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xxTIMURxx [149]
3 years ago
7

Drag and drop the words into the correct locations to label the layers of the atmosphere, with the lowest layer at the bottom. (

2 points)
Space

top layer: ___

fourth layer: ___

middle layer: ___

second layer: ___

lowest layer: ___

Earths surface


Word bank:

exosphere

thermosphere

mesophere

tropospere

stratosphere
Physics
2 answers:
professor190 [17]3 years ago
4 0

Answer:

The top layer is exosphere. The fourth is the Thermosphere. The middle layer is the mesophere. The second layer is the Stratospere. the lowest layer is the troposphere

Explanation:

Valentin [98]3 years ago
3 0

Answer:

All the answers are.

Explanation: 1 clouds and possible precipitation

2 each grid point on a three-dimensional get

3 it produces uneven heating of earths surface by the sun

4 C a warm surface current flows along the east coast of the United States and a cold water surface current flows along the West Coast of the United States

5 plate tectonics

6 wet soil absorbs heat more slowly, due to waters high heat capacity

7 top layer: exosphere, fourth layer:thermosphere, middle layer:mesosphere, second layer: stratosphere, lowest layer: troposphere

8 physics that describes how air moves and heat and moisture are exchanged.

I just took the test and these are the correct answers, I hope this helped you.

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Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.
Korolek [52]

Answer:

If the acceleration is constant, the movements equations are:

a(t) = A.

for the velocity we can integrate over time:

v(t) = A*t + v0

where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:

\int\limits^{10}_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)

Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.

4 0
3 years ago
You and a friend each hold a lump of wet clay. Each lump has a mass of 25 grams. You each toss your lump of clay into the air, w
solniwko [45]

Answer:

a) p(total) = <0.05, 0.1, 0.1 > kg m/s

b) p = <0.05, 0.1, 0.1 > kg.m/s

c) v_f = < 1, 2, 2 > m/s

Explanation:

a.)

Mass of each lump = 25 g = 0.025 kg

Velocity of lump 1 = < -2, 0, -7 > m/s

Momentum of lump 1 = Mass×Velocity

                                   = 0.025×< -2, 0, -7 >

                                   = < -0.05, 0, 0.175> kg m/s

Velocity of lump 2 = < 4, 4, -3 > m/s

Momentum of lump 2 = Mass×Velocity

                                    = 0.025×< 4, 4, -3 >

                                    = < 0.1, 0.1, -0.075> kg m/s

Total momentum before impact  =  < -0.05,  0,  0.175 > + < 0.1, 0.1, -0.075>

                                                      = < 0.05, 0.1, 0.1 > kg m/s

⇒p(total) = <0.05, 0.1, 0.1 > kg m/s

b)

As we know that,

By the law of conservation of linear momentum,

The total momentum will be the same before and after the collision.

⇒Momentum of the stuck together  after the collision = Total momentum of the lumps just before impact.

⇒ p = <0.05, 0.1, 0.1 > kg m/s

c)

Let the final velocity =  v_f

Total mass = 0.025 + 0.025 = 0.05 kg

As

Momentum = mass ×velocity

⇒ <0.05, 0.1, 0.1 > = 0.05 ×v_f

⇒ v_f = <0.05, 0.1, 0.1 > / 0.05

          = < 1, 2, 2 > m/s

⇒v_f = < 1, 2, 2 > m/s

7 0
3 years ago
Free charges do not remain stationary when close together. To illustrate this, calculate the magnitude of the instantaneous acce
ASHA 777 [7]

Answer:

a=2.304×10¹⁶m/s²

Explanation:

Given data

Distance d=2.5 nm=2,5×10⁻⁹m

Mass of proton m=1.6×10⁻²⁷kg

charge of proton q=1.6×10⁻¹⁹C

To find

acceleration a

Solution

Apply the Coulombs Law

F=k\frac{q_{1}q_{2}  }{r^{2} }

Where k is coulombs constant (k=9×10⁹Nm²/C²)

q=q₁=q₂

r=d

So

F=k\frac{|q^{2} |}{d^{2} }\\ as \\F=ma\\ma=k\frac{|q^{2} |}{d^{2} }\\a=\frac{k}{m} \frac{|q^{2} |}{d^{2} }\\a=\frac{(9*10^{9} )*(1.6*10^{-19} )^{2} }{(1.6*10^{-27} )*(2.5*10^{-9} )^{2} }\\ a=2.304*10^{16}m/s^{2}  

4 0
3 years ago
An animal cell placed in pure distilled 100% will
jasenka [17]

in a hypotonic solution like distilled water, a red blood cell would burst, because inside the cell has a higher solute concentration than outside.

In a hypertonic solution, there is a higher solute concentration on the outside of the cell than on the inside, causing the cell to shrivel.

i hope this helps

8 0
3 years ago
An inclined plane makes it easier by _____ the distance and decreasing the force
Flauer [41]

Answer:increasing

Explanation:

8 0
3 years ago
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