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Gennadij [26K]
3 years ago
14

A 616 g block is released from rest at height h0 above a vertical spring with spring constant k = 540 n/m and negligible mass. t

he block sticks to the spring and momentarily stops after compressing the spring 23.3 cm. how much work is done (a) by the block on the spring and (b) by the spring on the block? (c) what is the value of h0? (d) if the block were released from height 2h0 above the spring, what would be the maximum compression of the spring?
Physics
1 answer:
galben [10]3 years ago
5 0
A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs. 
Read more on Brainly.com - brainly.com/question/1581851#readmore
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A diagram labeled Before Collision shows two train cars moving toward each other. The car on the left is labeled m Subscript 1 B
OLEGan [10]

(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

#SPJ1

3 0
1 year ago
Which equation best describes the law of conservation of momentum?
NeX [460]
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>
8 0
3 years ago
Read 2 more answers
Why do objects that are thrown or shot follow a curved path
LenaWriter [7]
The initial force of the throw overcomes gravity quite easily. Then, gravity begins to bring it back down to earth, making a curved path.

8 0
3 years ago
Within a period of the periodic table, how do the properties of the elements vary?
Nonamiya [84]

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.

5 0
3 years ago
A jet transport has a weight of 2.25 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel
Rudik [331]

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

6 0
3 years ago
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