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castortr0y [4]
3 years ago
12

How do things rust? Is this chemical or mechanical weathering?

Physics
2 answers:
Sholpan [36]3 years ago
8 0
Rust is a type of iron oxide. When oxygen and a substance collide it creates oxides. This is a process called oxidation which is a type of chemical weathering.
IrinaK [193]3 years ago
6 0
I'll say that's mechanical weathering
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1. Describe what happens to the gravitational when you increase mass versus when you
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Answer: if the mass is doubled, the force of gravity is doubled, meaning it decreases. If the distance is doubled, the force of gravity is 1/4 as strong as before

Explanation:

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3 years ago
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A rock has a specific gravity of 2.32 and a volume of 8.64 in3 how much does it weigh
lianna [129]

Answer: 3.21 N

Specific\hspace{1mm} gravity = \frac {Density\hspace{1mm}of\hspace{1mm}substance}{Density\hspace{1mm} of \hspace{1mm}water}

\Rightarrow Density \hspace{1mm}of\hspace{1mm}substance= 2.32\times 1000\hspace{1mm} kg/m^3 = 2320\hspace{1mm}kg/m^3\\ Mass =Density\times volume\\ \Rightarrow 2320 \hspace{1mm} kg/m^3\times 8.64 \hspace{1mm}in^3\times \frac {1.64\times10^{-5} m^3}{1\hspace{1mm}in^3}=0.328 kg

For weight, we will multiply by g=9.8 m/s^{-2}

weight= 0.328\times9.8=3.21\hspace{1mm}N

Hence, the rock would weigh 3.21 N.

3 0
2 years ago
The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The heig
Mariulka [41]

Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

  • the upper and the lower channels are at the same pressure (the atmospheric pressure).
  • the velocity of water in the upper channel is zero (v₁ = 0 m/s).
  • y₁ = 2.00 m  (height of the upper channel)
  • y₂ = 0.00 m  (height of the lower channel)
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  • ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A  ⇒   A = Q/v ⇒   A = Q/v₂

⇒   A = (350 m³/s)/(6.264 m/s)

⇒   A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4   ⇒   D = 2*√(A/π)

⇒   D = 2*√(55.873 m²/π)

⇒   D = 8.434 m ≈ 8.45 m

Now, the length of the pipe can be obtained as follows

L² = x² + h²

⇒ L² = (100.00 m)² + (2.00 m)²

⇒ L ≈ 100.02 m

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2 years ago
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An electric motor rotating a workshop grinding wheel at a rate of 151 rev/min is switched off. Assume constant angular decelerat
tamaranim1 [39]

Answer:

Explanation:

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= 15.8  rad /s

final velocity = 0

Angular deceleration α = 2.23 rad / s

ω² = ω₀² -  2 α θ

0 = 15.8² - 2 x 2.23 θ

= 55.99  rad

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8 0
3 years ago
4. A car accelerates, from rest, at 2.4 m/s?. How fast is the car traveling
pantera1 [17]

Answer:

19.2m/s

Explanation:

Assuming that 2.4m/s^2 was the acceleration and not a typo, we can use the equation v=at, where v=velocity, a=acceleration, and t=time,

plug in known varibles,

v=2.4*8

v=19.2m/s

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