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3 years ago

Describe how to mix and apply body filler?

1 answer:

8 0

This is all you need to mix body filler.
Start with a golf-ball size glob of filler.
Squeeze a ribbon of hardner across the filler.
Stir the mixture quickly, but do not "whip" it. ...
Mix until overall color is even.
Pick up an appropriate amount of filler for the task at hand.

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A 650-kN column load is supported on a 1.5 m square, 0.5 m deep spread footing. The soil below is a well-graded, normally consol

insens350 [35]

<u>Explanation:</u>

Determine the weight of footing

$W_{f}=\gamma(L)(B)(D)$

Where $W_{f}$ is the weight of footing, γ is the unit weight of concrete, L is the length of footing is the width of footing, and D is the depth of footing

Substitute $2 m \text { for } L, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 23.6 kN / m ^{3}$ for γ in the equation

$\begin{aligned}W_{f} &=\left(23.6 kN / m ^{3}\right)(2 m )(1.5 m )(0.5 m ) \\&=35.4 kN\end{aligned}$

Therefore, the weight of the footing is 35.4 kN

Determine the initial vertical effective stress.

$\sigma_{z p}^{\prime}=\gamma(D+B)-u$

Here, $\sigma_{z^{p}}^{\prime}$ is initial vertical stress at a depth below ground surface γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute $18 kN / m ^{3} \text { for } \gamma, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 0$ for u in the equation

$\begin{aligned}\sigma_{z p}^{\prime} &=\left(18 kN / m ^{3}\right)(1.5+0.5) m -0 \\&=36 kPa\end{aligned}$

Therefore, the initial vertical stress is 36 kPa

Determine the vertical effective stress.

$\sigma_{z D}^{\prime}=\gamma D$

Here, $\sigma_{z^{p}}^{\prime}$ is initial vertical stress at a depth below ground surface γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute $18 kN / m ^{3}\) for $\gamma, 0.5 m$ for $D$ and 0 for \(u$ in the equation.

$\begin{aligned}\sigma_{z b}^{\prime} &=\left(18 kN / m ^{3}\right)(0.5 m )-0 \\&=9 kPa\end{aligned}$

Therefore, the vertical stress at a depth below the ground surface is

9 kPa

Determine the influence factor at the midpoint of soil layer,

$I_{e p}=0.5+0.1 \sqrt{\frac{q-\sigma_{s 0}^{\prime}}{\sigma_{z p}^{\prime}}}$

Here $I_{e p}$ is the influence factor at the midpoint of soil layer $\sigma_{z^{p}}^{\prime}$ is initial vertical stress, $\sigma_{z^{p}}^{\prime}$ is vertical effective stress, and $Q$ is bearing pressure

Substitute $36 kPa for $\sigma_{z p}^{\prime}, 228.47$ kPa for $q,$ and 9 kPa for $\sigma_{z D}^{\prime}$$ in the equation $\begin{aligned}I_{\epsilon P} &=0.5+0.1 \sqrt{\frac{228.47 kPa -9 kPa }{36 kPa }} \\&=0.75\end{aligned}$

Therefore the influence factor at the midpoint of the soil layer is 0.693

6 0

4 years ago

What are the minimum and maximum required footing projections for a concrete footing having a thickness of 8 inches

SpyIntel [72]

The minimum and maximum required footing projections for a concrete footing having a thickness of 8 inches will be 4 inches and 8 inches.

<h3>What is projection?</h3>

It should be noted that footing projections are important for construction purposes

In this case, the minimum and maximum required footing projections for a concrete footing having a thickness of 8 inches will be 4 inches and 8 inches.

Learn more about projection on:

brainly.com/question/3703881

#SPJ12

8 0

2 years ago

A ____________ is a term that originally was referring to a way to reproduce a technical drawing documenting an architectural or

larisa86 [58]

Answer:

The answer is blueprint.

Explanation:

Have a nice day or night!

7 0

2 years ago

Convert 850 nm wavelength into frequency, eV, wavenumber, joules and ergs.

BabaBlast [244]

Answer:

Frequency = 3.5294×10¹⁴ s⁻¹

Wavenumber = 1.1765×10⁶ m⁻¹

Energy = 2.3365x 10⁻¹⁹ J , 1.4579 eV , 2.3365x 10⁻¹² erg

Explanation:

Given the wavelength = 850 nm

1 nm = 10⁻⁹ m

So, wavelength is 850×10⁻⁹ m

The relation between frequency and wavelength is shown below as:

c = frequency × Wavelength

Where, c is the speed of light having value = 3×10⁸ m/s

So, Frequency is:

Frequency = c / Wavelength

$Frequency=\frac {3\times 10^8\ m/s}{850\times 10^{-9}\ m}$

Frequency = 3.5294×10¹⁴ s⁻¹

Wavenumber is the reciprocal of wavelength.

So,

Wavenumber = 1 / Wavelength = 1 / 850×10⁻⁹ m

Wavenumber = 1.1765×10⁶ m⁻¹

Also,

$Energy=h\times frequency$

where, h is Plank's constant having value as 6.62x 10⁻³⁴ J.s

So,

Energy = 6.62x 10⁻³⁴ J.s × 3.5294×10¹⁴ s⁻¹

Energy = 2.3365x 10⁻¹⁹ J

Also,

1 J = 6.24×10¹⁸ eV

So,

Energy = 2.3365x 10⁻¹⁹ × 6.24×10¹⁸ eV

Energy = 1.4579 eV

Also,

1 J = 10⁷ erg

So,

Energy = 2.3365x 10⁻¹⁹ × 10⁷ erg

Energy = 2.3365x 10⁻¹² erg

8 0

3 years ago

How can I solve this sequence problem?

Vinvika [58]

Answer:

Podes hablar en español?

Explanation:

8 0

3 years ago

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