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3 years ago

What are 3 types of contact forces

Physics

2 answers:

Andrej [43]3 years ago

6 0

FRICTION, TENSION, AND REACTION FORCE

Svetach [21]3 years ago

5 0

There are different types of contact forces like normal Force, spring force, applied force and tension force.

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Which of the best describes a solution

Andreyy89

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1).A mixture having a uniform composition where the components can't be seen separately and all components are in the same state best describes a solution. In chemistry, a solution is a homogeneous mixture composed of two or more substances.

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2 years ago

What should the Architect do to ensure Field-Level Security is enforced on a custom Visualforce page using the Standard Lead Con

OLEGan [10]

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3 years ago

Explain what nuclear fusion is in the role it plays in Stars. DO NOT LOOK UP ANSWERS

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Stars are powered by nuclear fusion in their cores, mostly converting hydrogen into helium. The production of new elements via nuclear reactions is called nucleosynthesis. A star's mass determines what other type of nucleosynthesis occurs in its core (or during explosive changes in its life cycle).

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3 years ago

What happens to energy and frequency of a wave if its wavelength increases?

Dahasolnce [82]

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2 years ago

A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are

Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, $\Sigma F_y$ = 0

The vertical component of the tension, $T_y$ , in each half of the rope is given as follows;

$T_y$ = T × sin(θ)

∴ $\Sigma F_y$ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0

2 years ago