Just explain the day of how you were shopping and there you have it
Answer:
The frictional force
6.446 N
The acceleration of the block a = 6.04 
Explanation:
Mass of the block = 3.9 kg
°
= 0.22
(a). The frictional force is given by


3.9 × 9.81 × 
29.3 N
Therefore the frictional force
0.22 × 29.3
6.446 N
(b). Block acceleration is given by

F = 30 N
= 6.446 N
= 30 - 6.446
= 23.554 N
The net force acting on the block is given by

23.554 = 3.9 × a
a = 6.04 
This is the acceleration of the block.
Answer:
he electron is directly transferred to NADP+ to NADPH, but electron flow is used to generate a proton gradient for ATP synthesis. Electron is not directly transferred to ATP
NADP= Nicotin amide adenine dinucleotide phosphate
NADPH =Nicotinamide adenine dinucleotide phosphate is the reduced form of NADP
therefore electron is not directly transferred to ATP
Answer: vf1/vf2= 1/ sqrt(2)
Explanation :on the moon no drag force so we have only the force of gravity. aceleration is g(moon)= 1.62m/s2.the rest is basic kinematics
if the rock travels H to the bottom we can calculate velocity:
vo=0m/s (drops the rock) , yo=0
vf*vf= vo*vo+2g(y-yo)
when the rock is halfway y = H/2 so:
vf1*vf1=2*g*H/2 so vf1 = sqrt(gH)
when the rock reach the bottom y=H so:
vf2*vf2=2*g*H so vf2 = sqrt(2gH)
so vf1/vf2= 1/ sqrt(2)
good luck from colombia
The final temperature is 83 K.
<u>Explanation</u>:
For an adiabatic process,


Given:-



(the gas is monoatomic)

T = 275
0.30
T = 83 K.