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3 years ago

What action within the heart creates the characteristic “lib-dub” heartbeat?

Physics

2 answers:

harkovskaia [24]3 years ago

8 0

Valves of the heart oppening and closing

balu736 [363]3 years ago

8 0

A heartbeat is the sound caused by the heart valves opening and closing while they pump blood. A properly working heart only has blood flowing in one direction. This is possible by the valves opening and closing in exact coordination with the heart’s pumping action. Assuming you meant lub-dub, the lub sound is the tricuspid and mitral valves closing. The dub sound is the aortic and pulmonary valves closing. Combined, they make the sound of a heartbeat/the sound of someone's heart working properly.

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What do you call rocxs that form under intense heat and pressure

kirill [66]

Metamorphic

Metamorphic rocks are formed under the surface of the earth from the metamorphosis (change) that occurs due to intense heat and pressure (squeezing).

5 0

3 years ago

Allen Aby sets up an Atwood Machine and wants to find the acceleration and the tension in the string. Please help him. Two block

Vitek1552 [10]

<u>Answers:</u>

In order to solve this problem we will use Newton’s second Law, which is mathematically expressed after some simplifications as:

This can be read as: The Net Force $F$ of an object is equal to its mass $m$ multiplied by its acceleration $a$ .

We will also need to <u>draw the Free Body Diagram of each block</u> in order to know the direction of the acceleration in this system and find the Tension $T$ of the string (<u>See figure attached). </u>

We already know<u> $m_{2}$ is greater than $m_{1}$ </u>, this means the weight of the block 2 $P_{2}$ is greater than the weight of the block 1 $P_{1}$ ; therefore <u>the acceleration of the system will be in the direction of $P_{2}$ </u>, as shown in the figure attached.

We also know by the information given in the problem that <u>the pulley does not have friction and has negligible mass</u>, and <u>the string is massless</u>.

This means that the tension will be the same along the string regardless of the difference of mass of the blocks.

Now that we have the conditions clear, let’s begin with the calculations:

1) Firstly, we have to find the weight of each block, in order to verify that block 2 is heavier than block 1.

This is done using equation (1), where the force of the weight $P$ is calculated using the <u>acceleration of gravity</u> $g=9.8\frac{m}{s^{2}}$ acting on the blocks:

<u>For block 1: </u>

$P_{1}=m_{1}g$ (3)

$P_{1}=1.5kg(9.8\frac{m}{s^{2}})$

<u>For block 2: </u>

$P_{2}=m_{2}g$ (5)

$P_{2}=2.4kg(9.8\frac{m}{s^{2}})$

Then, we are going to <u>find the acceleration $a$ of the whole system: </u>

$F_{r}=P_{1}+P_{2}$ (7)

Where the Resulting Force $F_{r}$ is equal to the sum of the weights $P_{1}$ and $P_{2}$ .

In the figure attached, note that $P_{1}$ is in opposite direction to the acceleration $a$ , this means it must <u>have a negative sing</u>; while $P_{2}$ is in the same direction of $a$ .

Here we only have to isolate $a$ from equation (8) and substitute the values according to the conditions of the system:

$-14.7N+23.52N=(1.5kg+2.4kg)a$

$8.82N=(3.9kg)a$

Then:

$a=\frac{8.82N }{3.9kg}$

<h2> $a=2.26\frac{m}{ s^{2}}$ </h2><h2>This is the acceleration of the system. </h2>

2) For the second part of the problem, we have to find the tension $T$ of the string.

We can choose either the Free Body Diagram of block A or block B to make the calculations, <u>the result will be the same</u>.

