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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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Using the initial momentum vector as a basis, the change in momentum vector Δp for the cart is drawn as shown in the attachment.

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

\large {\boxed {F = ma }

F = Force ( Newton )

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Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

Initial speed of cart = v_i = v

Final speed of cart = v_f = v

<u>Unknown:</u>

The change in momentum of cart = I  = ?

<u>Solution:</u>

I = \Delta p

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I = mv_f - mv_i

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I = m ( -v - v )

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<em>From the results above, we can conclude that the change in momentum vector Δp is twice the initial momentum vector p_i but in opposite direction.</em>

The vector <em>Δp could be drawn as shown </em><em>in the attachment.</em>

\texttt{ }

<h3>Learn more</h3>
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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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the mass of a brick is 4 kg, find the mass of water displaced by it when completely immersed in water.​
vova2212 [387]

Answer:

The correct answer is = 1.6

Explanation:

Density of water = 1000kg/m³ = d₁

Mass of brick = 4kg = m

Density of brick = 2.5 g/cm³ = 2.5 × 1000 =2500 kg/m³ = d₂

Volume of brick = m/d₂ = 4/2500 =16/10000 = 0.0016 L = v

Buoyant Force = v × d₁ × g          (g= acceleration due to gravity =9.8m/s²)

= 0.0016 × 1000 × 9.8 = 15.68 Newtons

By the Archimedes' Principle, the buoyant force is equal to the weight of the liquid displaced by an object.

Weight of the water displaced=Buoyant Force

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15.68/9.8 =Mass of water displaced

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4 0
3 years ago
In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10
sergij07 [2.7K]

Answer:

A)\Phi=83.84\times 10^{-9}

B)\Phi=0 Wb

C)emf=5.4090\times 10^{-4}V

Explanation:

Given that:

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  • time interval of rotation, t=3.1\times 10^{-2}\,s
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(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

\Phi=B.a\,cos \theta..................................(1)

\theta is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}

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(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴  \theta=90^{\circ}

From eq. (1)

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(C)

According to the Faraday's Law we have:

emf=n\frac{B.a}{t}

emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}

emf=5.4090\times 10^{-4}V

7 0
3 years ago
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