He preformed the first ever successful open heart surgery
Answer:
top speed = 17.25
Total height = 281.19 m
Explanation:
given data
mass = 75 kg
thrust = 160 N
coefficient of kinetic friction = 0.1
solution
we get here frictional force acting that is
frictional force =
.............1
frictional force = 0.1 × 75 × 9.8
frictional force = 73.5 N
and
Net force acting will be F = 160 - 73.5 N
F = 86.5 N
so
Acceleration in the First 15 second will be
F = ma .........2
86.5 = 75 × a
a = 1.15 m/s²
and
now After 15 second the velocity will be as
v = u + at ..........3
here u is 0
so v will be
V = 1.15 × 15
v = 17.25
and
now we get travels distance S in 15 s
s = u × t + 0.5 × a × t²
here u is 0
so distance s will be
s = 0.5 × a × t²
s = 0.5 × 1.15 × 15²
s = 129.37 m
and
now acceleration acting is
F =
m a =
a = 
a = - 0.98
here it is negative it mean downward nature of acceleration
and
now we get distance s by this formula
V² - u² = 2 a s
here v velocity is 0 and
u initial velocity is 17.25 m/s
put here value
0 - 17.25² = 2 × (-0.98) × s
solve it we get
s = 151.82 m
so
Total height is
Total height = 129.37 m + 151.82 m
Total height = 281.19 m
and closing
.
The heart has 4 valves. They are what makes the lub-dub lub-dub sounds that can be heard from the chest.
The mitral valve is located between the left atrium and the left ventricle. It closes the left atrium to collect oxygenated blood from the lungs and opens to pass it on to the left ventricle.
The tricuspid valve is located between the right atrium and the right ventricle. It closes the right atrium to hold unoxygenated blood and opens to pass it on to the right ventricle ensuring a one way flow.
The aortic valve is located between the aorta and the left ventricle. It closes the left ventricle and opens to the aorta to pass on the oxygen-rich blood to the body.
The pulmonary valve is located between the pulmonary artery and the right ventricle. It closes off the right ventricle and opens to pass on unoxygenated blood to the lungs.
Answer:
83,900 J
Explanation:
First, find the acceleration:
F = ma
1150 N = (1600 kg) a
a = 0.719 m/s²
Now find the final velocity.
Given:
Δx = 45.8 m
v₀ = 6.25 m/s
a = 0.719 m/s²
Find: v
v² = v₀² + 2aΔx
v² = (6.25 m/s)² + 2 (0.719 m/s²) (45.8 m)
v = 10.2 m/s
Now find the final KE:
KE = ½ mv²
KE = ½ (1600 kg) (10.2 m/s)²
KE = 83,920 J
Rounded to three significant figures, the final kinetic energy is 83,900 J.
Option A looks like the best definition