Coderz robotics 101: challenge mission 3 part 3

Is a diesel truck less expensive to drive than a gas truck?

MArishka [77]

Answer:

Typically, diesel trucks cost more than those with gas engines, especially when you're first buying them, as diesel is usually featured as an add-on for gas-powered cars. Diesel add-ons can cost over $5,000 for midsize trucks and around $10,000 for heavy-duty trucks.

Explanation:

Make me brain pls

3 0

2 years ago

A pump transfers water from a lake to a reservoir, which is located 29.2 m above the lake, at a rate of 11.5 L/s. Determine the

Greeley [361]

Answer:

$P=3290.84KW$

Explanation:

From the question we are told that:

Height $h=29.2m$

Rate $Q=11.5L/s=>11.5*10^{-3}$

Generally the equation for Power is mathematically given by

$P=\rho*g*Q*h$

$P=1000*9.81*11.5*10^{-3}*29.2$

$P=3290.84KW$

8 0

3 years ago

What is the vorticity?

katovenus [111]

Answer:

a field that describes the local spinning motion of a continuum near some point

Explanation:

7 0

3 years ago

Describe at least one way you take advantage or could take advantage of each of the different forms of energy

Scorpion4ik [409]

Answer:

Kinetic energy can be used to develop electric energy which can be used as electricity.

Explanation:

The kinetic energy can be harnessed; much like some hydro power technologies harness water movement. A way to convert this kinetic energy into electric energy is through piezoelectric. By applying a mechanical stress to a piezoelectric crystal or material an electric current will be created and can be harvested.

Kinetic energy is also generated by the human body when it is in motion. Studies have also been done using kinetic energy and then converting it to other types of energy, which is then used to power everything from flashlights to radios and more.

4 0

3 years ago

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