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3 years ago

Which of these represents three molecules of carbon dioxide? 3CO2 3C3O2 C3O6 3C2O

Physics

2 answers:

amid [387]3 years ago

7 0

Answer: $3CO_2$

Explanation: A molecule of compound is formed by the combination of atoms of different elements combined together in a fixed ratio by mass.

Carbon dioxide is a covalent compound formed by sharing of electrons between carbon and two oxygen atoms in which less electronegative atom is named first followed by the more electronegative element with a prefix indicating the number of atoms followed by an -ide.

The number is represented by a stochiometric coefficient in the beginning of chemical formula.

Thus the correct representation of 3 molecules of carbon dioxide is $3CO_2$

seraphim [82]3 years ago

5 0

............ Answer is 3C02

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A vertical spring stretches 9.6 cm when a 1.3 kg block is hung from its end. (a) Calculate the spring constant. This block is th

tatiyna

Answer:

F = - K x

a) K = 1.3 kg * 9.8 m/s^2 / .096 m = 133 kg/sec^2

b) ω = (K/m)^1/2 angular frequency of SHM

ω = (133 / 1.3)^1/2 = 10.1 / sec

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3 years ago

what will be the moment of inertia of a body if it rotates at a uniform rate of 10rad/sec^2 by a torque of 120Nm?

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Answer:

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Explanation:

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3 years ago

What is impulse? How does this relate to momentum?

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4 years ago

Hello guys! Can u please help me with physics. I translated it in English. Can yall help me please how much u can!!

DedPeter [7]

1. Since the body is thrown vertically upward, the only force acting on it as it rises and falls is gravity, which causes a constant downward acceleration with magnitude g = 9.8 m/s². Because this acceleration is constant, we can use the formula

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where

u = initial speed

v = final speed

a = acceleration

∆x = displacement

At its maximum height, some distance y above the point where the body is launched, the body has zero velocity, so

0² - (20 m/s)² = 2 (-9.8 m/s²) y

Solve for y :

y = (20 m/s)² / (2 (9.8 m/s²)) ≈ 20.4 m

2. Relative to the ground, the body's maximum height is 60 m + 20.4 m ≈ 80.4 m.

3. At any time t ≥ 0, the body's vertical velocity is given by

v = 20 m/s - gt

At the highest point, we have

0 = 20 m/s - (9.8 m/s²) t

and solving for t gives

t = (20 m/s) / (9.8 m/s²) ≈ 2.04 s

4. The body's height y above the ground at any time t ≥ 0 is given by

y = 60 m + (20 m/s) t - 1/2 gt²

Solve for t when y = 0 :

0 = 60 m + (20 m/s) t - 1/2 (9.8 m/s²) t²

Using the quadratic formula,

t = (-b + √(b² - 4ac)) / (2a)

(and omitting the negative root, which gives a negative solution) where a = -1/2 (9.8 m/s²), b = 20 m/s, and c = 60 m. You should end up with

t ≈ 6.09 s

5. At the time found in (4), the body's velocity is

v = 20 m/s - g (6.09 s) ≈ -39.7 m/s

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