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3 years ago

Label the part that make up the human heart. Drag the item on the left to the correct location on the right.

Physics

2 answers:

pantera1 [17]3 years ago

5 0

hey there! ;D

A) aorta - goes on the top right

E) pulmonary artery - goes below aorta

D) left atrium - goes below pulmonary artery

B) left ventricle - goes below left atrium

C) right atrium - goes on top left

F) right ventricle - goes below right atrium

hope this helps!

have an awesome day ;)

EleoNora [17]3 years ago

3 0

Answer:

Attached is the picture with the answers

Explanation:

I will describe the parts of the heart in the picture starting on the left side of the image, up-down, then right side, up-down.

C. Right Atrium - Is one of the chambers in the heart where blood enters the ventricles. It's located in the upper right part and it receives blood from the venae cavae (deoxygenated blood)

F. Right Ventricle - Is one of the two large chambers located in the right lower part of the heart. It receives blood from the right atrium and pumps it to the lungs.

A. Aorta - It's the biggest and principal artery in the body. It comes out of the left ventricle and goes to the lower abdomen, where it splits into 2 smaller arteries, distributing oxygenated blood to the body.

E. Pulmonary artery - This artery, part of pulmonary circulation, takes deoxygenated blood from the right part of the heart into the lungs.

D. Left atrium - Is one of the chambers in the heart where blood enters the ventricles. It's located in the upper left part and it receives blood from the lungs (pulmonary circulation)

B. Left ventricle - Is one of the two large chambers located in the left lower part of the heart. It receives blood from the left atrium and pumps oxygenated blood through the aorta.

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Rudiy27

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4 years ago

How much Power is needed to lift a 200 N block

vova2212 [387]

Answer:

10 W

Explanation:

P = F x v

F = 200N

v = x2 - x1 / t

= 6 - 3 / 60

= 3 / 60

= 1 / 20

= 0.05 m/s

P = 200 x 1/20

= 10 W

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2 years ago

A rocket rises vertically, from rest, with an acceleration of 3.6m/s^2 until it runs out of fuel at an altitude of 1500m. After

Montano1993 [528]

A. 103.9 m/s

The motion of the rocket until it runs out of fuel is an accelerated motion with constant acceleration a = 3.6 m/s^2, so we can use the following equation

$v^2 -u^2 = 2ad$

where

v is the velocity of the rocket when it runs out of fuel

u = 0 is the initial velocity of the rocket

a = 3.6 m/s^2 is the acceleration

d = 1500 m is the distance covered during this first part

Solving for v, we find

$v=\sqrt{u^2+2ad}=\sqrt{(0^2+2(3.6 m/s^2)(1500 m)}=103.9 m/s$

B. 28.9 s

We can calculate the time taken for the rocket to reach this altitude with the formula

$d=\frac{1}{2}at^2$

where

d = 1500 m

a = 3.6 m/s^2

Solving for t, we find

$t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(1500 m)}{3.6 m/s^2}}=28.9 s$

C. 2050.8 m

We can calculate the maximum altitude reached by the rocket by using the law of conservation of energy. In fact, from the point it runs out of fuel (1500 m above the ground), the rocket experiences the acceleration due to gravity only, so all its kinetic energy at that point is then converted into gravitational potential energy at the point of maximum altitude:

$K_i = U_f\\\frac{1}{2}mu^2 = mgh$

where h is distance covered by the rocket after it runs out of fuel, and v=103.9 m/s is the velocity of the rocket when it starts to decelerate due to gravity. Solving for h,

$h=\frac{v^2}{2g}=\frac{(103.9 m/s)^2}{2(9.8 m/s^2)}=550.8 m$

So the maximum altitude reached by the rocket is

$h' = d+h=1500 m +550.8 m=2050.8 m$

D. 39.5 s

The time needed for the second part of the trip (after the rocket has run out of fuel) can be calculated by

$h=\frac{1}{2}gt^2$

where

h = 550.8 m is the distance covered in the second part of the trip

g = 9.8 m/s^2

Solving for t,

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(550.8 m)}{9.8 m/s^2}}=10.6 s$

So the total time of the trip is

$t'=28.9 s+10.6 s=39.5 s$

E. 200.5 m/s

When the rocket starts moving downward, it is affected by gravity only. So the gravitational potential energy at the point of maximum altitude is all converted into kinetic energy at the instant the rocket hits the ground:

$\frac{1}{2}mv^2 = mgh$