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cestrela7 [59]
3 years ago
15

Why is it important to use the net force instead of the applied force when calculating the acceleration?

Chemistry
1 answer:
Agata [3.3K]3 years ago
4 0
I love helping others
it’s i’m portent bc u need to know why there is a net force
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Angelina_Jolie [31]

Answer:

A.

Explanation:

It’s just right

4 0
3 years ago
Why is the following molecule nonpolar and hydrophobic?
algol13

Explanation:

Hydrocarbon shows nonpolar

8 0
3 years ago
Help! please! hurry! I will mark brainliest if you get it correct!! NO SPAM!!!!!!
Troyanec [42]

Answer: 5.747 * 10^14 Hz

Explanation:

Convert 522nm to m = 522 * 10^-9 m (since 1nm=10^-9m)

If c = wavelength * frequency, where c is the speed of light (3.0 * 10^8 m/s), then you can manipulate the equation to solve for frequency (f).

f = c / wavelength

Plug in the given numbers:

f = (3.0 * 10^8) / (10^-9)

f = 5.747 * 10^14 Hz

4 0
2 years ago
Read 2 more answers
A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0
3 years ago
3. Jasmine pousse un objet sur une distance de
BARSIC [14]

Jasmine pushes an object a distance of  15 m at constant speed. It produces a  450J of work on the object. What strength Jasmine  does it exert on the object?

Answer:

30J

Explanation:

Given parameters:

Distance moved  = 15m

Work done  = 450J

Unknown:

Strength Jasmine applied  = ?

Solution:

The strength Jasmine applied or exerted on the object is the force of pull that cause the motion of the object and the distance it was moved.

Now;

  Work done  = Force x distance

So;

       450  = Force x 15

          Force  = \frac{450}{15}   = 30J

5 0
2 years ago
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