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MrRa [10]
3 years ago
9

Compare the strength of the bonds that hold the atoms in a molecule together with the forces that exist between different molecu

les
Chemistry
1 answer:
melomori [17]3 years ago
5 0
Hydrogen, ammonia, methane and water are also simple molecules with covalent bonds. All have very strongbonds between the atoms, but much weaker forces holding the molecules together. When one of these substances melts or boils, it is these weak 'intermolecular forces' that break, not the strong covalent bonds.
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Please help.This is due tomorrow.It's worth 2 grades.Please help.God bless u.Please and thankyou so much.
Daniel [21]

Answer:

1. False - compression

2. True

3. False - transform faults

4. False - horizontally

5. True

6. False- perpendicular

7. False - away from

8. False - increase

9. True

10. True

Explanation:

1. Mountains, oceanic trenches, and rift valleys are created by tension and compression stress. They are formed by divergent and convergent boundaries. Compression stress occurs when plates are pushing against each other, while tension stress occurs when the plates are pulling away from each other.

**Shear stress happens when the plates grind against each other. Often found in transform boundaries.

2. Transform faults happen when two plates glide or slide against each other. These areas are called transform boundaries. Transform faults occur in the ocean. When these boundaries are formed on land, they are called strike-slip faults.

3. Shear stress that occur in transform boundaries produce transform faults. These faults are usually identified by long faults and ridges. Sometimes small ponds form in the cracks due to deposition.

*** Rift valleys are produced by divergent boundaries or tension stress, when the plates are pulled apart.

4. Transform boundaries are formed when two plates slides against each other. Transform faults are formed in these boundaries and the movement of the plates are horizontal.

*** They do not move vertically.

5-6. Mid-oceanic ridges are segmented or divided by transform faults. The transform faults in the mid-oceanic ridges are perpendicular to the oceanic ridges. They separate them into distinct segments and can run across for hundreds of kilometers

7. New faults form as they move away from the ridges. Mid oceanic ridges are formed when the plates move apart, pushing the seafloor outwards and along with that, the transform faults. When new crust however overlaps the transform fault, they stop moving against each other, and start moving side by side, creating a crack.

8. Transform faults increase in size as long as the plates continue to move. The areas of transform faults, especially in the surface create earthquake faults.

9. Faults at the surface can be part of a larger underground system. Some faults can cut across continental crusts. These faults are created by different geological processes, like compression stress from convergent boundaries, tension stress from divergent boundaries, and shear stress from transform boundaries.

10. Fault zones are areas where you can find different faults formed, relatively close to each other. The faults in fault zones can be shallow or deeper like the fault zone Sierra Madre.

6 0
4 years ago
What is the value of the specific heat capacity of liquid water in j/mol·°c?
Scilla [17]
<span>he specific heat capacity of liquid water is 4.186 J/gm K.</span>
4 0
3 years ago
Read 2 more answers
What happens when an island volcano erupts and molten lava flows out?
qwelly [4]
It will explode together cause danger  
3 0
3 years ago
Read 2 more answers
What does quantities data mean?
aliina [53]
Data that can be measured, deals with numbers and length,height,area,volume etc.
8 0
3 years ago
A liquid of density 1290 kg/m 3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the
Gelneren [198K]

Answer:

114 kPa

Explanation:

By Bernoulli's equation when a fluid flows steadily through a pipe:

P + ρ*g*y + v² = constant in the pipe, where P is the pressure, ρ is the density of the fluid, g is the gravity acceleration (9.8 m/s²), y is the high, and v the velocity.

By the continuity equation, the liquid flow must be constant in the pipe, and then:

A1*v1 = A2*v2

Where A is the area, v is the velocity, 1 is the point 1, and 2 the point 2 in the pipe. The are is the circle area: π*(d/2)². So:

π*(0.105/2)²*9.91 = π*(0.167/2)²*v2

0.007v2 = 0.027

v2 = 3.9 m/s

Then:

P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²

ρ*g*y1 - ρ*g*y2 + v1² - v2² = P2 - P1

ρ*g*Δy + v1² - v2² = ΔP

ΔP = 1290*9.8*9.01 + 9.91² - 3.9²

ΔP = 113,987.42 Pa

ΔP = 114 kPa

8 0
3 years ago
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