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OLEGan [10]
3 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]3 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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HELP! FOR A TEST!
Sati [7]

Answer:

<h3>On a velocity vs time graph the slope of the line represents the acceleration of the object. With a slope of zero, the object is moving at a constant velocity in the positive (+) direction during this five minute interval. ... Displacement and distance can both be determine on a velocity vs.</h3>

4 0
3 years ago
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Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0
3 years ago
A cyclist rides 6.2 km east, then 9.28 km in a direction 27.27 degrees west of north, then 7.99 km west. A. What is the magnitud
Pani-rosa [81]

Answer:

Explanation:

given,

cyclist ride  6.2 km east and then 9.28 km in the direction of 27.27° west of north and then 7.99 km west.

vertical component = 9.28 cos∅

                                = 9.28 cos 27.27°

                                = 8.24 km

horizontal axis component = 9.28 sin ∅

                                             = 9.28 sin 27.27°

                                             = 4.5 km

distance of the final point from the origin

                            = 7.99 -(6.2-4.5)

                            = 6.29 km

displacement

d = \sqrt{6.29^2+8.24^2}

d = 10.37 km

b) tan \theta = \dfrac{6.29}{8.24}

θ = 37.36°

4 0
3 years ago
The Earth and Moon are separated by about 4.0 10^6 m. Suppose Mars is [{MathJax fullWidth='false' 2.9 \times 10^{11} }] m from E
Shalnov [3]

Answer:

a) α=7.9x10^-4 rad

b) θ=1.12x10^-4 rad

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4 0
3 years ago
The Sun is about 150 million km from earth. how long does it take light from the sun to reach earth? (Speed of light is 3x10^8m/
madam [21]

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Time  =  (150 x 10⁹ m) / (3 x 10⁸ m/s) =

                           50 x 10¹ sec  = 

                                 <em>500 sec</em>  =  8 min 20 sec


3 0
4 years ago
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