Correct Answer: A. 4° N and 20° N

Correct Answer: C. cumulonimbus

Correct Answer: A. 0.36J

Explanation

P.E at max = K.E on projection

= \(\frac{1}{2}\)mv² = \(\frac{1}{2} \times 0.02 \times 6^2 = 0.36J\)

Correct Answer: A) de-oxygenated blood from the right ventricle to the lungs

Explanation

The question is asking about what the pulmonary artery does. The pulmonary artery is a blood vessel that carries blood away from the heart and towards the lungs. It is important to note that there are two types of blood: oxygenated blood, which is rich in oxygen, and deoxygenated blood, which has less oxygen. The options given are: A. de-oxygenated blood from the right ventricle to the lungs (Correct) B. oxygenated blood from the right ventricle to the lungs C. oxygenated blood from the left ventricle to the right auricle D. de-oxygenated blood from the left ventricle to the right auricle To answer the question, we need to determine which option is correct. The correct answer is A, which says that the pulmonary artery carries de-oxygenated blood from the right ventricle to the lungs. This is because the right ventricle of the heart pumps deoxygenated blood to the lungs through the pulmonary artery. In the lungs, the blood picks up oxygen and releases carbon dioxide, which is then exhaled out of the body. It is important to understand the difference between oxygenated and deoxygenated blood, and how they are transported through the body. Oxygenated blood is carried by the arteries and veins to supply oxygen to the body's tissues, while deoxygenated blood is carried back to the heart and lungs to be replenished with oxygen. In summary, the pulmonary artery carries deoxygenated blood from the right ventricle of the heart to the lungs, where it picks up oxygen and releases carbon dioxide.

Correct Answer: B. 2.3V

Explanation

V = IR = 0.97 x 2.4 = 2.3V

Correct Answer: A. 16

Explanation

2x - 3y = -11 --- (i)

3x + 2y = 3 --- (ii)

Multiply equation (i) by 3 and equation (ii) by 2

6x - 9y = -33 --- (iii)

6x + 4y = 6 --- (iv)

Subtract equation (iii) from (iv)

13y = 39

y = \(\frac{39}{13}\) = 3

substitute (3) for y in equation (ii)

3x + 2(3) = 3

3x + 6 = 3

3x = 3 - 6

3x = -3

x = \(\frac{-3}{3}\) = - 1

Now,

\((y - x)^2 = (3 - (-1))^2\)

= \((3 + 1)^2\)

= \(4^2\)

= 16

Correct Answer: B. uniform velocity

Correct Answer: C. death

Correct Answer: B. trouble and shame

Correct Answer: A. 48