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Romashka-Z-Leto [24]
3 years ago
15

Technician A says that battery straps should be used whenever a battery is lifted. Technician B says that when a battery is remo

ved, the positive cable should be disconnected first. Who is correct?
Physics
1 answer:
Inessa [10]3 years ago
8 0

Answer:

The answer is <em>neither</em>.

Explanation:

The reason to why the <em>answer is neither</em> is because in order to jump start a vehicle you must:

  1. <em>Park the booster vehicle close to the hood of the vehicle that needs to be boosted.</em>
  2. <em>Set the parking brakes on both vehicles.</em>
  3. <em>Open the hoods of both vehicles to access the starting battery. On some vehicles this also in the trunk.</em>
  4. <em>Connect the clamp at the other end of the positive jumper cable (red) to the positive terminal on the battery or the jump-starting terminal.</em>
  5. <em>Connect the clamp at the other end of the positive jumper cable (red) to the positive terminal on the booster battery.</em>
  6. <em>Connect the clamp of the negative jumper cable (black) to the negative terminal on the booster battery.</em>
  7. <em>Connect the clamp at the other end of the negative jumper cable (black) to a piece of exposed metal part of the dead vehicle's engine and ensure that it is far away from any fuel or the battery.</em>
  8. <em>Run the engine of the booster vehicle for about 5 minutes.</em>
  9. <em>Now try to start the dead vehicle.</em>
  10. <em>If the vehicle does not start, check the connections of the different cables. If the engine does not start, then it should either be charged or replaced.</em>
  11. <em>Once the dead vehicle's engine has started, allow both vehicles to run for about 5 minutes.</em>
  12. <em>Then disconnect the negative cable from the vehicle and then from the booster battery.</em>
  13. <em>Then disconnect the positive cable from the vehicle and from the booster battery. </em>

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Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂  : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F  : Force ( N)

d  : Horizontal distance of the force to the point 2 ( N*m

Data

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Forces acting on the bar

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T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

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Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances  from forces the point 2

d₁₂ = (3-0.6) m = 2.4m  : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium  of moments at the point  2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

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