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OLEGan [10]
2 years ago
9

I need help please. I’m between B or C and many responses have been B but I think that’s for if NO would be added and not SO3...

..so idk I’m confused apex

Physics
1 answer:
krek1111 [17]2 years ago
8 0

Answer:

C. More NO₂ and SO₂ would form.

Explanation:

This is an example of Le Châtelier's principle.  The reaction will try to seek balance, so if you add something to the right side, the reaction will reverse until both sides are balanced again.  So adding SO₃ would cause more NO₂ and SO₂ to form.

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F - False.

The law of conservation of momentum states that the total momentum is conserved.
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Consider an object rolling off of your lab bench. Discuss how you might be able to make measure- ments to determine its initial
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Answer:

measure the position every so often with a stopwatch

Explanation:

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This is a method used in measurements of uniform movements of bodies

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The beginning of the Phanerozoic is marked by what occurrence
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The beginning of the Phanerozoic is marked by the development of hard body parts, such as shells and bones.
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3 years ago
A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se
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Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

\lambda = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

\vec{E} = 1800\ N/C     (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C

\vec{E'} = 1800\ N/C     (towards)

Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

7 0
3 years ago
Technician A says that module communications networks are used to reduce the number of wires in a vehicle. Technician B says tha
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Answer:

c) both technician A and B

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