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2 years ago

What is the difference between engineering controls and administrative controls?

1 answer:

5 0

Engineering controls could involve altering the weight of the products, adjusting the heights of the work surfaces, or getting lifting equipment. Administrative controls are workplace policies, practices, and processes that reduce the likelihood of risky situations for employees.

<h3>What are Engineering controls?</h3>

Engineering controls are methods for preventing dangerous situations for workers by erecting a barrier between them and the danger or by ventilating the area to remove any hazardous materials.
Instead of depending on employee behavior or mandating that employees wear safety gear, engineering controls entail making physical changes to the workplace itself.

<h3>What is Administrative controls?</h3>

Administrative controls are policies, procedures, shift designs, or training that reduce the threat of a hazard to a person.
Typically, administrative controls aim to alter people's behavior (such as that of manufacturing workers) rather than eliminating the real risk or provide personal protective equipment (PPE).

Learn more about Administrative controls here:

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A mass of 5 kg of saturated water vapor at 150 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate

yulyashka [42]

Answer:

The work done by the steam is 213 kJ.

Explanation:

Given that,

Mass = 5 kg

Pressure = 150 kPa

Temperature = 200°C

We need to calculate the specific volume

Using formula of work done

$W=Pm\DeltaV$

$W=Pm(\dfrac{RT_{1}}{P_{atm}}-\dfrac{RT_{2}}{P_{atm}}$

$W=\dfrac{PmR}{P_{atm}}(T_{2}-T_{1})$

Where,R = gas constant

T = temperature

P = pressure

$P_{atm}$ =Atmosphere pressure

m = mass

Put the value into the formula

$W=\dfrac{150\times10^{3}\times5\times287.05}{1.01\times10^{5}}\times(473-373)$

$W=213\ kJ$

Hence, The work done by the steam is 213 kJ.

6 0

3 years ago

At a certain instant, the earth, the moon, and a stationary 1160 kg spacecraft lie at the vertices of an equilateral triangle wh

Afina-wow [57]

Answer:

$W = 1.22 \times 10^9 J$

Explanation:

Initial potential energy of the given spacecraft is given as

$U = -\frac{GM_e m}{r} - \frac{GM_m m}{r}$

so we have

$U = - \frac{Gm}{r}(M_e + M_m)$

so we have

$M_e = 5.98 \times 10^{24} kg$

$M_m = 7.35 \times 10^{22} kg$

$m = 1160 kg$

$r = 3.84 \times 10^8 m$

$U = - \frac{(6.67 \times 10^{-11})(1160)}{3.84 \times 10^8}(5.98 \times 10^{24} + 7.35 \times 10^{22})$

$U = -1.22 \times 10^9 J$

now total work done to move it to infinite is given

W = 0 - U

$W = 1.22 \times 10^9 J$

6 0

3 years ago

A 4.40-kilogram hoop starts from rest at a height 1.70 m above the base of an inclined plane and rolls down under the influence

Anestetic [448]

Answer:

The linear velocity is $v=4.08m/s$

Explanation:

According to the law of conservation of energy

The potential energy possessed by the hoop at the top of the inclined plane is converted to the kinetic energy at the foot of the inclined plane

The kinetic energy can be mathematically represented as

$KE = \frac{mv^2}{2} + \frac{Iw}{2}$

Where $I$ is the moment of inertia possessed by the hoop which is mathematically represented as

$I = mr^2$

Here R is the radius of the hoop

$w$ is the angular velocity which the hoop has at the bottom of the lower part of the inclined plane which is mathematically represented as

$w = \frac{v}{r}$

Where v linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface

Now expressing the above statement mathematically

$potential \ energy = \frac{mv^2}{y} + \frac{Iw^2}{2}$

$mgh = \frac{mv^2}{y} + \frac{Iw^2}{2}$

=> $mgh =\frac{mv^2}{2} + \frac{(mr^2)(\frac{v}{r})^2 }{2}$

=> $mgh = \frac{mv^2}{2} + \frac{mv^2}{2}$

=> $mgh = mv^2$

=> $v = \sqrt{gh}$

Substituting values

$v = \sqrt{9.81 * 1.7}$

$v=4.08m/s$

4 0

3 years ago

Read 2 more answers

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x= 5.0 cos

lakkis [162]

Answer:

a) x = 4.33 m , b) w = 2 rad / s , f = 0.318 Hz , c) a = - 17.31 cm / s²,

d) T = 3.15 s, e) A = 5.0 cm

Explanation:

In this exercise on simple harmonic motion we are given the expression for motion

x = 5 cos (2t + π / 6)

they ask us for t = 0

a) the position of the particle

x = 5 cos (π / 6)

x = 4.33 m

remember angles are in radians

b) The general form of the equation is

x = A cos (w t + Ф)

when comparing the two equations

w = 2 rad / s

angular velocity and frequency are related

w = 2π f

f = w / 2π

f = 2 / 2pi

f = 0.318 Hz

c) the acceleration is defined by

a == d²x / dt²

a = - A w² cos (wt + Ф)

for t = 0 , we substitute

a = - 5,0 2² cos (π / 6)

a = - 17.31 cm / s²

d) El period is

T = 1/f

T= 1/0.318

T = 3.15 s

e) the amplitude

A = 5.0 cm

3 0

3 years ago

Help<br> A. 12,240<br> B. 6,120

nata0808 [166]

Answer:

A. 12,240.

Explanation:

1,530 times 8.0 = 12,240.

hope this helped :)

3 0

2 years ago