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Greeley [361]
3 years ago
9

What is the chemical name for a sand

Physics
1 answer:
vovikov84 [41]3 years ago
7 0
It is known as silicon dioxide or silica!

Hope this helps!
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Douglas has a segment with endpoints I(5, 2) and J(9, 10) that is divided by a point K such that IK and KJ form a 2:3 ratio. He
Rufina [12.5K]
For the answer to the question above, 
 the distance from i to j is 5 parts 
(2 parts from i to k and 3 parts from k to j) 

The y distance from i to j is 
10 - 2 = 8 

Each part is 8/5 = 1.6 
Therefore the distance between the 2 parts from i to k is 3.2 

From the y coordinate of I which is 2 plus the 3.2 to point k 
2 + 3.2 = 5.2 

Answer y =5.2 

Now just convert that to fraction and that will be the answer
3 0
3 years ago
The energy an object acquires when it is exposed to a force is called _____ energy
r-ruslan [8.4K]
I'm pretty sure the energy an object acquires when exposed to a force is known was potential energy. 
4 0
3 years ago
Does air resistance affect the motion of a falling object differently when the initial velocity of the object is greater?
DIA [1.3K]
Yes. Even greater. Air resistance or drag becomes harder the faster an object goes. This is why when cars reach their max speed they don't accelerate as fast, because they are pushing harder against the wind. If I take a tennis ball and shoot it down a bottomless pit, a 400 kph, the drag will slow the ball down till it reaches terminal velocity. 
4 0
3 years ago
Read 2 more answers
What instrument is used to expand burr holes?
k0ka [10]
Craniotomes are used
4 0
3 years ago
An infinitely long cylindrical insulating shell of inner radius a and outer radius b has a uniform volume charge density p. Dete
xenn [34]

Answer: The electric field is: a) r<a , E0=; b) a<r<b E=ρ (r-a)/εo;

c) r>b E=ρ b (b-a)/r*εo

Explanation: In order to solve this problem we have to use the Gaussian law in diffrengios regions.

As we know,

∫E.dr= Qinside/εo

For r<a --->Qinside=0 then E=0

for a<r<b er have

E*2π*r*L= Q inside/εo       in this case Qinside= ρ.Vol=ρ*2*π*r*(r-a)*L

E*2π*r*L =ρ*2*π*r* (r-a)*L/εo

E=ρ*(r-a)/εo

Finally for r>b

E*2π*r*L =ρ*2*π*b* (b-a)*L/εo

E=ρ*b* (b-a)*/r*εo

3 0
3 years ago
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