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irina1246 [14]
3 years ago
15

Interactive Solution 11.13 presents a model for solving this problem. A solid concrete block weighs 100 N and is resting on the

ground. Its dimensions are 0.400 m x 0.250 m x 0.130 m. A number of identical blocks are stacked on top of this one. What is the smallest number of whole bricks (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:
mash [69]3 years ago
3 0

Answer:

The value is }  N  =  66 \  blocks

Explanation:

From the question we are told that

The weight of the block is W_b  = 100 \  N

The dimension of the block is d =  0.400 m  \ X  \ 0.250 \  m  \  X  \ 0.130 \ m

Generally two atmosphere is equivalent to

P_{2atm} =  2 *  1.013 *10^{5} =  202600 \  N/m^2

Generally 1 atm = 1.013 *10^{5} N/m^2

The area of the block would be evaluated using width and height because we need for the smaller surface to be in contact with the ground in order to maximize the pressure and minimize number of blocks

So

A =  0.250 *  0.130

=> A =  0.0325 \  m^2

Generally the force due to this blocks is mathematically represented as

F =  N  *  W_b

Here N is the number of blocks

So

}  202600 =  \frac{N  *  100 }{ 0.0325}

=>   }  N  =  66 \  blocks

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Answer:

The correctly equation is D) N2+ 3H2 → 2NH3

Explanation:

Both nitrogen and hydrogen are presented in bimolecular form (N2 and H2), on one side and on the other side of the equation we have 2 Nitrogens and 6 Hydrogens (it is balanced)

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3 years ago
A wheel turns through 5.5 revolutions while being accelerated from rest at 20rpm/s.(a) What is the final angular speed ? (b) How
alina1380 [7]

Answer:

(a) The final angular speed is 12.05 rad/s

(b) The time taken to turn 5.5 revolutions is 5.74 s

Explanation:

Given;

number of revolutions, θ = 5.5 revolutions

acceleration of the wheel, α = 20 rpm/s

number of revolutions in radian is given as;

θ = 5.5 x 2π = 34.562 rad

angular acceleration in rad/s² is given as;

\alpha = \frac{20 \ rev}{min} *\frac{1}{s} *(\frac{2\pi \ rad}{1 \ rev } *\frac{1 \ min}{60 \ s}) \\\\\alpha = 2.1 \ rad/s^2

(a)

The final angular speed is given as;

\omega _f^2 = \omega_i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 +  2\alpha \theta\\\\\omega _f^2 =   2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta}\\\\ \omega _f  = \sqrt{2(2.1) (34.562)}\\\\ \omega _f = 12.05 \ rad/s

(b) the time taken to turn 5.5 revolutions is given as

\omega _f = \omega _i + \alpha t\\\\12.05 = 0 + 2.1t\\\\t = \frac{12.05}{2.1} \\\\t = 5.74 \ s

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2 years ago
Two students, Jenny and Cho, are investigating motion.
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Answer:

1: a measuring instruments the students should use for time is a stopwatch

2: a measuring instruments the students should use for distance is a measuring tape

Explanation:

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1 year ago
Three identical train cars, coupled together, are rolling east at 1.8 m/s . A fourth car traveling east at 4.5 m/s catches up wi
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Answer:

v = 1.98\ m/s

Explanation:

consider the mass of each train car be m

m₁ = m₂ = m₃ = m

speed of the three identical train

u₁ = u₂ = u₃ = 1.8 m/s

m₄ = m             u₄ = 4.5 m/s

m₅ = m              u₅ = 0 (initial velocity )

final velocity

v₁ = v₂ = v₃ = v₄ = v₅ = v

using conservation of momentum

m₁u₁ + m₂u₂ + m₃u₃ + m₄u₄ + m₅u₅ = m₁v₁ + m₂v₂ + m₃v₃ + m₄v₄ + m₅v₅

m (1.8 + 1.8 + 1.8 +4.5) = 5 m v

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Two 1-kg objects, C and D, increase in temperature by the same amount, but the
Ede4ka [16]

The object D is made up of material Lead. The correct option is D.

<h3>What is specific heat?</h3>

The specific heat is the amount of heat required to change the temperature by 1°C. It is denoted by C.

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The energy transfer is proportional to specific heat.

Specific heat of D must be less. The possible material with specific heat less than the given value is for Lead material.

Thus, the correct option is D.

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