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Lina20 [59]
3 years ago
13

You guys bad news... i failed nnn :c

Physics
2 answers:
abruzzese [7]3 years ago
6 0

Answer:

I am so sorry about that.

Explanation:

Darina [25.2K]3 years ago
5 0
Awwwwwwwwwwwwwwww dang
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Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4
dsp73

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

6 0
4 years ago
NEED HELP ASAP PLEASE!
katrin [286]

Answer:

Here is the solution hope it helps:)

5 0
2 years ago
25.16 what is the reaction product of acetic acid and ethylamine at room temperature?
Ne4ueva [31]

The ammonium salt of acetic acid is the reaction product of acetic acid and ethylamine at room temperature

<h3 /><h3>What is acetic acid ?</h3>

Acetic acid is a monofunctional carboxylic acid containing two carbon atoms. It acts as a protein solvent,  food acidity regulator, antibacterial food preservative. It is a conjugate acid of an acetate.

Acetic acid is used in the production  of acetic anhydride, cellulose acetate, vinyl acetate monomer, acetic ester, chloroacetic acid, plastics, dyes, insecticides, photographic chemicals, and rubber. Other commercial uses include the production of vitamins, antibiotics, hormones,  organic chemicals, and as a food additive. Typical concentrations of acetic acid found naturally in foods are 700 to 1200 milligrams/kg (mg/kg) in wine, up to 860 mg/kg in aged cheeses, and 2.8 mg/kg in aged cheeses. fresh orange juice.

 

learn more about acetic acid, visit;

brainly.com/question/16970860

#SPJ4

7 0
1 year ago
Come all ruj-jxgn-qua​
Shalnov [3]

Answer:

this app is for solving doubts not sending links ok

5 0
3 years ago
A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th
Mila [183]

Answer:

x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0

\frac{3}{(r + a)^2} = \frac{2}{r^2}

\frac{r+a}{r} = \sqrt{\frac{3}{2}}

1 + \frac{a}{r} =\sqrt{\frac{3}{2}}

\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}

so we will have

r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}

so the x coordinate of this position is given as

x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}

6 0
3 years ago
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