Pre-Test Active

Name two laboratory equipment in the picture . b ) Identify two mistakes that these people in the laboratory above are doing

pishuonlain [190]

The two laboratory equipment in the picture are beaker and chemicals.The people in the picture are not wearing any protective gears while (a) working in the laboratory (b) Throwing the chemicals on the floor.

Among the many items that would be considered general lab equipment are pipettes, scales, centrifuges, Bunsen burners, freezers, hot plates, incubators, coolers, stirrers, water baths, and fume hoods

The dangers of working in a laboratory setting include:

Chemical hazards: Handling toxic substances can cause irritation and carcinogenicity.

Biological hazards: Biological hazards include hazards from working with small animals, working with blood borne pathogens and working with biological agents, such as viruses and bacteria.

Physical hazards: Physical hazards include exposure to noise, poor posture and the explosibility and flammability of substances.

Safety hazards: Safety hazards include unbalanced centrifuges, danger when handling hot sterilized items and electrical hazards, such as shock, explosions, blasts and electrocutions.

Allergy hazards: A common allergy hazard in the laboratory setting is a latex allergy, as many of the materials used in a laboratory setting are latex.

Learn more about Laboratory experiments at brainly.com/question/8430576

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3 0

1 year ago

A 78 kg skydiver can be modeled as a rectangular "box" with dimensions 24 cm × 35 cm × 170 cm . If he falls feet first, his drag

ANTONII [103]

Answer:

The terminal speed of his is 137.68 m/s.

Explanation:

Given that,

Mass of skydiver = 78 kg

Area of box $A =24\times35=840\ cm$

Drag coefficient = 0.80

Density of air $\rho= 1.2\times kg/m^3$

We need to calculate the terminal velocity

Using formula of drag force

$F_{d} = \dfrac{1}{2}\rho v^2Ac$

Where,

$\rho$ = density of air

A = area

C= coefficient of drag

Put the value into the formula

$78\times9.8=\dfrac{1}{2}\times1.2\times v^2\times24\times10^{-2}\times35\times10^{-2}\times0.80$

$v^2=\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}$

$v=\sqrt{\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}}$

$v=137.68\ m/s$

Hence, The terminal speed of his is 137.68 m/s.

3 0

3 years ago

What is the net force acting on the object above?

photoshop1234 [79]

The answer would be C (the one you picked)

Hope this helps

Have a great day/night

6 0

3 years ago

A steel sphere of mass 0.5kg travelling at 2m/s collides with an identical sphere at rest. What is the total momentum after coll

PSYCHO15rus [73]

Answer:

The total momentum after the collision is 1 kg-m/s.

Explanation:

We have,

Mass of a steel sphere is 0.5 kg

It is travelling with a speed of 2 m/s

It collides with an identical sphere at rest.

The law of conservation of momentum states that the initial momentum is equal to the final momentum for an isolated system. Here, initial momentum is :

$p_i=p_f=mv\\\\p_i=p_f=0.5\times 2\\\\p_f=1\ kg-m/s$

So, the total momentum after the collision is 1 kg-m/s.

3 0

3 years ago

I was driving 150 km/hr for 20 km, then had to pull over for a quarter of an hour to replace my tire, and then drove the last 3.

Likurg_2 [28]

Answer:

ertyu

Explanation:

4 0

3 years ago