Capillaries unlike arteries and veins in mammals and other organisms due to their size and structure can branch out across multiple regions within the tissues, providing essential nutrients and other compounds.
The correct solution would be 2.
Answer:
180 W
Explanation:
The work done by the man against gravity is equal to its gain in gravitational potential energy:

where
(mg) = 720 N is the weight of the man
is the change in height
Substituting,

The power he must deliver is given by

where
W = 3600 J
t = 20 s is the time taken
Substituting,

Answer:
Current = 8696 A
Fraction of power lost =
= 0.151
Explanation:
Electric power is given by

where I is the current and V is the voltage.

Using values from the question,

The power loss is given by

where R is the resistance of the wire. From the question, the wire has a resistance of
per km. Since resistance is proportional to length, the resistance of the wire is

Hence,

The fraction lost = 
Answer:
It will take 15.55s for the police car to pass the SUV
Explanation:
We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:
1. 
2. 
Since both cars will travel the same distance x, we can equal both formulas and solve for t:

We simplify the fraction present and rearrange for our formula so that it equals 0:

In the very last step we factored a common factor t. There is two possible solutions to the equation at
and:

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at
(when the SUV passed the police car) and
(when the police car catches up to the SUV)