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Svetach [21]
2 years ago
13

PLS HURRY THIS IS DUE TODAY!!!

Physics
2 answers:
loris [4]2 years ago
8 0

Answer:

B. between carbon and oxygen

Explanation:

formaldehyde has a carbon atom that has single bonds with two hydrogen atoms on both sides, and it has a double bond with oxygen.

baherus [9]2 years ago
7 0

Answer:

The placement of the double bond is between carbon atom and oxygen atom

this same question was for me so I did and got it correct

please mark me as brainliest

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First Condition of Equilibrium is that it must be experiencing no acceleration or the external forces acting on the body should be zero.

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According to the barometer in the attachment below, which statement is true?
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Pls mark me brainiest and I sure hope this helps you

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PLEASE HELP!!!! The displacement vectors A and B, when added together, give the resultant vector R, so that R = A + B. Use the d
GuDViN [60]

Answer:

Ax = 0

Ay = 6 m

Bx = 8 cos phi = cos 34 = 6.63 m

By = 8 sin phi = 8 sin (-34) = -4.47 m

Rx = Ax + Bx = 0 + 6.63 = 6.63 m

Ry = Ay + By = 6 - 4.47 = 1.53 m

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7 0
3 years ago
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N
Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

r=\sqrt{(r_{1})^2+(r_{2})^2}

Put the value into the formula

r=\sqrt{(9.8)^2+(2.1)^2}

r=10.02\ cm

We need to calculate the force

Using formula of force

F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}

Force F₁₂,

F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}

F_{12}=13.33\ N

F_{21}=-13.33\ N

Force F₂₃,

F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}

We need to calculate the value of third charge

F_{net}=F_{21}+F_{23}

4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}

q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}

q_{3}=7.99\times10^{-7}\ C

q_{3}=0.8\times10^{-6}\ C

Hence, The value of third charge is 0.8μC.

4 0
3 years ago
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