Explanation:
Given that,
Mass of a freight car, 
Speed of a freight car, 
Mass of a scrap metal, 
(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :
So, the final velocity of the loaded freight car is 0.182 m/s.
(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy
![\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J](https://tex.z-dn.net/?f=%5CDelta%20K%3D%5Cdfrac%7B1%7D%7B2%7D%5B%28m_1%2Bm_2%29V%5E2-m_1u_1%5E2%29%5D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20%5B%2830%2C000%2B110%2C000%20%290.182%5E2-30000%280.85%29%5E2%5D%5C%5C%5C%5C%3D-8518.82%5C%20J)
Lost in kinetic energy is 8518.82. Negative sign shows loss.
The velocity of the boat after the package is thrown is 0.36 m/s.
<h3>
Final velocity of the boat</h3>
Apply the principle of conservation of linear momentum;
Pi = Pf
where;
- Pi is initial momentum
- Pf is final momentum
v(74 + 135) = 15 x 5
v(209) = 75
v = 75/209
v = 0.36 m/s
Thus, the velocity of the boat after the package is thrown is 0.36 m/s.
Learn more about velocity here: brainly.com/question/6504879
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Answer:
0.893 rad/s in the clockwise direction
Explanation:
From the law of conservation of angular momentum,
angular momentum before impact = angular momentum after impact
L₁ = L₂
L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)
L₂ = angular momentum of cylinder and angular momentum of bullet after collision.
L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision
So,
L₁ = L₂
L₁ = (I₁ + I₂)ω
ω = L₁/(I₁ + I₂)
ω = L₁/(1/2MR² + mR²)
ω = L₁/(1/2M + m)R²
substituting the values of the variables into the equation, we have
ω = L₁/(1/2M + m)R²
ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²
ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)
ω = + 9 kgm²/s/(2.52 kg)(4 m²)
ω = +9 kgm²/s/10.08 kgm²
ω = + 0.893 rad/s
The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.
Answer:
Proper weighting
Explanation:
Proper weighing involves the condition of a scuba diver that is fully geared having a near empty tank and the BCD emptied with a held breadth is expected to float at eye level
The fundamental of adequate or good buoyancy of a scuba diver is to ensure proper weighting when diving, With proper weighting, there is more control for the diver when a safety stop is required. There is less need to carry excess weight that increases drag and gas consumption.
Answer:
D. is greater for turbulent flow than for laminar flow
Explanation:
what is friction drag?
- friction drag is a phenomenon experienced when a body moves through a fluid. A practical example can be seen in the mild warmth we experience rubbing the palm's of one's hand together only in this case we are dealing with a solid body and a fluid (e.g air, water). friction drag is directly proportional to the area of the surface in contact with the fluid and increases as velocity increases. We see a practical example of this when the rate at which one rubs the palms together is fast but we use the word turbulent when we are dealing with fluids. Turbulent flow creates more friction drag than laminar flow( Flow between a smooth body and fluid) due to its greater interaction with the surface of the body
- it is important to know that friction is also called viscous drag or skin drag
- I recommend Richardson and coulson vol 2 textbook, page 149, Chemical enginering fluid mechanics textbook by Ron dardy, page 341 for clearer explanation
