The total power with the current source equal to 4A of the following cir

Martha has been running a small business for two years. She now seeks additional investment to finance her business. She has fou

Dafna11 [192]

Answer:

The correct option is B) Balance Sheet

Explanation:

A Balance Sheet offers a description of a company's obligations, assets, and investments as well as net income over a given span of time such as a period of 6 months or 12 months, for instance.

Also known as the Statement of Financial Position, it contains sufficient information for investors and business owners to determine the company's financial performance in that period as well as to compare the performance of that company with industry norms or competition.

Cheers

8 0

3 years ago

Which answer best describes the relationship between the left-hand and right-hand panels in both File Explorer and Finder?

Art [367]

Answer: I think The left-hand panel shows like a high level list of folders. And Right-hand panel shows the contents of selected folder.

4 0

2 years ago

A polyethylene rod exactly 10 inches long with a cross-sectional area of 0.04 in2 is used to suspend a weight of 358 lbs-f (poun

Nadya [2.5K]

Answer:

Final length of the rod = 13.90 in

Explanation:

Cross Sectional Area of the polythene rod, A = 0.04 in²

Original length of the polythene rod, l = 10 inches

Tensile modulus for the polymer, E = 25,000 psi

Viscosity, $\eta = 1*10^{9} psi -sec$

Weight = 358 lbs - f

time, t = 1 hr = 3600 sec

Stress is given by:

$\sigma = \frac{Force}{Area} \\\sigma = \frac{358}{0.04} \\\sigma = 8950 psi$

Based on Maxwell's equation, the strain is given by:

$strain = \sigma ( \frac{1}{E} + \frac{t}{\eta} )\\Strain = 8950 ( \frac{1}{25000} + \frac{3600}{10^{9} } )\\Strain = 0.39022$

Strain = Extension/(original Length)

0.39022 = Extension/10

Extension = 0.39022 * 10

Extension = 3.9022 in

Extension = Final length - Original length

3.9022 = Final length - 10

Final length = 10 + 3.9022

Final length = 13.9022 in

Final length = 13.90 in

7 0

3 years ago

Carbon dioxide initially at 50 kPa, 400 K, undergoes a process in a closed system until its pressure and temperature are 2 MPa a

Elena L [17]

Answer:

$S_2 - S_1 = -0.1104 \: \: kJ/kg.K$

The entropy change of the carbon dioxide is -0.1104 kJ/kg.K

Explanation:

We are given that carbon dioxide undergoes a process in a closed system.

We are asked to find the entropy change of the carbon dioxide with the assumption that the specific heats are constant.

The entropy change of the carbon dioxide is given by

$S_2 - S_1 = C_p \ln (\frac{T_2}{T_1}) - R\ln (\frac{P_2}{P_1})$

Where Cp is the specific heat constant

Cp = 0.846 kJ/kg.K

R is the universal gas constant

R = 0.1889 kJ/kg.K

T₁ and T₂ is the initial and final temperature of carbon dioxide.

P₁ and P₂ is the initial and final pressure of carbon dioxide.

$S_2 - S_1 = 0.846 \ln (\frac{800}{400}) - 0.1889\ln (\frac{2000}{50})$

$S_2 - S_1 = 0.846(0.69315) - 0.1889(3.6888)$

$S_2 - S_1 = 0.5864 - 0.6968$

$S_2 - S_1 = -0.1104 \: \: kJ/kg.K$

Therefore, the entropy change of the carbon dioxide is -0.1104 kJ/kg.K

6 0

3 years ago

9. A box contains (4) red balls, and (7) white balls ,we draw( two) balls with return , find 1. Show the sample space & n(s)

zzz [600]

Answer:

The answers to your questions are given below.

Explanation:

The following data were obtained from the question:

Red (R) = 4

White (W) = 7

1. Determination of the sample space, S.

The box contains 4 red balls and 7 white balls. Therefore, the sample space (S) can be written as follow:

S = {R, R, R, R, W, W, W, W, W, W, W}

nS = 11

2. Determination of the probability of all results that appeared in the sample space.

From the question, we were told that the two balls was drawn with return. There, the probability of all results that appeared in the sample space can be given as follow:

i. Probability that the first draw is red and the second is also red.

P(R1) = nR/nS

Red (R) = 4

Space space (S) = 11

P(R1) = nR/nS

P(R1) = 4/11

P(R2) = nR/nS

P(R2) = 4/11

P(R1R2) = P(R1) x P(R2)

P(R1R2) = 4/11 x 4/11

P(R1R2) = 16/121

Therefore, the Probability that the first draw is red and the second is also red is 16/121.

ii. Probability that the first draw is red and the second is white.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(R) = nR/nS

P(R) = 4/11

P(W) = nW/nS

P(W) = 7/11

P(RW) = P(R) x P(W)

P(RW) = 4/11 x 7/11

P(RW) = 28/121

Therefore, the probability that the first draw is red and the second is white is 28/121.

iii. Probability that the first draw is white and the second is also white.

White (W) = 7

Space space (S) = 11

P(W1) = nW/nS

P(W1) = 7/11

P(W2) = nW/n/S

P(W2) = 7/11

P(W1W2) = P(W1) x P(W2)

P(W1W2) = 7/11 x 7/11

P(W1W2) = 49/121

Therefore, the probability that the first draw is white and the second is also white is 49/121.

iv. Probability that the first draw is white and the second is red.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(W) = nW/nS

P(W) = 7/11

P(R) = nR/nS

P(R) = 4/11

P(WR) = P(W) x P(R)

P(WR) = 7/11 x 4/11

P(WR) = 28/121

Therefore, the probability that the first draw is white and the second is red is 28/121.

7 0

3 years ago