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vodka [1.7K]
3 years ago
7

Pick 3 safety rules in science that you think are important and explain what would happen if the rules were no followed. Used yo

ur own word.
Physics
1 answer:
IgorLugansk [536]3 years ago
7 0

Answer:

Answer down below!

Explanation:

  1. Always wear goggles! This is to prevent chemicals or any other substance to get into your eyes. If this wasn't followed, more people could get hurt.
  2. Clean up afterwards! Its always better to clean up so someone else couldn't get hurt! If people don't follow this, it could be a mess and someone could get seriously injured.
  3. When smelling something, waft it to your face, and not get real close. This is to prevent getting chemicals into your body that could be poisonous. If people didn't folloe this, someone could be seriously sick.
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Un paracaidista desciende desde 6 000 m de altura. Si la masa, con su equipo, es de 65 kg, ¿cuánto valdrá su energía potencial e
iris [78.8K]

Explanation:

PRIMERO ENCUENTRAS EL PESO DEL PARACAIDISTA

F_{peso} = 65 kg(9.80 m/s^{2}) = 637 N

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3 years ago
You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an
Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}

\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}

\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}

Substitute the given values to find "terminal speed"

\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}

\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}

\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}

\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}

The “terminal speed” of the ball bearing is 5.609 m/s

7 0
3 years ago
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