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katovenus [111]

2 years ago

11

Designing a quiet generator to reduce the amount of hazardous noise is an example of what type of risk control

1 answer:

kirza4 [7]2 years ago

4 0

Designing a quiet generator to reduce the amount of hazardous noise is an

example of Engineering risk control.

<h3>What is Engineering risk control?</h3>

Engineering risk control involves the process of designing or modification

of a product in order to reduce the risks or hazards associated with it. The

methods involved include:

Process control

Enclosure

Isolation

Ventilation

In this scenario, a quiet generator was designed to reduce the amount of

hazardous noise which correctly represents an Engineering risk control.

Read more about Engineering risk control here brainly.com/question/21213968

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A conducting spherical shell with inner radius ‘a’ and outer radius ‘b’ has a positive charge Q located at its center. The total

kifflom [539]

Answer:deeeznuts

Explanation:

4 0

3 years ago

In anticipation of a long 10 upgrade, a bus driver accelerates at a constant rate of 5 ft/s2 while still on a level section of a

fgiga [73]

Answer:

$S = 20903.4$ ft

Explanation:

First we will determine the acceleration of the bus while it is moving upward

The equilibrium equation would be

$F - W sin\theta = ma'\\ma - W sin\theta = ma'\\m = \frac{W}{g} \\\frac{W}{g}*a - W sin\theta = \frac{W}{g} * a'\\\frac{a}{g} - sin\theta = \frac{a'}{g}\\\frac{x}{y} \frac{5}{32.2} - sin 10 = \frac{a'}{32.2} \\\a' = -0.59$

Let the displacement be $S_0$

As per newton's third law of motion

$v^2 -u^2 = 2as\\u = 80 \frac{mi}{h} = 117.33\frac{ft}{s}\\v = 50 \frac{mi}{h} = 73.33\frac{ft}{s}\\73.33 ^ 2 - 117.33^2 = 2 * (-0.59) * (S-S_0)\\\\S_0 = 0\\S = 20903.4$

3 0

3 years ago

Two particles are fixed to an x axis: particle 1 of charge −8.00 ✕ 10⁻⁷ C at x = 6.00 cm, and particle 2 of charge +8.00 ✕ 10⁻⁷

victus00 [196]

Answer:

0 N/C

Explanation:

Parameters given:

$q_1 = -8.00 * 10^{-7} C$

$x_1 = 6.00 cm$

$q_2 = +8.00 * 10^{-7} C$

$x_2 = 21 cm$

The distance between $q_1$ and $q_2$ is

$21 - 6 = 15cm$

Electric field is given as

$E = \frac{kq}{r^2}$

r = 15/2 = 7.5cm = 0.075m

The electric field at their midpoint due to $q_1$ is:

$E = \frac{9 * 10^9 * -8.0 * 10^{-7}}{0.075^2}$

$E_1$ = $-1.28 * 10^6 N/C$

The electric field at the midpoint due to $q_2$ is:

$E = \frac{9 * 10^9 * 8.0 * 10^{-7}}{0.075^2}$

$E_2$ = $1.28 * 10^6 N/C$

The net electric field will be:

E = $E_1$ + $E_2$

E = $-1.28 * 10^6$ + $1.28 * 10^6$

E = 0 N/C

7 0

4 years ago

Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68

lions [1.4K]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, $y_0=1500\ m$

Initial horizontal velocity is, $u_x=688\ m/s$

Initial vertical velocity is, $u_y=0(\textrm{Horizontal velocity only initially)}$

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is $S=y-y_0=0-1500=-1500\ m$

Acceleration in the vertical direction is due to gravity, $g=-9.8\ m/s^2$

Therefore, the displacement of the bomb is given as:

$S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s$

So, the bomb will remain in air for 17.5 s before hitting the ground.

5 0

3 years ago

Two astronauts, each having a mass of 74 kg, are connected by a 8.53 m rope of negligible mass. They are isolated in space, orbi

AveGali [126]

Answer:

a) 2575 kgm²/s

b) 1.23 kJ

c) 0.478 rad/s

Explanation:

Given

Mass of astronauts, m = 74 kg

Length of rope, l = 8.53 m

Speed of orbit, v = 4.08 m/s

L = m1v1.x1(i) + m2v2.x2(i) = 2mv(d/2)

Thus, L = 2.m.v.(d/2)

L = 2 * 74 * 4.08 * (8.53/2)

L = 2 * 74 * 4.08 * 4.265

L = 2575.38 kgm²/s

Rotational Energy of the system

K(i) = 1/2m1v1(i)² + 1/2m2v2(i)²

K(i) = 2(1/2) * 74 * 4.08²

K(i) = 74 * 16.6464

K(i) = 1231.83 J = 1.23 kJ

Angular momentum is conserved, thus, angular velocity, w = v/r

w = 4.08 / 8.53

w = 0.478 rad/s

7 0

3 years ago