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mamaluj [8]
3 years ago
4

Declare a variable named thisTime containing a Date object for February 3, 2018 at 03:15:28 AM. Use the toLocaleString() method

to save the text of the thisTime variable in the timeStr variable. Change the inner HTML code of the page element with the ID "timeStamp" to the value of the timeStr variable.
Engineering
1 answer:
Ivahew [28]3 years ago
4 0

Answer:

test.js

  1. let thisTime = new Date(2018, 1, 3, 3, 15, 28, 0);
  2. let timeStr = thisTime.toLocaleString();
  3. document.getElementById("timeStamp").innerHTML = timeStr;

Explanation:

In test.js file, create a Date object with specified date and time. We can specify seven numbers: year, month, day, hour, minute, second and miliseconds for the Date object (Line 1) and assign it to thisTime variable.

Next, use toLocaleString method to convert the date to string and assign it to timeStr variable (Line 2).

Lastly, use document getElementById method to create a reference object and point it to HTML element with id "timeStamp" and set the timeStr to its innerHTML property (Line 3). This will display the time string to the web page. We should see 2/3/2018, 3:15:28 AM displayed on the web page.

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Consider a 1.80-m-tall man standing vertically in water and completely submerged in a pool. Determine the difference between the
defon

Answer:

17.658 kPa

Explanation:

The hydrostatic pressure of a fluid is the weight of a column of that fluid divided by the base of that column.

P = \frac{weight}{base}

Also, the weight of a column is its volume multiplied by it's density and the acceleration of gravity:

weight = \delta * v * g

Meanwhile, the volume of a column is the area of the base multiplied by the height:

V = base * h

Replacing:

P = \frac{\delta * base * h * g}{base}

The base cancels out, so:

P = \delta * h * g

The pressure depends only on the height of the fluid column, the density of the fluid and the gravity.

If you have two point at different heights (or depths in the case of objects submerged in water) each point will have its own column of fluid exerting pressure on it. Since the density of the fluid and the acceleration of gravity are the same for both points (in the case of hydrostatics density is about constant for all points, it is not the case in the atmosphere), we can write:

\Delta P = \rho * g * (h1 - h2)

We do not know at what depth the man of this problem is, but it doesn't matter, because we know the difference in height of the two points of interes (h1 - h2) = 1.8 m. So:

\Delta P = 9.81 \frac{m}{s^{2} } * 1000 \frac{kg}{m^3} * 1.8 m = 17658 Pa = 17.658 kPa

4 0
3 years ago
If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical
lubasha [3.4K]

Answer:critical stress= 20.23 MPa

Explanation:

Since there was an internal crack, we will divide the length of the internal crack by 2

Length of internal crack, a = 0.7mm,

Half length = 0.7mm/2= 0.35mm  changing to meters becomes

0.35/ 1000= 0.35 x 10 ^-3m

The formulae for critical stress is calculated using

σC = (2Eγs /πa) ¹/₂

σC = critical stress=?

Given

E= Modulus of Elasticity= 225GPa =225 x 10 ^ 9 N/m²

γs= Specific surface energy = 1.0 J/m2 = 1.0 N/m

a= Half Length of crack=0.35 x 10 ^-3m

σC= (2 x 225 x 10 ^ 9 N/m² x 1.0 N/m /π x 0.35 x 10 ^-3m)¹/₂

=(4.5 x 10^11/π x 0.35 x 10 ^-3)¹/₂

=(4.0920 x10 ^14)¹/₂

σC=20.23 x10^6 N/m² = 20.23 MPa

​  

​

8 0
3 years ago
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