust and the upper mantle that behave as brittle solids.

The Mantle is made up of rock containing silicon, iron, magnesium, aluminum, oxygen and other minerals.

The Core has two parts. The solid inner core made up of iron. The outer core is surrounded by a liquid composed of a nickel-iron alloy.

I hope this helps!

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I need help thanks :)))))))))

dsp73

°C = (5/9) · (°F-32)

The "wet" thermometer is the upper one ... you can see the wet cloth wrapped around the bulb at the end. It's reading 70° F.

°C = (5/9) · (38) = 21.1° C

The "dry" thermometer is the lower one. It's reading 80° F.

°C = (5/9) · (48) = 26.7° C

So it looks like choice-A is your answer.

6 0

3 years ago

in the primitive yo-yo apparatus (figure 1), you replace the solid cylinder with a hollow cylinder of mass m , outer radius r ,

kirza4 [7]

The magnitude of the downward acceleration of the hollow cylinder is 6m/s^2.

Z = I α

T.R =1/2 M ( $R^{2}$ + $(R/2)^{2}$ )α

T.R = 1/2M 5 $R^{2}$ /4 α

T = 5Ma/8

Mg - T = Ma

Mg - 5Ma/8 = Ma

Mg= 5Ma/8 + Ma = 13Ma / 8

acceleration = 8g/13 = 6 m/s^2

The rate at which an object's velocity with respect to time changes is called its acceleration. The direction of the net force imposed on an item determines its acceleration in relation to that force. According to Newton's Second Law, the magnitude of an object's acceleration is the result of two factors working together

The size of the net balance of all external forces acting on that item is directly proportional to the magnitude of this net resultant force; the magnitude of that object's mass, depending on the materials from which it is built, is inversely related to its mass.

Learn more about acceleration here:

brainly.com/question/2303856

#SPJ4

4 0

1 year ago

An observer stands 24.7 m behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of

GaryK [48]

Explanation:

The given data is as follows.

Velocity of bullet, $c_{p}$ = 814.8 m/s

Observer distance from marksman, d = 24.7 m

Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.

t = $\frac{24.7}{343}$ (velocity in air = 343 m/s)

= 0.072 sec

Now, before the observer hears the report the distance traveled by the bullet is as follows.

$d_{b} = c_{b} \times t$

= $814.8 \times 0.072$

= 58.66

= 59 (approx)

Thus, we can conclude that each bullet will travel a distance of 59 m.

8 0

3 years ago

A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how

Montano1993 [528]

<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity $V_{o}$ has two components, because the brick was thrown at an angle $\alpha=61\º$ :

$V_{ox}=V_{o}cos\alpha$ (1)

$V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}$ (2)

$V_{oy}=V_{o}sin\alpha$ (3)

$V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}$ (4)

As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

$X=V_{ox}.t$ (5)

Where:

$X=10.1m$ is the distance where the brick landed

$t$ is the time in seconds

If we already know $X$ and $V_{ox}$ , we have to find the time (we will need it for the following equation):

$t= \frac{X}{ V_{ox}}$ (6)

$t=2.42s$ (7)

<h2>In the y-axis: </h2>

$-y=V_{oy}.t+\frac{1}{2}g.t^{2}$ (8)

Where:

$y$ is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

$g=-9.8\frac{m}{s^{2}}$ is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

$-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}$ (9)

$-y=-10.52m$ (10)

Multiplying by -1 each side of the equation:

$y=10.52m$ >>>>This is the height of the building

3 0

3 years ago

A golfer takes two putts to get his ball into the hole he is on green. The first putt displaces the ball 4.5m east, and the seco

LenaWriter [7]

I think its the one going it at 4.5m

8 0

3 years ago