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2 years ago

There are a number of requirements that employers must do to protect their workers from caught-in or

Engineering

1 answer:

GrogVix [38]2 years ago

6 0

Answer:

B.Provide guards on power tools and other equipment with moving parts

Explanation:

A guard/ safeguard is a machine feature that is on/around an engineering equipment which can prevent the contact of a user with hazardous parts of the machines such as point of operation and rotating parts.The guard can be on power tools to enable an alternataive ways of switching off the equipment when a danger scenario occurs.When using a tool/equipment, safeguards must be on/active.

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A car is traveling at 36 km/h on an acceleration lane to a freeway. What acceleration is required to obtain a speed of 72 km/h i

dmitriy555 [2]

First, write down the information given and the change units if necessary (we must have similar units to operate on).

Initial speed, u = 36 km/h = 10 m/s

Final speed, v = 72 km/h = 20 m/s

Distance, s = 100 m

We know that

${v}^{2} - {u}^{2} = 2as \\ {20}^{2} - {10}^{2} = 2 \times a \times 100 \\ 400 - 100 = 200 \times a \\ a = \frac{300}{200 } = \frac{3}{2} \: m {s}^{ - 2}$

Now, we substitute v, u, and a in the formula

$v = u + at \\ 20 = 10 + \frac{3}{2} t \\ \frac{3}{2} t = 10 \\ 3t = 20 \\ t = \frac{20}{3} = 6.67 \: seconds$

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6 0

3 years ago

La Patrulla Fronteriza de los Estados Unidos analiza la compra de un helicóptero nuevo para la vigilancia aérea de la frontera d

patriot [66]

Answer:

The Border Patrol of the United States analyzes the purchase of a new helicopter for the aerial surveillance of the border of New Mexico and Texas with the Mexican Republic. 4 years ago a similar helicopter was purchased at a cost of $ 140,000.00. with an interest rate of 7% per year. Calculate the single payment factor and the present value factor with the above data with the table and formula. Draw the flow chart.

Explanation:

7 0

2 years ago

The water of a 14’ × 48’ metal frame pool can drain from the pool through an opening at the side of the pool. The opening is abo

Tresset [83]

Answer:

Explanation:

Height h = 1.03m

Volume v = 3780 gallons = 3780 * 0.0037851m^3 = 14.3073m^3

Time t = 13.5 mins = 13.5 * 60 = 810 seconds

Length of pool L = 14 inch = 14 * 2.54 = 35.56cm

width of pool b = 48 inch = 48 * 2.54 = 121.92 cm

a.) Consider the bernoulli's equation is given as:

$P_1+\rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2 ...(1)$

consider the equation of bernoulli at the top of the pool

$P_0+\rho gh_1 + \frac{1}{2}\rho v_1^2 =constant ...(2)$

where $P_1=P_0$ atm pressure

At the top of the pool $v_1=0m/s$ , substitute in V_1 in equation (2)

$P_0+\rho gh_1 =constant ...(3)$

Hence equation (3) serves as the bernoullis equation at the top.

b.) Consider the equation of bernoulli's at the opening of the pool

$P_2+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(4)\\P_0+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(5)$

where $P_2=P_0$ atm pressure and $h_2=0m$

$P_0+\rho v_1^2 =constant ...(6)$

Hence equation (6) serves as the bernoullis equation of water at the opening of the pool.

c.) Consider the equation (3) and (4)

$P_0+\rho gh_1 =P_0+\rho v_1^2\\\\\frac{1}{2}\rho v_2^2=\rho gh_1\\v_2^2=2gh_1\\v_2=(\sqrt{2gh_1})m/s...(7)$

Hence velocity is $v_2=(\sqrt{2gh_1})m/s$

d.) consider (7)

$v_2=(\sqrt{2(9.81)(1.03)})=4.4954m/s(approx)$

This is the norminal value of velocity

e.) consider the equation of flow rate interval of v and t

flow(t)= $\frac{dv}{dt}(m^3/s)$ hence this is the flow rate

f.) Consider the equation cross sectional area in terms of V,v2 and t

$AV_2=\frac{v}{t}\\\\A=\frac{v}{v_2t}(m^2)...8$

hence this serves as the cross sectional area.

g.) Consider the equation of area from equation (8)

$A=\frac{v}{v_2t}\\=\frac{14.3073}{4.4954\times 810}=0.003929=0.00393m^2=39.3cm^2$

6 0

2 years ago

A fair die is thrown, What is the probability gained if you are told that 4 will

Semmy [17]

Answer:

1/6

Explanation:

A dice has 6 sides, the probability of 4 appearing is 1/6.

7 0

2 years ago

A liquid refrigerant (sg=1,2) is flowing at a weight flow rate of 20,9 N/h. Refrigerant flashes into a vapor and its specific we

Iteru [2.4K]

Answer:

Explanation:

volume of 20.9 N

= 20.9 / 11.5 m³

= 1.8174 m³

In one hour 1.8174 m³ flows

in one second volume flowing = 1.8174 / 60 x 60

= 5 x 10⁻⁴ m³

Rate of volume flow = 5 x 10⁻⁴ m³ / s .

5 0

3 years ago