Answer:
I think its heat but im not sure
Explanation:
Answer:
Ig =7.2 +j9.599
Explanation: Check the attachment
Answer:
Glycogen is the primary energy source for muscle and liver cells.
Explanation:
Glycogen is a readily mobilized storage form of glucose. It is a very large, branched polymer of glucose residues that can be broken down to yield glucose molecules when energy is needed. Most of the glucose residues in glycogen are linked by α-1,4-glycosidic bonds. Branches at about every tenth residue are created by α-1,6-glycosidic bonds.
Glycogen is not as reduced as fatty acids are and consequently not as energy rich. Why do animals store any energy as glycogen? Why not convert all excess fuel into fatty acids? Glycogen is an important fuel reserve for several reasons. The controlled breakdown of glycogen and release of glucose increase the amount of glucose that is available between meals. Hence, glycogen serves as a buffer to maintain blood-glucose levels. Glycogen's role in maintaining blood-glucose levels is especially important because glucose is virtually the only fuel used by the brain, except during prolonged starvation. Moreover, the glucose from glycogen is readily mobilized and is therefore a good source of energy for sudden, strenuous activity. Unlike fatty acids, the released glucose can provide energy in the absence of oxygen and can thus supply energy for anaerobic activity.
Answer:
See explaination
Explanation:
int RED=10; int BLUE=11; int GREEN=12; int BUTTON1=8; int BUTTON2=9; void setup() { pinMode(RED, OUTPUT); pinMode(BLUE, OUTPUT); pinMode(GREEN, OUTPUT); pinMode(BUTTON1, INPUT); pinMode(BUTTON2, OUTPUT); } void loop() { int BTN1_STATE=digitalRead(BUTTON1); int BTN2_STATE=digitalRead(BUTTON2); if(BTN1_STATE==HIGH) { digitalWrite(BLUE, HIGH); delay(1000); // Wait for 1 second digitalWrite(BLUE, LOW); } if(BTN2_STATE==HIGH) { digitalWrite(RED, HIGH); delay(4000); // Wait for 4 seconds digitalWrite(RED, LOW); } if(BTN1_STATE==HIGH && BTN2_STATE==HIGH) { digitalWrite(GREEN, HIGH); delay(2000); // Wait for 2 second digitalWrite(GREEN, LOW); } }
Answer:
See attached image for diagrams and solution