Let’s prove it:

For $m_{1}$

we see in the free body diagram that the <u>acceleration is in the same direction of the tension of the string</u>, so:

$F_{r}=T-P_{1}$ (9)

$T-P_{1}=m_{1}a$ (10)

$T-14.7N=(1.5kg)( 2.26\frac{m}{ s^{2}})$

Then;

<h2> $T=18.09N$ This is the tension of the string </h2><h2> </h2>

For $m_{2}$

we see in the free body diagram that the acceleration is in opposite direction of the tension of the string and must <u>have a negative sign,</u> so:

$F_{r}=T-P_{2}$ (9)

$T-P_{2}=m_{2}a$ (10)

$T-23.52N=(2.4kg)(-2.26\frac{m}{ s^{2}})$

Then;

<h2> $T=18.09N$ This is the same tension of the string </h2>

6 0

3 years ago

It has been proposed that we could explore Mars using inflated balloons to hover just above the surface. The buoyancy of the atm

klio [65]

Answer:

a) mb = 0.0596 kg ; r = 0.974 m

b) a = 754 m/s^2 .. (Upward)

c) mL = 5.96 kg

Explanation:

Given:-

- The density of Mars atmosphere , ρ = 0.0154 kg/m^3

- The surface density of ballon, σ = 5.0g/m^2

Solution:-

(a) What should be the radius and mass of these balloons so they just hover above the surface of Mars?

- We will first isolate a balloon in the Mar's atmosphere and consider the forces acting on the balloon. We have two forces acting on the balloon.

- The weight of the balloon - "W" - i.e ( Tough plastic weight + Gas inside balloon). Since, the balloon is filled with a very light gas we will assume the weight due to gas inside to be negligible. So we have:

W = mb*g

Where, mb: Mass of balloon

g: Gravitational constant for Mars

- The mass of the balloon can be determined by using the surface density of the tough plastic given as "σ" and assuming the balloon takes a spherical shape when inflated with surface area "As".

As = 4πr^2

Where, r: The radius of balloon

So, mb = 4σπr^2

- Substitute the mass of balloon "mb" in the expression developed for weight of the balloon:

W = 4*σ*g*πr^2 ......... Eq1

- The weight of the balloon is combated by the buoyant force - "Fb" produced by the volume of Mars atmosphere displaced by the balloon acting in the upward direction:

Fb = ρ*Vs*g

Where, Vs : Volume of sphere = 4/3 πr^3

So, Fb = ρ*g*4/3 πr^3 ....... Eq 2

- Apply the Newton's equilibrium conditions on the balloon in the vertical direction:

Fb - W = 0

Fb = W

ρ*g*4/3 πr^3 = 4*σ*g*πr^2

r = 3σ / ρ

r = 3*0.005 / 0.0154

r = 0.974 m .... Answer

- Use the value of radius "r" and compute the "mb":

mb = 4σπr^2

mb = 4*0.005*π (0.974)^2

mb = 0.0596 kg ... Answer

(b) If we released one of the balloons from part (a) on earth, where the atmospheric density ρ = 1.20kg/m^3, what would be its initial acceleration assuming it was the same size as on Mars? Would it go up or down?

- The similar analysis is to be applied when the balloon of the same size i.e r = 0.974 m and mass mb = 0.0596 kg is inflated on earth with density ρ = 1.20kg/m^3.

- Now see that the buoyant force acting on the balloon due to earth's atmosphere is different from that found on Mars. So the new buoyant force Fb using Eq2 is:

Fb = ρ*g*4/3 πr^3

Where, g: Gravitational constant on earth = 9.81 m/s^2

Fb = (1.20)*(9.81)*(4/3)* π*(0.974)^3

Fb = 45.5 N

- Apply the Newton's second law of motion in the vertical direction on the balloon:

Fb - W = mb*a

Where, a: The acceleration of balloon

a = (Fb - W) / mb

a = Fb/mb - g

a = 45.5/0.0596 - 9.81

a = 754 m/s^2 (upward) ..... Answer

c), d) If on Mars these balloons have five times the radius found in part (a), how heavy an instrument package could they carry?

- The new radius of the balloon - "R" -is five times what was calculated in part (a):

- Apply the Newton's equilibrium conditions in the vertical direction on the balloon with the addition of downward weight of load "WL":

Fb - W - WL = 0

WL = Fb - W

mL*g = ρ*g*4/3 πR^3 - 4*σ*g*πR^2

Where, mL : The mass of load due to instrument package

mL = ρ*4/3 πR^3 - 4*σ*πR^2

mL = 0.0154*4/3*π*(5*0.974)^3 - 4*(0.005)*π*(5*0.974)^2

mL = 7.45 - 1.45

mL = 5.96 kg ..... Answer

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