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3 years ago

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3. When performing overhead work on scaffolding, what protective measures must be taken to prevent objects

1 answer:

hjlf3 years ago

7 0

Answer:

Toeboards, debris nets, or canopies

Explanation:

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Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of

earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

$\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant$

Since the losses are neglected thus applying this theorm between upper and lower porion we have

$\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}$

Now by continuity equation we have

$A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s$

Applying the values in the Bernoulli's equation we get

$\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars$

6 0

3 years ago

den301095 [7]

Jae pain seems the most off

4 0

3 years ago

3. In order to obtain your commercial driver's license (CDL) you must first:

Murljashka [212]

A and C is the answer to the question. Be 15 years old & get a permit

8 0

3 years ago

A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027

Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = $\frac{d}{2}$ = 3 inches

density, $\rho}$ = 493 lbm/ $ft^{3}$

S.G = 1.0027

g = 9.8 m/ $m^{2}$ = 386.22 inch/ $s^{2}$

Solution:

Using the formula for terminal velocity,

$v_{T}$ = $\sqrt{\frac{2V\rho g}{A \rho C_{d}}}$ (1)

$(Since, m = V\times \rho)$

where,

V = volume of sphere

$C_{d}$ = drag coefficient

Now,

Surface area of sphere, A = $4\pi r^{2}$

Volume of sphere, V = $\frac{4}{3} \pi r^{3}$

Using the above formulae in eqn (1):

$v_{T}$ = $\sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho g}{4\pi r^{2} \rho C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2gr}{3C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}$

Therefore, terminal velcity is given by:

$v_{T}$ = $\frac{27.79}{\sqrt{C_d}}$ inch/sec

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3 years ago

Plis 3 conclusiones de este video

vazorg [7]

No hay videos? de cual video estás hablando?

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3 years ago