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3 years ago

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

Physics

2 answers:

Mnenie [13.5K]3 years ago

7 0

Answer:

Explanation:

I got it right

Kipish [7]3 years ago

4 0

Answer:

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

Explanation:

A protective apron or lab coat is important when working with acids because acids break down fabrics and can cause burns if the acids are strong.

An acid is a substance that interacts with water to produce excess hydroxonium ions in an aqueous solution.
A strong acid ionizes completely in solution.
When they come in contact with a fabric, they break them down violently.
So, if they come in contact with the skin, it causes a violent break down of body tissues.
The apron acts a protective layer.

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If a force acts constantly on a stationary object, the object will

wolverine [178]

Answer:

accelerate in the direction of the force

Explanation:

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3 years ago

A sphere has a net charge of 8.11 nC, and a negatively charged rod has a charge of −6.13 nC. The sphere and the rod undergo a pr

malfutka [58]

Answer:

Charge of the sphere = $7.44 * 10^{-9} C$

Charge of the rod = $-5.46 * 10^{-9} C$

Explanation:

Parameters given:

Initial charge of sphere = 8.11 nC = $8.11 * 10^{-9} C$

Initial charge of rod = -6.13 nC = $-6.13 * 10^{-9} C$

In the process, the rod lost $4.20 * 10^{9} C$ . We have to find the charge of these electrons. Then, we subtract it from the rod's initial charge to get its final charge and add it to the sphere's initial charge to get its final charge.

Charge of $4.20 * 10^{9} C$ electrons = electronic charge * number of electrons

= $4.20 * 10^9 * (-1.6023 * 10^{-19}) C\\$

= $-6.73 * 10^{-10} C$

FINAL CHARGE OF ROD

Final charge of rod = $-6.13 * 10^{-9} C$ - ( $-6.73 * 10^{-10} C$ )

= $-5.46 * 10^{-9} C$

FINAL CHARGE OF SPHERE

Final charge of sphere = $8.11 * 10^{-9} C$ + ( $-6.73 * 10^{-10} C$ )

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3 years ago

The top and bottom surfaces of a metal block each have an area of A = 0.030 m 2, and the height of the block is d = 0.11 m. At t

Marat540 [252]

Answer:

Shear modulus is equal to $9.82\times 10^{10}N/m^2$

Explanation:

We have given area $A=0.030m^2$

Force is given $F_1=F_2=30\times 10^6N$

Height of the block d = 0.11 m

Change in height of the block $\Delta d=1.12\times 10^{-3}m$

Stress is given by

$stress=\frac{force}{area}$

$stress=\frac{30\times 10^6}{0.030}=10^9N/m^2$

Strain is equal to

$strain=\frac{\Delta d}{d}$

$strain=\frac{1.12\times 10^{-3}}{0.11}=10.18\times 10^{-3}$

Shear modulus is equal to

Shear modulus $=\frac{stress}{strain}$

$=\frac{10^9}{10.18\times 10^{-3}}=9.82\times 10^{10}N/m^2$

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3 years ago

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The correct option is d

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