(a) The total momentum of the system before the train cars collide is 1,600 kgm/s.
(b) The total momentum of the system be after the train cars collide is 1,600 kgm/s.
<h3>What is the total momentum of the car system before the collision?</h3>
The total momentum of the car system before the collision is determined by applying the formula for linear momentum.
Pi = m₁u₁ + m₂u₂
where;
- m₁ is the mass of the car on the right
- m₂ is the mass of the car on the left
- u₁ is the initial velocity of the right
- u₂ is the initial velocity of the car on the left
Let the rightward direction = positive
Let the leftward direction = negative
Pi = (600 kg x 4 m/s) + (400 kg) x (-2 m/s)
Pi = 2,400 kgm/s - 800 kgm/s
Pi = 1,600 kgm/s
Based on the law of conservation of linear momentum, the sum of the initial momentum of an isolated system is <u>equal</u> to the sum of the final momentum of the system
Pf = Pi = 1,600 kgm/s.
Learn more about conservation of linear momentum here: brainly.com/question/7538238
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The law of conservation of momentum<span> states that for two objects colliding in an isolated system, the total </span>momentum<span> before and after the collision is equal. Momentum should be conserved. Hope this answers the question. Have a nice day.</span>
The initial force of the throw overcomes gravity quite easily. Then, gravity begins to bring it back down to earth, making a curved path.
Answer:
Elements in the same period have the same number of electron shells; moving across a period (so progressing from group to group), elements gain electrons and protons and become less metallic. This arrangement reflects the periodic recurrence of similar properties as the atomic number increases.
Explanation:
The Periodic Table can predict the properties of new elements, because it organizes the elements according to their atomic numbers. ... They hope that the two nuclei at the centre of these atoms will fuse and form a heavier nucleus. When these heavy elements form, they are usually highly unstable.
Answer:
Explanation:
Given that,
Weight of jet
W = 2.25 × 10^6 N
It is at rest on the run way.
Two rear wheels are 16m behind the front wheel
Center of gravity of plane 10.6m behind the front wheel
A. Normal force entered on the ground by front wheel.
Taking moment about the the about the real wheel.
Check attachment for better understanding
So,
Clock wise moment = anti-clockwise moment
W × 5.4 = N × 16
2.25 × 10^6 × 5.4 = 16•N
N = 2.25 × 10^6 × 5.4 / 16
N = 7.594 × 10^5 N
B. Normal force on each of the rear two wheels.
Using the second principle of equilibrium body.
Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces
ΣFy = 0
Nr + Nr + N — W = 0
2•Nr = W—N
2•Nr = 2.25 × 10^6 — 7.594 × 10^5
2•Nr = 1.491 × 10^6
Nr = 1.491 × 10^6 / 2
Nr = 7.453 × 10^5 N
