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Julli [10]
2 years ago
10

If the equation on the board had shown 3 atoms of carbon on the reactants side, how many atoms of carbon would need to be repres

ented on the products side? 2 3 4
Physics
2 answers:
Sholpan [36]2 years ago
6 0

3

Explanation:

If the equation on the board had shown 3 atoms of carbon on the reactants side, it would need 3 atoms of carbon on the product side to balance it.

This is in accordance to the law of conservation of mass which states that "matter is neither created nor destroyed during the course of chemical reaction, they are only transformed from one form to another".

We expect the same number of atoms on both sides of the equation. For example:

                    Fe + S  →   FeS

               1 iron atom combines with 1 sulfur atom to produce FeS which contains 1 iron and sulfur atoms.

Learn more:

Chemical equation brainly.com/question/2924195

#learnwithBrainly

EastWind [94]2 years ago
5 0

Answer:

3

Explanation:

i did the test and confirmed this answer

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To water the yard you use a hose with a diameter of 3.2 cm. Water flows from the hose with a speed of 1.1 m/s. If you partially
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Answer:

The speed of water flow inside the pipe at point - 2 = 34.67 m / sec

Explanation:

Given data

Diameter at point - 1 = 3.2 cm

Velocity at point - 1 = 1.1 m / sec = 110 cm / sec

Diameter at point - 2 = 0.57 cm

Velocity at point - 2 = ??

We know that from the continuity equation the rate of flow is constant inside  a pipe between two points.

Thus

⇒ A_{1} × V_{1} = A_{2} × V_{2}

⇒  \frac{\pi }{4} × d_{1} ^{2} × V_{1} =

⇒  d_{1} ^{2} × V_{1} =  d_{2} ^{2}  × V_{2}

⇒  (3.2)^{2} × 110 = (0.57)^{2} × V_{2}

⇒ V_{2} = 3467 cm / sec

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3 0
3 years ago
5) A 20.0 kg cart with no friction wheels sits on a table. A light string is attached to it and runs over a low friction pulley
ella [17]

Answer:

1) Please find attached, created with Microsoft Visio

2) The acceleration of the masses connected by the light string is 0.00735 m/s²

3) The tension in the cord is 0.147 N

4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s

5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s

Explanation:

1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio

2) The acceleration of the masses connected by the light string is given as follows;

F = Mass, m × Acceleration, a

The mass of the truck, M = 20.0 kg

The mass attached to the string, hanging rom the pulley, m = 0.0150 kg

The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N

The pulling force acting on the cart, F = M × a

∴ F = 0.147 N = 20.0 kg × a

a = 0.147 N/(20.0 kg) = 0.00735 m/s²

The acceleration of the truck = a = 0.00735 m/s²

3) The tension in the cord = F = 0.147 N

4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²

Where;

s = The distance to the edge of the table = 1.2 m

u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)

a = The acceleration of the cart = 0.00735 m/s²

t = The time taken

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s = u·t + 1/2·a·t²

1.2 = 0 × t + 1/2 ×0.00735 × t²

1.2 = 1/2 ×0.00735 × t²

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v² = u² + 2·a·s

u = 0 m/s

v² = 0² + 2 × 0.00735 × 1.2 = 0.001764

v = √(0.001764) = 0.042

The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.

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Answer:

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