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alex41 [277]
2 years ago
8

Which notations are commonly used on force diagrams to show forces acting on an object? Check all that apply. Fp, Fz,

Chemistry
2 answers:
VLD [36.1K]2 years ago
6 0

Answer:

fn, ff, fg, and fp are the answeres for edginuity test

Explanation:

jeka57 [31]2 years ago
4 0
Force diagrams are used in physics in order to help solve problems involving mechanics. There are a number of forces that can be labelled upon a force diagram and this includes FN (Normal force), Ff (Friction force) and Fg (Gravitational force). When a system is in equilibrium, forces within a force diagrams should add to zero.
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Rewrite and Balance the Equation <br> N₂ + H₂ -&gt; NH3
Mars2501 [29]

Answer:N

2

+ 3

H

2

-----> 2N

H

3

Explanation:

N

2

+

H

2

-----> N

H

3

Let us balance this equation by counting the number of atoms on both sides of the arrow.

N

2

+

H

2

-----> N

H

3

N=2 , H=2 N=1, H=3

To balance the number of N atom on Right Hand Side (RHS) , I will add one molecule of N

H

3

on RHS

N

2

+

H

2

-----> 2N

H

3

N=2 , H=2 N=2 , H= 6

To balance the number of H atoms on Left Hand Side (LHS) , I will add two molecules of

H

2

on LHS

N

2

+ 3

H

2

-----> 2N

H

3

N=2 , H=6 N=2 , H= 6

Answer link

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2 years ago
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castortr0y [4]

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Explanation:

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2 years ago
The 10 short columns in the middle of the periodic table make up the
nekit [7.7K]

Answer:

The transition elements or transition metals occupy the short columns in the center of the periodic table, between Group 2A and Group 3A.Explanation:

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2 years ago
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Características de la báscula granataria porfi
saveliy_v [14]

Answer:

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Explanation:

3 0
3 years ago
Calculate the following quantity: molarity of a solution prepared by diluting 45.45 mL of 0.0404 M ammonium sulfate to 550.00 mL
dybincka [34]

Answer:

M_2=3.34x10^{-3}M

Explanation:

Hello!

In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

M_1V_1=M_2V_2

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M

Best regards!

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