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Elina [12.6K]
3 years ago
7

A solid concrete block weighs 150. N and is resting on the ground. Its dimensions are 0.400 m ✕ 0.200 m ✕ 0.100 m. A number of i

dentical blocks are then stacked on top of this one. What is the smallest number of whole blocks (including the one on the ground) that can be stacked so that their weight creates a pressure of at least two atmospheres on the ground beneath the first block?
Physics
1 answer:
N76 [4]3 years ago
5 0

Answer:

27 blocks

Explanation:

First, the expression to use here is the following:

P = F/A

Where:

P: pressure

F: Force exerted

A: Area of the block.

Now , we need to know the number of blocks needed to exert a pressure that equals at least 2 atm. To know this, we should rewrite the equation. We know that certain number of blocks, with the same weight and dimensions are putting one after one over the first block, so we can say that:

P = W/A

P = n * W1 / A

n would be the number of blocks, and W1 the weight of the block.We have all the data, and we need to calculate the area of the block which is:

A = 0.2 * 0.1 = 0.02 m²

Solving now for n:

n = P * A / W1

The pressure has to be expressed in N/m²

P = 2 atm * 1.01x10^5 N/m² atm = 2.02x10^5 N/m²

Finally, replacing all data we have:

n = 2.02x10^5 * 0.02 / 150

n = 26.93

We can round this result to 27. So the minimum number of blocks is 27.

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Explanation:

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3 years ago
Two horizontal metal plates, each 10.0 cm square, are aligned 1.00 cm apart with one above the other. They are given equal-magni
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a) motion is PARABOLIC, b) positive particle is accelerated towards the negative plate,  c)  x = 6.19 10⁹ m

Explanation:

This exercise looks at the motion of a positively charged particle in an electric field.

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on the x-axis there are no relationships because there are no forces.

Since the particle has velocities in both axes, its motion is PARABOLIC,

b) the positive particle is accelerated towards the negative plate,

The field is descending, for which the event is down

c) where  hit the particle on the x-axis

they indicate that the particle leaves the center of the negative plate, for which we will fix our reference system at this point.

Let's find the components of the initial velocity.

           sin θ = v_{oy} / v

           cos θ = v₀ₓ / v

           v_{oy} = v₀ sin θ

           v₀ₓ = v₀ cos θ

           v_{oy) = 1.02 10⁵ sin 37 = 0.6139 10⁵ m / s

           v₀ₓ = 1.02 10⁵ cos 37 = 0.8146 10⁵ m / s

Let's find the time it takes to hit the negative plate

            y = y₀I + v_{oy} t + ½ a and t2

in this case the positions are y = y₀ = 0 and the accelerations

a = - 1,616m/s2,

we substitute

            0 = 0 + v_{oy} t - ½ a_y t²

            v_{oy}= ½ a_y t

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           x = 0.8145 10⁵ 7.597 10⁴

           x = 6.19 10⁹ m

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