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3 years ago

The lithosphere is made up of the ____

Physics

2 answers:

umka21 [38]3 years ago

8 0

The lithosphere is Earth's outermost layer

Alex17521 [72]3 years ago

6 0

The lithosphere is Earth's outermost layer, composed of rocks in the crust and upper mantle that behave as brittle solids. The rigid lithosphere sits on top of the asthenosphere, a layer of the mantle in which rocks are hot and deformable.

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Two cars, initially at rest and 5 km apart at t=0 , simultaneously move toward each other. Car A travels at a constant speed of

Anastasy [175]

Answer:

d. 268 s

Explanation:

Constant Speed Motion

An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

$\displaystyle v=\frac{d}{t}$

Where

v = Speed of the object

d = Distance traveled

t = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

And we can also solve it for t:

$\displaystyle t=\frac{d}{v}$

Two cars are initially separated by 5 km are approaching each other at relative speeds of 55 km/h and 12 km/h respectively. The total speed at which they are approaching is 55+12 = 67 km/h.

The time it will take for them to meet is:

$\displaystyle t=\frac{5}{67}$

t = 0.0746 hours

Converting to seconds: 0.0746*3600 = 268.56

The closest answer is d. 268 s

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3 years ago

A hockey ball accelerates from 0.m/s to 25m/s in 0.05 seconds what is the acceleration of the ball ?

S_A_V [24]

Answer:

500

Explanation:

25/0.05 .

I think this ?

3 0

3 years ago

Describe a well-known hypothesis that was discarded because it was found to be untrue.earth-centered model of the universe. the

polet [3.4K]

The widely accepted hypothesis before that turned out wrong was the Earth-Centered theory or the Geocentric Theory. This was proposed by the philosopher Ptolemy. He came about to this hypothesis from hi observation that from the Earth's perspective, the celestial bodies like the Sun, stars and the moon, look like they rotate around the Earth each day and night. However, this was disproved by Galileo Galelei by his Heliocentric Theory. He observed through the telescope that the Venus also changes phases like the moon. However, he deduced that this is not possible from the positions of the Venus, Earth, Moon and Sun.

6 0

3 years ago

What are non-metals? explain briefly

aliya0001 [1]

Answer:

In chemistry, a nonmetal is a chemical element that is mechanically weak in its most stable form, brittle if solid, and usually gains or shares electrons in chemical reactions. There is no universal agreement on which elements are nonmetals; the numbers generally range from fourteen to twenty-three, depending on the criterion or criteria of interest.

3 0

2 years ago

Read 2 more answers

An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ outward

3) y-component of the electric field at r =8.7 cm: 0

4) x-component of the electric field at r = 1.15 cm: $1.28\cdot 10^7 N/C$ outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

The linear charge density of the cylindrical insulating shell can be found by using

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

in the outward direction.

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

$E_y=0$

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0

3 years ago