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 Ans 1: Human erythrocytes lack a nucleus, but frog erythrocytes have nuclei and are capable of dividing.

2: Haemoglobin – a specialised protein that binds to oxygen in the lungs so that the oxygen can be transported to the rest of the body. At high altitudes, oxygen molecules are further apart because there is less pressure to "push" them together. This effectively means there are fewer oxygen molecules in the same volume of air as we inhale. So because of low availability of oxygen the hemoglobin level in the blood is rises rt high altitude in order to increase the binding efficiency for oxygen molecules and vice versa.

3: An elevated blood sugar level can determine diabetes and pre-diabetes.

4: Atherosclerosis, sometimes called "hardening of the arteries," occurs when fat, cholesterol, and other substances build up in the walls of arteries. These deposits are called plaques. Over time, these plaques can narrow or completely block the arteries. This results in the decrement in the blood flow to the heart which eventually causes chest pain, shortness of breath or other signs and symptoms of coronary artery disease. Uncontrolled coronary artery disease or a complete blockage of arteries can cause a heart attack.

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Question: A 250-V, 4-pole, wave-wound d.c. series motor has 782 conductors on itsarmature. It has armature and series field resistance of 0.75 ohm. The motor takesa current of 40 A. Estimate its speed and gross torque developed if it has a flux per pole of 25 mWb Answer: Step 1 Mechanical Engineering homework question answer, step 1, image 1 Mechanical Engineering homework question answer, step 1, image 2 Step 2 Mechanical Engineering homework question answer, step 2, image 1 Step 3 Mechanical Engineering homework question answer, step 3, image 1

Question: A 250-V, 4-pole, wave-wound d.c. series motor has 782 conductors on itsarmature. It has armature and series field resistance of 0.75 ohm. The motor takesa current of 40 A. Estimate its speed and gross torque developed if it has a flux per pole of 25 mWb Answer: Step 1 Step 2 Step 3

QUESTION 6 (a) The bar shown in Figure Q2(a) is subjected to tensile load of 150 Kn. If the stress in the middle portions is limited to 160 N/mm², determine the diameter of the middle portion. Find also the length of the middle portion if the total elongation of the bar is to be 0.25 mm. E E = 2.0 + 105N/mm². (12 marks) 150 KN 10 cm DIA 10 cm DIA 150 KN 45 cm Figure Q6(a) (b) A brass bar, having cross-section area of 900 mm², is subjected to axial forces as shown in Figure Q2(b), in which AB = 0.6 m, BC = 0.8 m, and CD = 1.0 m. Find the total elongation of the bar. E = 1.0 + 105N/mm2 . (8 marks) Page 4 of 5 B D 40 KN 70 KN 20 KN 10 KN Figure Q6(b) (TOTAL = 20 MARKS)

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