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Andreas93 [3]
3 years ago
7

Corrosive substances are rarely harmful to human skin.

Physics
2 answers:
Shtirlitz [24]3 years ago
5 0

Corrosive substances are rarely harmful to human skin.   This statement is <em>False.</em>  Corrosive substances are ALMOST ALWAYS harmful to the skin.

Most problems addressed by the technological design process have only one solution.  This statement is also <em>False.</em>  There is more than one way to skin a cat.

Anna007 [38]3 years ago
5 0

Answer:

False. Corrosive substances are harmful to human skin.

False. Most problems addressed by the technological design process have more than one solution.

Explanation:

Corrosive substances have a high tendency to damage other substances  which it comes in contact with. They could either by concentrated acid or a base which are very harmful to human skin. These have the capacity to destroy, disfigure or burn any part of human body that comes in contact with the substance.

Technological design is the process of developing appropriate knowledge into physical solutions to solve various problems. In technology, there are always more than a solution to a problem, as there are different ways to solve technological problems.

Thus, the answers to the two questions is false.

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Fofino [41]

Answer:

A)     K / K₀ = 4   b)     v / v₀ = 4

Explanation:

A) For this exercise we can use the conservation of mechanical energy

in the problem it indicates that the displacement was doubled (x = 2xo)

starting point. At the position of maximum displacement

      Em₀ = Ke = ½ k (2x₀)²

final point. In the equilibrium position

      Em_{f} = K = ½ m v²

        Em₀ = Em_{f}

        ½ k 4 x₀² = K

        (½ K x₀²) = K₀

         K = 4 K₀

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B) the speed value

          ½ k 4 x₀² = ½ m v²

          v = 4 (k / m) x₀

if we call

           v₀ = k / m x₀

          v = 4 v₀

         v / v₀ = 4

3 0
3 years ago
I'm walking 1.6m/s to 7-11 and it started to rain so I sped up to 2.7m/s in 1.2
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Answer:

Explanation:

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5 0
3 years ago
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olga2289 [7]

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3 years ago
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Ilia_Sergeevich [38]
Work done = 0.5*m*[(v2)^2 - (v1)^2]
where m is mass,
v2 and v1 are the velocities.

Given that m = 1.50 x 10^3 kg, v2 = -15 m/s (decelerates), v1 = 25 kg,

Work done = 0.5 * 1.50 x 10^3 * ((-15)^2 - 25^2) = 3 x 10^5 joules

Just ignore the negative value for the final result because work is a scalar quantity.
8 0
3 years ago
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