Answer:
Explanation:
Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .
velocity of approach = 1.5 - 0 = 1.5
velocity of separation = v₁ + v₂
coefficient of restitution = velocity of separation / velocity of approach
.8 = v₁ + v₂ / 1.5
v₁ + v₂ = 1.2
applying law of conservation of momentum
m x 1.5 + 0 = mv₂ - mv₁
1.5 = v₂ - v₁
adding two equation
2 v ₂= 2.7
v₂ = 1.35 m /s
v₁ = - .15 m / s
During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.
For second collision ,
coefficient of restitution = velocity of separation / velocity of approach
.5 = v₃ + v₄ / 1.35
v₃ + v₄ = .675
applying law of conservation of momentum
m x 1.35 + 0 = mv₄ - mv₃
1.35 = v₄ - v₃
adding two equation
2 v ₄= 2.025
v₄ = 1.0125 m /s
v₃ = - 0 .3375 m / s
Answer:
532 millimeters of mercury
Explanation:
In order to convert the pressure from atm to millimeters of mercury (mm Hg), we should remind the conversion factor between the two units:
1 atm = 760 mm Hg
Therefore, we can solve the problem by setting up the following proportion:

Solving for x, we find

Answer:
The correct answer is -
A (the entire green box): Chemical Equation
B (the blue box): Reactants
C (the arrow): Reacts to Form
D (the number): Coefficient
E (the purple box): Products
Explanation:
The chemical reaction of burning methane and oxygen is as follows;
Here, the green part A is the chemical equation that includes various parts that are reactants B, methane, and oxygen, C is an arrow that indicates the formation of products.
2 is here coefficient that indicates the moles of the oxygen which forms carbon dioxide and water in box E is products
About 9.81 m/s^2
. Hope that helps
Answer:
I have no clue I'm just trying to get points
Explanation:
:) sorry