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Anna007 [38]
2 years ago
9

Metal workers and theory causes of accident​

Physics
1 answer:
ser-zykov [4K]2 years ago
4 0

The Domino principle Heinrich supplied a fixed of theorems referred to as 'the axioms of industrial protection'. is the theory causes of the accident.

The primary axiom treated coincidence causation, declaring that 'the prevalence of damage continuously results from a complex sequence of things, the final one among which being the twist of fate itself.

In keeping with the natural hazard theory, everyone in any given set of workers has the same danger of being concerned about a coincidence. It similarly means that there's no unmarried discernible sample of occasions that results in a coincidence

Main factors to an accident.

  • Ancestry & social environment.
  • The fault of the person.
  • unsafe act/mechanical or bodily chance.
  • accident.
  • damage.

Learn more about accidents​ here:-brainly.com/question/8430576

#SPJ9

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Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha
mel-nik [20]

Answer:

(a) 0.17 m

(b) 5.003 m

(c) 6.38 × 10^{-26} N

(d) 7.37 ×10^{-29} N

Explanation:

(a) The minimum value of x will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

h^{2} = b^{2} + p^{2}

h^{2} = 5^{2} + 0.17^{2}   \\h = \sqrt{} 25.03\\h= 5.002 m

<em>Hence, the maximum distance is 5.002 m</em>

(c) For minimum magnitude we use the minimum distance calculated in (a)

Minimum Distance = 0.17 m

For electrostatic force=     F=\frac{kq1q2}{x^{2} }

F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19}  }{0.17^{2} }

F= 6.38×10^{-26} N

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F =  \frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19}   }{5.002^{2} } }

F= 7.37×10^{-29 N

4 0
3 years ago
The average mass of a car in the US is 1.440 x 10^6 g. Express this mass in kg.
wolverine [178]

Answer:

Average mass of acar in the US (in kg) = 1440 kg

Explanation:

Average mass of a car in the US (in g) = 1.440 × 10⁶ g

Mass in kg:

\rm 1 \: g =  {10}^{ - 3}  \: kg \\  \\  \rm 1.440 \times 10^6 \ g = 1.440 \times 10^6  \times  {10}^{ - 3} \ kg \\  \\  \rm = 1.440 \times 10^{6 - 3} \ kg \\  \\  \rm = 1.440 \times 10^3 \ kg \\  \\ \rm = 1.440 \times 1000 \ kg \\  \\ \rm = 1440 \ kg

5 0
3 years ago
Which of the following are examples of projectile motion?
liubo4ka [24]

Answer: A. a basketball being shot toward the basket

Explanation: The definition of projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. So, the basketball is the object being thrown and the person throwing the ball is aiming it to go into the basket making that the path of trajectory. Hope that makes sense and helps!

5 0
2 years ago
A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the
Ierofanga [76]

Answer:

The discharge rate is Q = 0.0192 \  m^3 /s

Explanation:

From the question we are told that

   The  diameter is  d =  60 \ mm   =  0.06 \ m

    The  head is  h  =  6 \ m

     The  coefficient of contraction is  Cc  =  0.68

     The  coefficient of  velocity is  Cv  =  0.92

The radius is mathematically evaluated as

         r =  \frac{d}{2}

substituting values

        r =  \frac{ 0.06 }{2}

        r =  0.03 \ m

The  area is mathematically represented as

      A =  \pi r^2

substituting values

      A =  3.142 *  (0.03)^2

      A = 0.00283 \ m^2

 The  discharge rate is mathematically represented as

        Q =  Cv *Cc  *  A  *  \sqrt{ 2 * g *  h}

substituting values

       Q = 0.68 *  0.92*   0.00283  *  \sqrt{ 2 * 9.8 *  6}

       Q = 0.0192 \  m^3 /s

6 0
3 years ago
8. An airplane is flying at 200 m/s when it touches the ground at the airport. It has a constant negative acceleration, and slow
Temka [501]

Answer:

Explanation:

Given

Initial velocity u = 200m/s

Final velocity = 4m/s

Distance S = 4000m

Required

Acceleration

Substitute the given parameters into the formula

v² = u²+2as

4² = 200²+2a(4000)

16 = 40000+8000a

8000a = 16-40000

8000a = -39,984

a = - 39,984/8000

a = -4.998m/s²

Hence the acceleration is -4.998m/s²

8 0
3 years ago
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