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nlexa [21]
2 years ago
13

When students work in a chemistry lab, the location of which item would be the most important for each student to know?

Physics
2 answers:
siniylev [52]2 years ago
7 0
Fire extinguisher and eye wash station? hope that helps
denis-greek [22]2 years ago
4 0

Answer: safety shower

Explanation: edge2020 answer

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Define torque qualitatively and quantitatively.
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When you use a wrench to tighten or loosen a nut on a bolt, you are applying torque. It is measured in units of force times distance.  A force of F newtons pulling on a handle of L meters in length would supply a torque of F L newton-meters. More technically, torque is the vector cross product of force times perpendicular distance from the object, F x r = F r sin @
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Which of these is NOT a chemical property of matter?
marshall27 [118]

C. Magnetism

It is a physical property, not a chemical property.

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2 years ago
A light bulb uses 0.5 amperes from a source of 120 volts. How much power is used by the bulb? 
Sladkaya [172]

Power = voltage(V) * current(I)

          = 120 * 0.5

Power = 60 watts

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2 years ago
What is the maximum number of unpaired d electrons that an atom or ion can possess?
Sonbull [250]

Answer:

5

Explanation:

The d subshell has 5 orbitals, each capable of holding a maximum of two electrons. Hund's rule tells us that every orbital in a sub-level must first be singly occupied by electrons before any orbital is doubly occupied. Therefore five electrons will fill the five orbitals within the d subshell.

3 0
3 years ago
For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103
Alona [7]

Answer:

a) P = 44850 N

b) \delta l =0.254\ mm

Explanation:

Given:

Cross-section area of the specimen, A = 130 mm² = 0.00013 m²

stress, σ = 345 MPa = 345 × 10⁶ Pa

Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa

Initial length, L = 76 mm = 0.076 m

a) The stress is given as:

\sigma=\frac{\textup{Load}}{\textup{Area}}

on substituting the values, we get

345\times10^6=\frac{\textup{Load}}{0.00013}

or

Load, P = 44850 N

Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>

b)The deformation (\delta l) due to an axial load is given as:

\delta l =\frac{PL}{AE}

on substituting the values, we get

\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}

or

\delta l =0.254\ mm

3 0
2 years ago
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