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3 years ago

The decomposition of dinitrogen tetraoxide into nitrogen gas and oxygen gas is shown by which balanced chemical equation?

Physics

1 answer:

Nataliya [291]3 years ago

5 0

Answer: C. N2O4(g) → N2(g) + 2O2(g)

Explanation: The products produced are Nitrogen gas and 2 moles of Oxygen gas.

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Which event is an example of condensation?

ale4655 [162]

Answer: D

If the fog disappears when the Sun comes out, then this is an example of condensation because:

the Sun actually dries up the fog, and it makes it into higher clouds.

Hope this helps you!

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3 years ago

State two applications of electrostatic

kaheart [24]

Atmospheric electricity and storms, electrostatic control filters, and industrial electrostatic seperation as well as spark discharge. these are just a few. hope it helps.

6 0

3 years ago

Urgente!!!!! <br> Necesito ayuda con esto!!!!

ahrayia [7]

media.discordapp.net/attachments/782414373888458783/826224189828366377/video0.mp4

5 0

3 years ago

Draw the following vector quantity Using the coordinate system.

DiKsa [7]

The given vectors quantities can be described by their properties of both

magnitude and direction.

a. The drawing of the vector extending from point (0, 0) to (190, 0) on the coordinate plane is attached.

b. The velocity vector extending from (0, 0) to (108.76, 50.714) on the coordinate plane is attached.

c. The displacement vector extending from (0, 0) to (30·√2, 30·√2) is attached.

Reasons:

a. The magnitude of the vector = 190 N

The direction in which the vector acts = East

Therefore, in vector form, we have;

$\vec{F}$ = 190 × cos(0)·i + 190 × sin(0)·j = 190·i

The vector can be represented by an horizontal line, 190 units long

Coordinate points on the vector = (0, 0) and (190, 0)

The drawing of the vector with the above points using MS Excel is attached.

b. Magnitude of the velocity vector = 120 km/hr. 25° North of east

Solution;

The vector form of the velocity is; $\vec{v}$ = 120 × cos(25)·i + 120×sin(25)·j, which gives;

$\vec{v}$ = 120 × cos(25)·i + 120×sin(25)·j ≈ 108.76·i + 50.714·j

$\vec{v}$ ≈ 108.76·i + 50.714·j

Therefore, points that define the vector are; (0, 0) and (108.76, 50.714)

The drawing of the vector is attached

c. The magnitude of the vector = 60 m

The direction of the vector is southwest = West 45° south

The vector form of the displacement is $\vec{d}$ = 60 × cos(45°)·i + 60 × sin(45°)·j

Which gives;

$\vec{d}$ = 30·√2·i + 30·√2·j

Points on the vector are therefore; (0, 0), and (30·√2, 30·√2)

The drawing of the vector is attached

Learn more about vectors here:

brainly.com/question/10409036

4 0

2 years ago

A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, $v_{Ci}=25\ m/s$

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, $v_{Ti}=20\ m/s$

After the collision, velocity of the car, $v_{Cf}=18\ m/s$

Let $v$ is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

$initial\ momentum=final\ momentum$

$1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v$

$210000-21600=9000\ kg\times v$

$v=20.93\ m/s$

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

8 0

2 years ago