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2 years ago

What is the best reason for a student to remove a dangling bracelet when heating test tubes of acidic solutions in a hot water

Physics

2 answers:

jarptica [38.1K]2 years ago

7 0

Answer: To minimize the chance the bracelet pulling over a test tube, causing a spill.

Explanation:

A chemistry lab is a laboratory where the potential hazards related to the chemicals may result if safety precautions are neglected. The chances of the cause of chemical splash, burns from chemicals or fire hazards can possibly happen in a chemistry lab. The use of ornaments should be avoided in the chemical lab. Proper safety measures are to be taken to protect the body from chemical hazards by wearing close toed shoes, lab coat and gloves.

A dangling bracelet worn during the heating process is likely to spill off the acid solutions by pulling over a test tube. The test tube neck may get caught in the dangling bracelet. The test tube may collapse and break. Thus the dangling bracelet should be avoided during the process of heating.

soldi70 [24.7K]2 years ago

4 0

Answer:

Explanation:

this equestion is about the safety

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A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = $\frac{27}{60} h$

Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

Solution:

Now, the displacement, 's' in east direction is given by:

$\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km$

Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

Now, displacement in the west direction for 37 min:

$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

(a) Now, average velocity, $v_{avg}$ is given:

$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

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3 years ago

You are attempting to row across a stream in your rowboat. Your paddling speed relative to still water is 3.0 m/s (i.e., if you

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Answer:

Please check the attached file for the diagram

Explanation:

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