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vivado [14]
4 years ago
12

Which of the following describes the electron sharing between hydrogen and fluorine A.Hydrogen and fluorine share one electron w

ith each other B.
Hydrogen and fluorine share seven electrons with each other C.Hydrogen shares seven electrons, while fluorine shares one electron. D.Hydrogen shares one electron, while fluorine shares seven electrons
Physics
1 answer:
atroni [7]4 years ago
5 0
Hydrogen has one electron in its outermost shell, while fluorine has seven electron in its outermost shell, hence both hydrogen and fluorine needs a single electron to complete its outermost shell. 
That's why there is a single bond between hydrogen and fluorine.
Hence both hydrogen and fluorine share one electron with each other, so option "A" is correct.
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As the distance between two charged objects increases, the strength of the electrical force between the objects
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Answer:

I believe the answer is It increases

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3 years ago
From a hot air balloon that is at rest at a certain height, a projectile is launched horizontally at 30m / s, how fast will it h
Yuki888 [10]

Answer:

A. 50 m/s

Explanation:

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 4 s

Find: v

v = at + v₀

v = (10 m/s²) (4 s) + 0 m/s

v = 40 m/s

In the x direction, the velocity is constant at 30 m/s.

The overall speed is:

v² = (30 m/s)² + (40 m/s)²

v = 50 m/s

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3 years ago
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3 years ago
What is the main force that must be overcome in order to push an object
My name is Ann [436]

Answer: Friction

Explanation:

Friction and the normal force would be the two initial forces to overcome.

4 0
3 years ago
A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o
masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, F_{f2} = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = F_{f2} × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27

Frictional \ force, F_{f2} = \left (981 -  \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8}  \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times  0.25 \right) \approx 219.92

The frictional force on the block W₂, F_{f2} ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ F_{w2} = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + \mathbf{F_{w2}}

The frictional force from the ground, \mathbf{F_{f1}} = N×μ + \mathbf{F_{f2}} = P

Where;

P = The horizontal force applied to the block

P = (W₁ + \mathbf{F_{w2}}) × μ + \mathbf{F_{f2}}

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

brainly.com/question/18038995

3 0
3 years ago
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