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Vanyuwa [196]
3 years ago
11

T

Engineering
1 answer:
solniwko [45]3 years ago
3 0

Answer:

Common Uses: Boxwood is well-suited for carving and turning, and the tree's diminutive size restricts it to smaller projects. Some common uses for Boxwood include: carvings, chess pieces, musical instruments (flutes, recorders, woodwinds, etc.), rulers, handles, turned objects, and other small specialty items.If you want a small, compact, low-growing shrub to form a hedge that serves as an accent or border along your walkway, fence line or planting beds, dwarf boxwood varieties are the best pick. The "Dwarf English" boxwood (Buxus sempervirens “Suffruticosa”) creates a border hedge approximately 1 to 2 feet in height.

Explanation:

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Which of the following requires formwork?
vekshin1

Answer:

D

Explanation:

Masonry uses stone work, making a stone wall requires perfect masonry.

3 0
3 years ago
Which term describes a Cloud provider allowing more than one company to share or rent the same server?
svlad2 [7]

Answer:

multitenancy is the term.

6 0
2 years ago
Can crushers help us recycle in a space efficient way which is good for saving the earth and for giving you more room in your ap
777dan777 [17]

Answer:

Force magnitude = 296.7 N

Explanation:

Detailed illustration is given in the attached document.

6 0
3 years ago
Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,
Lyrx [107]

Answer:

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0

Explanation:

v_0 = Initial velocity of ball A

v_A=v_0\cos45^{\circ}

v_B = Initial velocity of ball B = 0

(v_A)_n' = Final velocity of ball A

v_B' = Final velocity of ball B

e = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

Coefficient of restitution is given by

e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

Adding the above two equations we get

2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0

\boldsymbol{\therefore v_B'=0.6364v_0}

(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0

From the conservation of momentum along the plane of contact we have

(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}

v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0.

5 0
3 years ago
Water is flowing at a rate of 0.15 ft3/s in a 6 inch diameter pipe. The water then goes through a sudden contraction to a 2 inch
Georgia [21]

Answer:

Head loss=0.00366 ft

Explanation:

Given :Water flow rate Q=0.15 \frac{ft^{3}}{sec}

         D_{1}= 6 inch=0.5 ft

        D_{2}=2 inch=0.1667 ft

As we know that Q=AV

A_{1}\times V_{1}=A_{2}\times V_{2}

So V_{2}=\frac{Q}{A_2}

     V_{2}=\dfrac{.015}{\frac{3.14}{4}\times 0.1667^{2}}

     V_{2=0.687 ft/sec

We know that Head loss due to sudden contraction

           h_{l}=K\frac{V_{2}^2}{2g}

If nothing is given then take K=0.5

So head lossh_{l}=(0.5)\frac{{0.687}^2}{2\times 32.18}

                                    =0.00366 ft

So head loss=0.00366 ft

4 0
3 years ago
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