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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

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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A wagon is pulled at a speed of 0.40metets/seconds by a horse exerting an 1,800 Newton's horizontal force. what is the power of

Tatiana [17]

Given:

speed of 0.40meters/seconds

1,800 Newton's horizontal force

Required:

Power of the horse

Solution:

P = F(D/T) where P is power in watts, F is the force, D is the distance and T is time

P = (1,800N) (0.40 meters/seconds)

P = 720 Watts

6 0

4 years ago

A 2kg mass is initially at rest on a frictionless surface. A 45N force acts at an angle of 300

raketka [301]

Answer:

a) The work done by the force is 136.400 joules.

b) The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

Explanation:

The correct statement is shown below:

<em>A 2-kg mass is initially at reat on a frictionless surface. A 45 N force acts at an angle of 30º to the horizontal for a distance of 3.5 meters:</em>

<em>a)</em><em> Find the work done by the force.</em>

<em>b)</em><em> Find the speed of the mass at the end of the 3.5 meters.</em>

a) Given that a external constant force is acting on the mass on a frictionless surface, the work done by the force ( $W_{F}$ ), measured in joules, is:

$W_{F} = F\cdot \Delta s \cdot \cos \theta$ (1)

Where:

$F$ - External constant force exerted on the mass, measured in newtons.

$\Delta s$ - Horizontal travelled distance, measured in meters.

$\theta$ - Direction of the external force regarding the horizontal, measured in sexagesimal degrees.

If we know that $F = 45\,N$ , $\Delta s = 3.5\,m$ and $\theta = 30^{\circ}$ , then the work done by the force is:

$W_{F} = (45\,N)\cdot (3.5\,m)\cdot \cos 30^{\circ}$

$W_{F} = 136.400\,J$

The work done by the force is 136.400 joules.

b) The final speed of the mass at the end ot the 3.5 meter is calculated by means of the Work-Energy Theorem, which means that the work done on the mass is transformed into a translational kinetic energy. That is:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$ (2)

Where:

$m$ - Mass, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the mass, measured in meters per second.

If we know that $W_{F} = 136.400\,J$ , $m = 2\,kg$ and $v_{1} = 0\,\frac{m}{s}$ , then the final speed of the mass is:

$\frac{2\cdot W_{F}}{m} = v_{2}^{2}-v_{1}^{2}$

$v_{2}^{2} =v_{1}^{2}+\frac{2\cdot W_{F}}{m}$

$v_{2} =\sqrt{v_{1}^{2}+\frac{2\cdot W_{F}}{m} }$

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\frac{2\cdot (136.400\,J)}{2\,kg} }$

$v_{2} \approx 11.679\,\frac{m}{s}$

The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

3 0

3 years ago

Itzel travels south 4 miles and then goes east 2 miles and then north 10 miles. What is her displacement? Round your answer to t

expeople1 [14]

The displacement of Itzel according to the question is 6.3 miles SW

Displacement is defined as the distance moved by a body in a specified direction

Find the diagram attached

From the diagram given, we can see that AB is the displacement

To get the length AB, we will have to use the Pythagoras theorem:

$AB^2=2^2+ 6^2\\AB^2 ^2=4+36\\AB^2=40\\AB=\sqrt{40}\\AB= 6.3 miles\\$

From the diagram, we can also se that the direction of the displacement in the South West direction.

Hence the displacement of Itzel according to the question is 6.3 miles SW

Learn more here: brainly.com/question/19108075

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