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Valentin [98]
3 years ago
12

What is a problem that technology can help solve that problem?

Engineering
2 answers:
Maslowich3 years ago
8 0
Seeing what the other side of the world is doing right now
Margarita [4]3 years ago
5 0
Small technologies can solve big problems. From famine to poverty, water scarcity to business management, or healthcare to education, technology has all the answers…just ask any question!
You might be interested in
How does the human body use phospholipids ?
Shtirlitz [24]
Phospholipids are crucial for building the protective barrier, or membrane, around your body's cells. In fact, phospholipids are synthesized in the body to form cell and organelle membranes. In blood and body fluids, phospholipids form structures in which fat is enclosed and transported throughout the bloodstream.
6 0
3 years ago
Read 2 more answers
The force of T = 20 N is applied to the cord of negligible mass. Determine the angular velocity of the 20-kg wheel when it has r
horrorfan [7]

Image of wheel is missing, so i attached it.

Answer:

ω = 14.95 rad/s

Explanation:

We are given;

Mass of wheel; m = 20kg

T = 20 N

k_o = 0.3 m

Since the wheel starts from rest, T1 = 0.

The mass moment of inertia of the wheel about point O is;

I_o = m(k_o)²

I_o = 20 * (0.3)²

I_o = 1.8 kg.m²

So, T2 = ½•I_o•ω²

T2 = ½ × 1.8 × ω²

T2 = 0.9ω²

Looking at the image of the wheel, it's clear that only T does the work.

Thus, distance is;

s_t = θr

Since 4 revolutions,

s_t = 4(2π) × 0.4

s_t = 3.2π

So, Energy expended = Force x Distance

Wt = T x s_t = 20 × 3.2π = 64π J

Using principle of work-energy, we have;

T1 + W = T2

Plugging in the relevant values, we have;

0 + 64π = 0.9ω²

0.9ω² = 64π

ω² = 64π/0.9

ω = √64π/0.9

ω = 14.95 rad/s

4 0
4 years ago
Assume that array1 and array2 are both 25-element integer arrays. Indeicate whether each of the following statements is legal or
vfiekz [6]

Answer:

a. array1 = array2; Illegal

b. cout<< array1; Illegal

c. cin >>array2; Illegal

Explanation:

a. array1 = array2;

Illegal, since the assignment operator doesn't work to directly assign the values into other array.

b. cout<< array1;

Illegal, since the index number in not defined, if you want to print all the elements in the array1 then run a for loop to iterate over the entire loop.

c. cin >>array2;

Illegal, since the index number is not defined, if you want get input from the user then run a for loop to iterate over the entire loop.

3 0
3 years ago
From the list below, select four situations in which casting is the preferred fabrication technique. (1) For alloys having low d
Masteriza [31]

Answer:

(1) , (2) , (4) ,(6).

Explanation:

Use of casting process:

1.When the size of the mold is more or we can say that when the size of casting is more.

2.When the mechanical strength of the casting product is not important.

3.When the required ductility of the product is not to high .Generally casting product is more brittle with high strength.

4.Generally cost of casting product is not too high or we can say that casting product is economically.

Therefore the following options are correct -

(1) , (2) , (4) ,(6).

5 0
4 years ago
A 10-ft-long simply supported laminated wood beam consists of eight 1.5-in. by 6-in. planks glued together to form a section 6 i
ruslelena [56]

Answer:

point B where Q_B = 101.25 \ in^3  has the largest Q value at section a–a

Explanation:

The missing diagram that is suppose to be attached to this question can be found in the attached file below.

So from the given information ;we are to determine the  point that  has the largest Q value at section a–a

In order to do that; we will work hand in hand with the image attached below.

From the image attached ; we will realize that there are 8 blocks aligned on top on another in the R.H.S of the image with the total of 12 in; meaning that each block contains 1.5 in each.

We also have block partitioned into different point segments . i,e A,B,C, D

For point A ;

Let Q be the moment of the Area A;

SO ; Q_A = Area \times y_1

where ;

y_1 = (6 - \dfrac{1.5}{2})

y_1 = (6- 0.75)

y_1 = 5.25 \  in

Q_A =(L \times B)  \times y_1

Q_A =(6 \times 1.5)  \times 5.25

Q_A =47.25 \ in^3

For point B ;

Let Q be the moment of the Area B;

SO ; Q_B = Area \times y_2

where ;

y_2 = (6 - \dfrac{1.5 \times 3}{2})

y_2= (6 - \dfrac{4.5}{2}})

y_2 = (6 -2.25})

y_2 = 3.75 \ in

Q_B =(L \times B)  \times y_1

Q_B=(6 \times 4.5)  \times 3.75

Q_B = 101.25 \ in^3

For point C ;

Let Q be the moment of the Area C;

SO ; Q_C = Area \times y_3

where ;

y_3 = (6 - \dfrac{1.5 \times 2}{2})

y_3 = (6 - 1.5})

y_3= 4.5 \  in

Q_C =(L \times B)  \times y_1

Q_C =(6 \times 3)  \times 4.5

Q_C=81 \ in^3

For point D ;

Let Q be the moment of the Area D;

SO ; Q_D = Area \times y_4

since there is no area about point D

Area = 0

Q_D =0 \times y_4

Q_D = 0

Thus; from the foregoing ; point B where Q_B = 101.25 \ in^3  has the largest Q value at section a–a

3 0
4 years ago
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