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dlinn [17]
3 years ago
11

100 POINTS! PLEASE HELP!

Physics
2 answers:
bagirrra123 [75]3 years ago
8 0

Answer:

By the information given, I am not sure how the event has played out. However, using common knowledge, I can give a scenario of how it may play out.

When removing the first block (the foundation), gravity will pull the other blocks down, as the strength of gravity exceeds that of the strength of air holding it up. This means that all the blocks on top would fall.

By definition of Newton's 1st - 3rd laws, the blocks would not move unless a force is given to it (i.e., removing the bottom block, which causes gravity). Gravity would pull the other blocks downwards. This would lead to the falling of the stack of blocks.

~

vaieri [72.5K]3 years ago
8 0

Answer:

Explanation:

A stack of blocks sits in equilibrium. That means, by Newton's 1st law, all blocks will remain stationary. The weight of each block is balanced by the reaction force from the block underneath it. The bottom block experiences a reaction force equals to the total weight of all blocks from the ground.

When the bottom block is removed, there will not be any reaction force from the ground. That causes a disturbance in the equilibrium. Without the balancing reaction, the blocks will have the gravity force from their weights pulling them down. By Newton 2nd law, the unbalance force will cause the blocks to accelerate towards the ground until they all crash-land.

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Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = m

Mass of the heavier cart = 2m

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ KE =  \frac{mv^2}{2}

⇒ KE =  \frac{mv^2}{2}\times \frac{m}{m}

⇒ KE =  \frac{m^2v^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(mv)^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(\triangle p)^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

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KE (K1) of mass 2m.

⇒ K_1=\frac{(F\times t)^2}{2\times 2m}

⇒ K_1=\frac{(F\times t)^2}{4m}         ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ \frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}

⇒ \frac{K_1}{K} =\frac{2m}{4m}

⇒ \frac{K_1}{K} =\frac{2}{4}

⇒ \frac{K_1}{K} =\frac{1}{2}

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The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

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Answer:

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Explanation:

Given;

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Final Kinetic energy of the box = initial elastic potential energy of the spring

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¹/₂ x 8 x 10² = Ux

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Therefore, the energy of the spring at the start is 400 J.

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