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3 years ago

7

In testing thousands of different materials for use as lightbulb filaments, Thomas Edison best illustrated a problem-solving app

roach known as:
Group of answer choices
a. fixation.
b. belief perseverance.
c. trial and error.
d. the confirmation bias.
e. the representativeness heuristic.

1 answer:

Aleksandr [31]3 years ago

6 0

Answer:

Trial and error.

Explanation:

This approach of Thomas Edison where in testing of thousand of different material for light bulb filament are used to find the most suitable material is called trial and error approach of problem solving.

Trial and error pursue a method, seeing if it performs, and not trying a new mechanism. This mechanism will be repeated until a solution or success is attained. Assume moving a large object like a couch into your house for example.

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Vikentia [17]

Answer: B superconductors

Explanation:

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3 years ago

Alice and Bob sit on a see saw, that consists of a uniform beam of mass 20kg and a fulcrum rock, that acts as a pivot point. The

tankabanditka [31]

Answer:

Explanation:

Given

Mass of alice $M_a=50 kg$

Mass of bob $M_b=70 kg$

Mass of beam $M=20 kg$

Suppose fulcrum is at distance of d from bob thus From diagram

Equating torque i.e. torque of Bob,Alice and beam must cancel out each other to be balanced

$500\times (8-d)+20\times (4-d)=70\times d$

$400-50d+80-2d=70d$

$480=140d$

$d=\frac{480}{140}$

$d=3.43 m$

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4 years ago

A box of mass m1 = 20.0 kg is released from rest at a warehouse loading dock and slides down a 3.0 m-high frictionless chute to

Harrizon [31]

´To develop this problem we will use the concepts related to the conservation of momentum and the application of energy conservation equations to find the velocity of the mass after the collision, like this:

Velocity of the mass $m_1$ just before the collision

$v_1 = \sqrt{2gh}$

$v_1 = \sqrt{2(9.8)(3)}$

$v_1 = 7.67m/s$

Therefore the momentum just before collision would be

$p_2 = m_1v_1+40(0)\\p_1 = 20*7.67+40(0)\\p_1 = 153.36kg \cdot m/s$

Momentum after the collision

$p_1 = 20*u_1+40u_1\\p_1 = 60u_1$

Since the momentum is conserved we have that

$153.36= 60u_1$

$u_1 = \frac{153.36}{60}$

$u_1 = 2.56m/s$

The velocity of mass $m_2$ after the collision is given by

$v_2 = \frac{2m_1}{m_1+m_2} u_1$

$v_2 = \frac{2(20)}{20+40}(2.56)$

$v_2 = 1.71m/s$

Therefore the change in momentum of mass 2 is

$p_2 = m_2v_2$

$p_2 = 40*1.71$

$p_2 = 68.4kg\cdot m/s$

Therefore the impulse acting on m2 during the collision between the two boxes is $p = 68.4kg\cdot m/s$

8 0

4 years ago

Why does it matter if a material is amorphous or crystalline? How does the arrangement of atoms affect the properties of materia

Rashid [163]

Answer:

<em>It matters because crystalline and amorphous materials have different properties. The arrange affects the melting point (defined in crystals and a larger range in amorphous) and shape (geometrical in crystals, no geometrical in amorphous). </em>

Explanation:

The particles that compose a solid material are held in place by strong tractive forces between them when we analyze solids we consider the position of the atoms (molecules or ions) rather than their motion (which is important in liquids and gases). This positioning can be arranged in two general ways:

Crystalline solids have internal structures that in turn lead to distinctive flat surfaces or face, these faces intersect at angles that are characteristic of the substance, crystals tend to have sharp, well defined and high melting points because of the same distance from the same number and type of neighbors. They generally have geometric shapes, some examples are diamonds, metals, salts.
Amorphous solids produce irregular or curved surfaces when broken and they have poorly defined patterns when exposed to x rays because of their irregular array. In contrast with crystal solids, amorphous solids soften over a wide temperature range due to the different amounts of thermal energy needed to overcome different interactions. Some examples of these solids are gels, plastics, and some polymers.

I hope you find this information useful and interesting! Good luck!

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4 years ago

Energy is defined as

stira [4]

the strength and vitality required for sustained physical or mental activity.

5 0

4 years ago

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