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2 years ago

Who is the mystical body of Christ

2 answers:

8 0

If you're talking ab jesus christ , some religions believe it was god in human form and some believe it was gods son it all depends on the religion and how it's taught

iVinArrow [24]2 years ago

5 0

Answer:look at explanation

Explanation:

The mystical body of Christ is The Church is called body, because it is a living entity; it is called the body of Christ, because Christ is its Head and Founder; it is called mystical body, because it is neither a purely physical nor a purely spiritual unity, but supernatural. hope this helps

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Bulk wind shear is calculated by finding the vector difference between the winds at two different heights. Using the supercell w

ivanzaharov [21]

Answer:

See explaination

Explanation:

2. 0-1 km shear value: taking winds at 1000mb and 850 mb

15 kts south easterly and 50 kts southerly

Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts

3. 0-6 km shear value: taking winds at 1000 mb and 500 mb

15 kts south easterly and 40 kts westerly

Vector difference 135/15 and 270/40 will be 281/51 kts

please see attachment

5 0

3 years ago

Which rules of the road apply to people riding bicycles, under Illinois law? *

bulgar [2K]

Answer:

C-People biking must follow all rules and laws applicable to a motorist, with some minor exceptions

Explanation:

People biking should ride on the right side of the right lane when safe, except to pass or make a left turn. When there is only one lane for traffic traveling in each direction and passing is permitted, the center of the street is marked with a broken yellow stripe

~Hope this helps!

4 0

3 years ago

2. Write a Java program that generates a new string by concatenating the reversed substrings of even indexes and odd indexes sep

Nana76 [90]

Answer:

public class Main {
public static void main(String[] args) {
String testString = "abscacd";
String evenStr = "";
String oddStr = "";
for(int i=testString.length() - 1; i >= 0; i--){
if(i % 2 == 0){
evenStr += testString.charAt(i);
}
else{
oddStr += testString.charAt(i);
}
}
System.out.println(evenStr + oddStr);
}
}

Explanation:

Firstly, let declare a variable testString to hold an input string "abscacd" (Line 1).

Next create another two String variable, evenStr and oddStr and initialize them with empty string (Line 5-6). These two variables will be used to hold the string at even index and odd index, respectively.

Next, we create a for loop that traverse the characters of the input string from the back by setting initial position index i to testString.length() - 1 (Line 8). Within the for-loop, create if and else block to check if the current index, i is divisible by 2, (i % 2 == 0), use the current i to get the character of the testString and join it with evenStr. Otherwise, join it with oddStr (Line 10 -14).

At last, we print the concatenated evenStr and oddStr (Line 18).

4 0

2 years ago

. Air at 200 C blows over a 50 cm x 75 cm plain carbon steel (AISI 1010) hot plate with a constant surface temperature of 2500 C

MrRissso [65]

Answer:

The inside temperature, $T_{in}$ is approximately 248 °C.

Explanation:

The parameters given are;

Temperature of the air = 20°C

Carbon steel surface temperature 250°C

Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²

Convection heat transfer coefficient = 25 W/(m²·K)

Heat lost by radiation = 300 W

Assumption,

Air temperature = 20 °C

Hot plate temperature = 250 °C

Thermal conductivity K = 65.2 W/(m·K)

Steady state heat transfer process

One dimensional heat conduction

We have;

Newton's law of cooling;

q = h×A×( $T_s$ - $T_{\infty$ ) + Heat loss by radiation

= 25×0.325×(250 - 20) + 300

= 2456.25 W

The rate of energy transfer per second is given by the following relation;

$P = \dfrac{K \times A \times \Delta T}{L}$

Thermal conductivity K = 65.2 W/(m·K)

Therefore;

$2456.25 = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}$

$T_{in} = 250 - \dfrac{2456.25 \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C$

The inside temperature, $T_{in}$ = 247.99 °C ≈ 248 °C.

3 0

3 years ago

A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The

AleksAgata [21]

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = $\frac{60+180}{2}$ = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

4 0

3 years ago