It is d because the law was passed around 1976 I am not sure but this is the best estimate
<em>The continental crust is up to 70 km thick.</em>
<u>Explanation</u>:
Upto 40% of the earth is made up of continental crust. <em>Sedimentary, igneous and metamorphic ricks </em>make up the continental crust. It makes up the upper layer of <em>continents and shallow seabed</em> areas near the shores.
<em>Continental crust</em> is thicker than oceanic crust with a thickness ranging from <em>25 km to 70km continental crust</em> makes up 70% of earth’s volume and is usually located above sea level with a few exceptions like the <em>Zealand continental crust</em> region which is largely under water.
Answer:
The oxygen dissociation curve represents the percentage saturation of Hb with oxygen at different partial pressure of oxygen. The different partial pressures gives sigmoid shapes to the curve. When this curves shifts to right, it indicates low affinity or binding of oxygen by the Hb. it also indicates the unloading or releases of Oxygen by Hb molecules at condition of low pressure. e,g in the muscles during strenuous exercise.However, when the curve shifts to the left, this indicate high affinity for oxygen, great binding, at high partial pressure of oxygen.e,g in the lungs to take oxygen and releases CO2.
Therefore in this scenario, the statement -. <u>During strenuous exercise, the oxygen-hemoglobin dissociation curve shifts to the right.</u> is correct. because oxygen is needed by the muscles therefore ,oxygen should be less binded by Hb, decrease affinity and easily unloaded to muscles.
<u>The statement </u>This rightward shift reflects an increase in the affinity of hemoglobin for oxygen and favors loading of O2 into hemoglobin in the lungs is wrong.
As explained above the rightwards shift indicated low affinity of Hb for oxygen(unloading)and favours unloading at the muscles because during strenuous exercise the partial pressure of oxygen is very low(but that of CO2 high) in the muscles which favours low oxygen molecules binding by Hb, and easy release to respiring cells.
Explanation:
Answer:
Q(0) = 0C, Q(1) = 264nC, Q(2) = 952C Q(3) = 2088nC, Q(4) = 3696C Q(5) = 5800nC
Explanation:
I = 4t³ + 200t² + 60t
But charge of an object =》 Q = IT
Charge of an object is the product of the current and the time in which the current passes through the membrane.
When t = 0
Q = 4(0)³ + 200(0)² + 60(0) = 0C
When t = 1
Q = 4(1)³ + 200(1)² + 60(1) = 264nC
When t = 2
Q = 4(2)³ + 200(2)² + 60(2) = 952nC
When t= 3
Q = 4(3)³ + 200(3)² + 60(3) = 2088nC
When t= 4
Q = 4(4)³ + 200(4)² + 60(4) = 3696nC
When t = 5
Q = 4(5)³ + 200(5)² + 60(5) = 5800nC
