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2 years ago

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The compression strength of concrete (the maximum pressure that can be exerted on the base of the structure) is 1.7 × 107 Pa. Wh

at is the height of the tallest cylindrical concrete pillar with a radius of 46 cm that will not collapse under its own weight?

1 answer:

yarga [219]2 years ago

8 0

<h2>Height of cylinder = 679.70 m</h2>

Explanation:

Compression strength = 1.7 x 10⁷ Pa

Normal force = Compression strength x Area

Radius = 46 cm = 0.46 m

Area of cylinder = π x 0.46² = 0.665 m²

Normal force = 1.7 x 10⁷ x 0.665 = 1.13 x 10⁷ N

So weight of cylinder = 1.13 x 10⁷ N

Density of concrete = 25 kN/m³ = 25000 N/m³

Volume of cylinder = Base area x Height = 0.665 x h

Weight of cylinder = 25000 x 0.665 x h

Equating

25000 x 0.665 x h = 1.13 x 10⁷

h = 679.70 m

Height of cylinder = 679.70 m

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Answer:

7560 Joules

Explanation:

$m_1$ = Mass of first car = $1.5\times 10^5\ kg$

$m_2$ = Mass of second car = $2\times 10^5\ kg$

$u_1$ = Initial Velocity of first car = 0.3 m/s

$u_2$ = Initial Velocity of second car = -0.12 m/s

v = Velocity of combined mass

As linear momentum of the system is conserved

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{1.5\times 10^5\times 0.3 + 2\times 10^5\times -0.12}{1.5\times 10^5 + 2\times 10^5}\\\Rightarrow v=0.06\ m/s$

Energy lost is

$\Delta E=\Delta E_i-\Delta E_f\\\Rightarrow \Delta=\frac{1}{2}(m_1u_1^2 + m_2u_2^2-(m_1+m_2)v^2)\\\Rightarrow \Delta=\frac{1}{2}(1.5\times 10^5\times 0.3^2 + 2\times 10^5\times (-0.12)^2-(1.5\times 10^5 + 2\times 10^5)\times 0.06^2)\\\Rightarrow \Delta=7560\ J$

The Energy lost in the collision is 7560 Joules

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2 years ago

erastovalidia [21]

Answer:11.5m

Explanation:

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2 years ago

EastWind [94]

Answer:

Elastic potential energy, $E=2.35\times 10^{-8}\ J$

Explanation:

Charge, $q=9.4\times 10^{-10}\ C$

Potential, V = 50 V

It is required to find the electric potential energy in a capacitor stored in it. The formula of the electric potential energy in a capacitor is given by :

$E=\dfrac{1}{2}qV\\\\E=\dfrac{1}{2}\times 9.4\times 10^{-10}\times 50\\\\E=2.35\times 10^{-8}\ J$

So, the electric potential energy stored in the capacitor is $2.35\times 10^{-8}\ J$

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3 years ago

What is the resistance of a light bulb if a potential difference of 120 V will produce a current of 0.5 A in the bulb?

Yakvenalex [24]

Explanation:

Ohm's law:

V = IR

120 V = (0.5 A) R

R = 240 Ω

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3 years ago

Read 2 more answers

If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu

Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

$\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}$ .

Volume of water:

$V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}$ .

Mass of water:

$m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}$ .

Amount of heat that the 15 L water absorbed:

$Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}$ .

What's the mass of the hot steel tool?

The specific heat of carbon steel is $0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}$ .

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

$Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}$ .

$\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}$ .

$m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}$ .

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3 years ago