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Inga [223]
2 years ago
15

Why is a protective apron or lab coat important to use when working with acids?

Physics
2 answers:
OLga [1]2 years ago
5 0

Answer:

Acids break down fabrics and can cause burns if the acids are strong.

Explanation:

When working with acids, it is advisable for a scientist for wear a protective apron or lab coat because acids break down fabrics and can cause burns.

Acids are chemical substances that produces excess hydroxonium ions in solutions.

  • Due to their unique properties they are highly corrosive and are used to remove rusts from any surface.
  • We they come in contact with the human body, they can cause severe burns especially if they are very strong acid.
  • If an apron is worn, then the body will be shielded.
yanalaym [24]2 years ago
5 0

Answer:

C. Acids break down fabrics and can cause burns if the acids are strong.

Explanation:

i did the quiz on edg

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What is the speed of sound for a noise that travels 2km in 5.8s?​
Gnesinka [82]

Answer:

Explanation:

Speed is defined as the rate at which an object covers a particular distance. So the formula for determining speed is given as the ratio of distance to time taken for covering that distance.

Speed = Distance/Time

As here the distance is given in km units and time in s units, so the units of any one parameter should be changed. Since we know that speed of sound is always about 300 m/s. So it is better to convert the unit of distance from km to m.

Hence, now the distance traveled by the noise is 2000 m and time taken is 5.8 s.

So the speed of noise = Distance/Time = 2000/5.8=345 m/s.

Thus, the speed of noise is slightly greater than the speed of sound and it is found to be 345 m/s.

5 0
3 years ago
What is glucose made of
tigry1 [53]

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3 0
3 years ago
A chevrolet corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. what
pychu [463]
By definition we have that the final speed is:
 Vf² = Vo² + 2 * a * d
 Where,
 Vo: Final speed
 a: acceleration
 d: distance.
 We cleared this expression the acceleration:
 a = (Vf²-Vo²) / (2 * d)
 Substituting the values:
 a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
 a = -77268 mi / h ^ 2
 its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
 First you must make a free body diagram and see the acceleration of the car:
 g = 32.2 feet / sec ^ 2
 a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
 a = -31.48 feet / sec ^ 2
 A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
 A = -22.07 feet / sec ^ 2
 Clearing the braking distance:
 Vf² = Vo² + 2 * a * d
 d = (Vf²-Vo²) / (2 * a)
 Substituting the values:
 d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
 d = 175.44 feet
 answer:
 its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
5 0
3 years ago
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Strike441 [17]

Answer:

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If a male runner in the fourth starting position ran the 400 m race in 44.40 s, how would you
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