Ans is <span>I. Flowing groundwater that dissolves rock , such as limestone</span>
Answer:
Kf > Ka = Kb > Kc > Kd > Ke
Explanation:
We can apply
E₀ = E₁
where
E₀: Mechanical energy at the beginning of the motion (top of the incline)
E₁: Mechanical energy at the end (bottom of the incline)
then
K₀ + U₀ = K₁ + U₁
If v₀ = 0 ⇒ K₀
and h₁ = 0 ⇒ U₁ = 0
we get
U₀ = K₁
U₀ = m*g*h₀ = K₁
we apply the same equation in each case
a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J
d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J
e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J
f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J
finally, we can say that
Kf > Ka = Kb > Kc > Kd > Ke
Answer:
a) 5.63 atm
Explanation:
We can use combined gas law
<em>The combined gas law</em> combines the three gas laws:
- Boyle's Law, (P₁V₁ =P₂V₂)
- Charles' Law (V₁/T₁ =V₂/T₂)
- Gay-Lussac's Law. (P₁/T₁ =P₂/T₂)
It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.
P₁V₁/T₁ =P₂V₂/T₂
where P = Pressure, T = Absolute temperature, V = Volume occupied
The volume of the system remains constant,
So, P₁/T₁ =P₂/T₂
a) 
The temperature of the lithosphere is around 300<span>°C</span> - 500<span>°<span>C</span></span>
Explanation:
Total mass=100+10=110
Total weight=mass×gravitational field strength
=110×10
=1100N
Work done=force×distance
=1100×10
=11000J
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