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vazorg [7]
3 years ago
8

If you increase the surface area of a rock, how will it affect the rate at which it weathers? a. It will weather more quickly. b

. It will have no effect because surface area is not a factor in weathering. c. It will weather more slowly. d. It will have no effect because a rock does not have a surface area. Please select the best answer from the choices provided A B C D
Physics
1 answer:
scoundrel [369]3 years ago
6 0

a. It will weather more quickly.

Because more the surface area, more the friction and more the weathering

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A man has 2 spheres A and B. He gently drops sphere A vertically down and throws the sphere B horizontally at the same time. Whi
anyanavicka [17]

Answer:

The answer is c

Explanation:

5 0
2 years ago
In 1923, the United States Army (there was no U.S. Air Force at that time) set a record for in-flight refueling of airplanes. Us
dybincka [34]

Answer:

1.95m/s

Explanation:

Please view the attached file for the detailed solution.

The following were the conversion factors used in order to express all quatities in SI units:

1 gallon=0.00378541m^3\\1 inch=0.0254m\\1 minute=60s

6 0
3 years ago
two objects are thrown from the top of a tall building and experience no appreciable air resistance. one is thrown up and the ot
Iteru [2.4K]

Answer:

The two objects are traveling at the same speed.

Explanation:

Neglecting air resistance, an object that is thrown up from the top of a tall building has the same speed as the second object thrown down from the top of the same tall building since the initial speed is the same.

The object thrown up is not traveling faster neither is the object thrown down traveling faster.  

Therefore, the two objects will have the same speed when they hit the ground but their time of landing might be different.

3 0
3 years ago
Read 2 more answers
Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially
AnnZ [28]

Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.

Th magnitude of object A's momentum change is thus 25.71 kgm/s

6 0
3 years ago
Anne releases a stone from a height of 2 meters. She measures the kinetic energy of the stone at 9.8 joules at the exact point i
insens350 [35]
A. 0.5kg

To get this answer you need to follow the equation of KE=0.5*mv^2 
But we don't have the m part in the equation. So just plug in the numbers to see which works best, though I can tell you before we do that the answer would be a. 

As you may know, gravity, is a force of 9.8 m/s. And we want to get 9.8 Joules. So if we take a half a kg stone, release it at one meter, we get half of the normal gravity pull, 4.90 Joules. That means if we take half a kg stone and drop it at a doubled height, we get 9.8 Joules.

That is also to say that if we have a 1kg stone and drop it at one meter you will get the normal pull of gravity in Joules, 9.8J. 

Be careful though, this does not mean if you drop a 1kg stone and a .5 kg stone the 1kg will hit first. This simply means that the 1kg stone will have twice the Joules that the .5kg stone has.
7 0
3 years ago
Read 2 more answers
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