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zhuklara [117]
2 years ago
13

Describe an experiment to demonstrate the principles of the parallelogram of forces

Physics
1 answer:
Tcecarenko [31]2 years ago
5 0

Answer:

There are 4 forces because a parallelogram has 4 sides.

It is a quadilateral. Let me know if I am right or not.

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When did the object have the highest average
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Between 120 and 180 seconds

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Molecular orbital theory correctly predicts paramagnetism of oxygen gas. true or false
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Answer: Yes.

Explanation:

Oxygen has a bond order of two. The bond order of Oxygen molecule is calculated, where the [<em>eight valence electrons in bonding molecular orbitals</em> minus (-) <em>four valence electrons in antibonding molecular orbitals</em>]/2 in the electron configuration.

Atoms/molecules where electrons are paired are diamagnetic (repelled by both poles of a magnetic); while atoms/molecules that have one or more unpaired electrons are paramagnetic (attracted to magnetic field).

The two unpaired electrons of dioxygen molecules has made it <u>paramagnetic</u>. By pouring liquid oxygen between the poles of a strong magnet, the liquid stream will be contained by the filed and fills up the space between the poles.

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3 years ago
When velocity is graphed with respect to time, what does the area under
Juliette [100K]

Answer:

the answer is D

Explanation:

Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object.

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3 years ago
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A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the
kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂  : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F  : Force ( N)

d  : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the  bar

mm = 1.5 kg :  mass of the  monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey  (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances  from forces the point 2

d₁₂ = (3-0.6) m = 2.4m  : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium  of moments at the point  2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0
3 years ago
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