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3 years ago

Using models helps scientists conduct research. How else can research using models save lives?

Engineering

1 answer:

Nina [5.8K]3 years ago

6 0

Answer: find the answer in the explanation.

Explanation:

Some of the methods of science are hypothesis, observations and experiments.

Scientific research and findings cut across many areas and field of life. Field like engineering, astronomy and medicine e.t.c

Using models in research is another great method in which researchers and scientists can master a process of a system and predict how it works.

Wilbur and Orville Wright wouldn't have died if the first airplane built by these two wonderful brothers was first researched by using models and simulations before embarking on a test.

Modelling in science involves calculations by using many mathematical methods and equations and also by using many laws of Physics. In this present age, many of these are computerised.

In space science, it will save lives, time and money to first model and simulate if any new thing is discovered in astronomy rather than people going to the space by risking their lives.

Also in medicine, drugs and many medical equipment cannot be tested by using people. This may lead to chaos and loss of lives.

So research using models in astronomy, medicine, engineering and many aspects of science and technology can indeed save lives.

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A gear train has two gears. The driver gear has 8 teeth and a diametral pitch of 6 teeth/inch. the follower gear has 24 teeth. W

Sliva [168]

Answer:

18 teeth/inch

Explanation:

Given that: i. driver gear has 8 teeth and diametral pitch of 6 teeth/inch.

ii. follower gear has 24 teeth.

Let the followers diametral pitch be represented by x.

Then,

8 teeth ⇒ 6 teeth/inch

24 teeth ⇒ x teeth/inch

So that;

x = $\frac{24 x 6}{8}$

= $\frac{144}{8}$

= 18 teeth/inch

The diametral is 18 teeth/inch

3 0

3 years ago

DRIVERS ED

forsale [732]

Answer:

Explanation:

only if there signal is turned on

8 0

3 years ago

Read 2 more answers

16 . You are turning onto a two-lane road divided by a broken yellow line. You know immediately that:

Over [174]

When a person is turning onto a two-lane road divided by a broken yellow line, you know immediately that you are on a two-way road.

<h3>What is the road about?</h3>

Note that a Yellow centerlines can be seen in roads and it is one that is often used to separate traffic moving in different directions.

Note also that Broken lines can be crossed to allow slower-moving traffic and as such, When a person is turning onto a two-lane road divided by a broken yellow line, you know immediately that you are on a two-way road.

See full question below

You are turning onto a two-lane road divided by a broken yellow line. You know immediately that:

Answers

You are on a two-way road.

You are on a one-way road.

The road is under repair.

You must stay to the left of the broken yellow lines.

Learn more about two-way road from

brainly.com/question/13123201

#SPJ2

5 0

1 year ago

An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40MPa. It has been dete

Nataly [62]

Answer:

Yes, fracture will occur

Explanation:

Half length of internal crack will be 4mm/2=2mm=0.002m

To find the dimensionless parameter, we use critical stress crack propagation equation

$\sigma_c=\frac {K}{Y \sqrt {a\pi}}$ and making Y the subject

$Y=\frac {K}{\sigma_c \sqrt {a\pi}}$

Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, $\sigma_c$ is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain

$Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682$

When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m

$\sigma_c=\frac {K}{Y \sqrt {a\pi}}$ and making K the subject

$K=\sigma_c Y \sqrt {a\pi}$ and substituting 260 MPa for $\sigma_c$ while a is taken as 0.003m and Y is already known

$K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa$

Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material

6 0

3 years ago

a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8

djverab [1.8K]

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = $\frac{MC}{I}$ = $\frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}$ = 159.07 MPa

B) Determine max transverse shear stress

attached below

ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

hence max stress in the beam = 159.07 / 2 = 79.535 MPa

attached below is the remaining solution

6 0

3 years ago