The speed of cart b is 6m/s while the total momentum of the systmen is 4200 kg m/s
<h3>Conservation of Linear Momentum</h3>
Given Data
- Mass of cart one M1 = 150kg
- Initial Velocity U1 = 8m/s
Mass of cart two M2 = 150kg
Velocity U2 = 6m/s
Applying the principle of conservation of linear momentum we have
M1U1+M2U2 = M1V1+ M2V2
a. what is the speed of cart b after collision
substituting our given data we have
150*8+ 150*6 = 150*5+150*V2
1200 + 900 = 1200+ 150V2
2100 - 1200 = 150V2
900 = 150V2
Divide both sides by 150
V2 = 900/150
V2 = 6m/s
b. what is the total momentum of the system before and after collision
Total Momentum in the system is
Total momentum = Momentum before Impact+ Momentum after Impact
Total momentum = M1U1+M2U2 + M1V1+ M2V2
Total momentum = 1200 + 900 + 1200+ 900
Total momentum = 4200 kg m/s
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The velocity of the pitcher at the given mass is 0.1 m/s.
The given parameters:
- <em>Mass of the pitcher, m₁ = 50 kg</em>
- <em>Mass of the baseball, m₂ = 0.15 kg</em>
- <em>Velocity of the ball, u₂ = 35 m/s</em>
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Let the velocity of the pitcher = u₁
Apply the principle of conservation of linear momentum to determine the velocity of the pitcher as shown below;
m₁u₁ = m₂u₂
Thus, the velocity of the pitcher at the given mass is 0.1 m/s.
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Answer:
m=ρV
V=4/3 * pi * r3
V=1.3 * 3.14 * 3.9^3
V=242.14 cm^3
m=7.58 * 242.14
m=1.8 kG
Explanation:
1. We calculate volume for sphere.
2. Then we calculate mass of sphere.
Based on my information, this would actually be representing
"the average kinetic energy of water particles". So, as you take notice that where this temperature is being located, and also, how this would be
°C, this would make more sense for this to be representing as <span>the
average kinetic energy of water particles.</span>
The carnot cycle attempts to model the most efficient possible process by avoiding irreversible processes.
In essence, the Carnot cycle is a reversible cycle made up of four other reversible processes. A reversible process is one that can be thought of as consisting of a sequence of equilibrium stages because it is carried out endlessly slowly.
Essentially, this means that any reversible cycle can be performed in reverse and that the amount of work or heat exchanged along the forward and backward pathways is the same.
It goes without saying that such reversible processes are not possible because they would take an unlimited amount of time. Therefore, the Carnot Engine is described as an idealized heat engine that uses the Carnot Cycle, a reversible cycle.
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