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Harlamova29_29 [7]
3 years ago
9

5. One of the major road blocks for direct DC power in residences is the ___. A) generation of DC B) lack of standardization C)

difficulty of installation D) quantity of components required for this type of system
Engineering
1 answer:
Montano1993 [528]3 years ago
6 0

Answer:

D) quantity of components required for this type of system

Explanation:

Electricity can be transmitted using the alternating and direct currents. The alternating current is one in which the flow of the current diverts at certain time intervals whereas, the direct current is one in which there is a constant one-directional flow of current. The DC is used in batteries and solar panels. Residential areas and business places make use of the AC current.

One of the several reasons why the DC is not used in homes is because unlike the AC it is not easy to build and sustain. Moreso, it has more components compared to the AC. For example, its motor consists of brushes and commutators . The components, example, switches are also large compared to the AC components.

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A cylindrical bar of metal having a diameter of 20.2 mm and a length of 209 mm is deformed elastically in tension with a force o
Rus_ich [418]

Answer:

A) ΔL = 0.503 mm

B) Δd = -0.016 mm

Explanation:

A) From Hooke's law; σ = Eε

Where,

σ is stress

ε is strain

E is elastic modulus

Now, σ is simply Force/Area

So, with the initial area; σ = F/A_o

A_o = (π(d_o)²)/4

σ = 4F/(π(d_o)²)

Strain is simply; change in length/original length

So for initial length, ε = ΔL/L_o

So, combining the formulas for stress and strain into Hooke's law, we now have;

4F/(π(d_o)²) = E(ΔL/L_o)

Making ΔL the subject, we now have;

ΔL = (4F•L_o)/(E•π(d_o)²)

We are given;

F = 50500 N

L_o = 209mm = 0.209m

E = 65.5 GPa = 65.5 × 10^(9) N/m²

d_o = 20.2 mm = 0.0202 m

Plugging in these values, we have;

ΔL = (4 × 50500 × 0.209)/(65.5 × 10^(9) × π × (0.0202)²)

ΔL = 0.503 × 10^(-3) m = 0.503 mm

B) The formula for Poisson's ratio is;

v = -(ε_x/ε_z)

Where; ε_x is transverse strain and ε_z is longitudinal strain.

So,

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

v = - [(Δd/d_o)/(ΔL/L_o)]

v = - [(Δd•L_o)/(ΔL•d_o)]

Making Δd the subject, we have;

Δd = -[(v•ΔL•d_o)/L_o]

We are given v = 0.33; d_o = 20.2mm

So,

Δd = -[(0.33 × 0.503 × 20.2)/209]

Δd = -0.016 mm

8 0
3 years ago
LINKS GET BODIED ON SITE! RAWR
bulgar [2K]

Answer:

False

Explanation:

MRK ME BRAINLIEST PLZZZZZZZZZZZZZZZZZZ

7 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
3 years ago
Hey answr this sajida Yusof
Allisa [31]
What do u need? Rusbaisuwvwbs
5 0
2 years ago
Read 2 more answers
The velocity field of a flow is given by where x and y are in feet. Determine the fluid speed at points along the x axis; along
lora16 [44]

Answer:

Using the formula

V =20y/(x^2+y^2)^1/2 - 20x/(x^2+y^2)^1/2

Hence fluid speed at x axis =20x/(x^2+y^2)^1/2

While the fluid speed at y axis =20y/(x^2+y^2)^1/2

Now the angle at 1, 5

We substitute into the formula above

V= 20×5/(1+25)^1/2= 19.61

For x we have

V = 20× 1/(1+25)^1/2= 3.92

Angle = 19.61/3.92= 5.0degrees

Angel at 5, and 2

We substitute still

V = 20×5/(2+25)^1/2=19.24

At 2 we get

V= 20×2/(2+25)^1/2=7.69

Dividing we get 19.24/7.69=2.5degrees

At 1 and 0

V = 20/(1)^1/2=20

At 0, v =0

Angel at 2 and 0 = 20degrees

At 5 and 2

V = 100/(25+ 4)^1/2=18.56

At x = 2

40/(√29)=7.43

Angle =18.56/7.43 = 2.49degrees.

6 0
2 years ago
Read 2 more answers
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