Ask question

Ask question

All categories

3 years ago

13

WILL MARK BRAINLIEST The Northern and Southern Hemispheres face the sun evenly. What do the hemispheres experience at that tim

e?

1 answer:

olga_2 [115]3 years ago

6 0

An equinox .............

You might be interested in

6. Find the heat flow in 24 hours through a refrigerator door 30.0" x 58.0" insulated with cellulose fiber 2.0" thick. The tempe

Ilia_Sergeevich [38]

Answer:

The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU

Explanation:

The given parameters are;

The duration of the heat transfer, t = 24 hours = 86,400 seconds

The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²

The material of the insulator in the door = Cellulose fiber

The thickness of the insulator in the door, d = 2.0" = 0.0508 m

The temperature inside the fridge = 38° F = 276.4833 K

The temperature of the room = 78°F = 298.7056 K

The thermal conductivity of cellulose fiber = 0.040 W/(m·K)

By Fourier's law, the heat flow through a by conduction material is given by the following formula;

$\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}$

$Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t$

Therefore, we have;

$Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522$

The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU

7 0

2 years ago

A helical compression spring is made with oil-tempered wire with wire diameter of 0.2 in, mean coil diameter of 2 in, a total of

Naya [18.7K]

Answer:

a. Solid length Ls = 2.6 in

b. Force necessary for deflection Fs = 67.2Ibf

Factor of safety FOS = 2.04

Explanation:

Given details

Oil-tempered wire,

d = 0.2 in,

D = 2 in,

n = 12 coils,

Lo = 5 in

(a) Find the solid length

Ls = d (n + 1)

= 0.2(12 + 1) = 2.6 in Ans

(b) Find the force necessary to deflect the spring to its solid length.

N = n - 2 = 12 - 2 = 10 coils

Take G = 11.2 Mpsi

K = (d^4*G)/(8D^3N)

K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in

Fs = k*Ys = k (Lo - Ls )

= 28(5 - 2.6) = 67.2 lbf Ans.

c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.

For C = D/d = 2/0.2 = 10

Kb = (4C + 2)/(4C - 3)

= (4*10 + 2)/(4*10 - 3) = 1.135

Tau ts = Kb {(8FD)/(Πd^3)}

= 1.135 {(8*67.2*2)/(Π*2^3)}

= 48.56 * 10^6 psi

Let m = 0.187,

A = 147 kpsi.inm^3

Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi

Ssy = 0.50 Sut

= 0.50(198.6) = 99.3 kpsi

FOS = Ssy/ts

= 99.3/48.56 = 2.04 Ans.

7 0

3 years ago

Design an op amp circuit to average the input of six sensors used to measure temperature in restaurant griddles for a large fast

Marina CMI [18]

Answer:

See the attached file for the design.

Explanation:

Find attached for the explanation.

3 0

3 years ago

Write a program that adds the numbers 1 through 5 into the numbers ArrayList and then prints out the first element in the list.

nekit [7.7K]

Answer:

Answer code is given below along with comments to explain each step

Explanation:

Method 1:

// size of the array numbers to store 1 to 5 integers

int Size = 5;

// here we have initialized our array numbers, the type of array is integers and size is 5, right now it is empty.

ArrayList<Integer> numbers = new ArrayList<Integer>(Size);

// lets store values into the array numbers one by one

numbers.add(1)

numbers.add(2)

numbers.add(3)

numbers.add(4)

numbers.add(5)

// here we are finding the first element in the array numbers by get() function, the first element is at the 0th position

int firstElement = numbers.get(0);

// to print the first element in the array numbers

System.out.println(firstElement);

Method 2:

int Size = 5;

ArrayList<Integer> numbers = new ArrayList<Integer>(Size);

// here we are using a for loop instead of adding manually one by one to store the values from 1 to 5

for (int i = 0; i < Size; i++)

{

// adding the values 1 to 5 into array numbers

numbers.add(i);

}

//here we are finding the first element in the array numbers

int firstElement = numbers.get(0);

// to print the first element in the array

System.out.println(firstElement);

6 0

3 years ago

Batteries don’t run forever. Why do you think batteries run down?

Marina86 [1]

Because there is no such thing as immortal batteries.

When a battery is connected to a circuit, the charge moves through the circuit, and a chemical reaction occurs inside that separate the charges. The strength of this reaction diminishes over time and the battery eventually dies.

3 0

3 years ago

Read 2 more answers