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oksian1 [2.3K]
2 years ago
6

Problematic substances mechanical engeneering???

Engineering
1 answer:
Sonja [21]2 years ago
6 0

Answer:

The climate crisis. ...

Making water clean and accessible. ...

Providing enough food. ...

Personalised and relevant education. ...

Improving health care. ...

The refugee crisis. ...

Cyber security. ...

Enlisting the youth

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Explain by Research how a basic generator works ? using diagram<br>​
natulia [17]
Correcto no se muy bien de que se trata el tema porque está en inglés.
Sorry
8 0
3 years ago
What is the benefit of theories? they are easily disproven they are written by famous people they were applying every circumstan
OlgaM077 [116]

Answer:

A scientific theory is an explanation of an aspect of the natural world and universe that has been repeatedly tested and verified in accordance with the scientific method, using accepted protocols of observation, measurement, and evaluation of results. Where possible, theories are tested under controlled conditions in an experiment. In circumstances not amenable to experimental testing, theories are evaluated through principles of abductive reasoning. Established scientific theories have withstood rigorous scrutiny and embody scientific knowledge.

A scientific theory differs from a scientific fact or scientific law in that a theory explains "why" or "how": a fact is a simple, basic observation, whereas a law is a statement (often a mathematical equation) about a relationship between facts. For example, Newton’s Law of Gravity is a mathematical equation that can be used to predict the attraction between bodies, but it is not a theory to explain how gravity works. Stephen Jay Gould wrote that "...facts and theories are different things, not rungs in a hierarchy of increasing certainty. Facts are the world's data. Theories are structures of ideas that explain and interpret facts.

mark me as brainlist

3 0
2 years ago
I'll mark brainliest plz help
Citrus2011 [14]

Answer:

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

7 0
3 years ago
Q2: The average water height of an ocean area is 2.5 m high and each wave lasts for an average period of 7 s. Determine (a) the
navik [9.2K]

Answer:

(a) 561.12 W/ m² (b) 196.39 MW

Explanation:

Solution

(a) Determine the energy and power of the wave per unit area

The energy per unit are of the wave is defined as:

E = 1 /16ρgH²

= 1/16 * 1025 kg/ m3* 9.81 m/s² * (2.5 m )²

=3927. 83 J/m²

Thus,

The power of the wave per unit area is,

P = E/ t

= 3927. 83 J/m² / 7 s = 561.12 W/ m²

(b) The average and work power output of a wave power plant

W = E * л * A

= 3927. 83 J/m² * 0.35 * 1 *10^6 m²

= 1374.74 MJ

Then,

The power produced by the wave for one km²

P = P * л * A

= 5612.12 W/m² * 0.35 * 1* 10^6 m²

=196.39 MW

8 0
3 years ago
An aluminum oxide component must not fail when a tensile stress of 12.5 MPa is applied. Determine the maximum allowable surface
aivan3 [116]

Answer:

1.44 mm

Explanation:

Compute the maximum allowable surface crack length using

C=\frac {2E\gamma}{\pi \sigma_c^{2}} where E is the modulus  of elasticity, \gamma is surface energy and \sigma_c is tensile stress

Substituting the given values

C=\frac {2\times 393\times 10^{9}\times 0.9}{\pi\times (16\times 10^{6})^{2}= 0.001441103 m\approx 1.44mm

The maximum allowable surface crack is 1.44 mm

4 0
3 years ago
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