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3 years ago

If the heart becomes damaged or weakened, how will this affect the body’s systems?

Physics

2 answers:

mr Goodwill [35]3 years ago

3 0

Answer:

As your heart works harder, it becomes weaker and the damage increases. Your body gets less oxygen, and you might notice symptoms like shortness of breath, swelling in your legs, and fluid buildup. Your body tries to keep the blood it has to supply your heart and brain.

Explanation:

Bad White [126]3 years ago

3 0

Answer:

well when your heart works harder, it becomes weaker and the damage increases. Your body gets less oxygen, and you might notice symptoms like shortness of breath, swelling in your legs, and fluid buildup. Your body tries to keep the blood it has to supply your heart and brain

Explanation:

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When looking at the chemical symbol, the charge of the ion is displayed as the

Black_prince [1.1K]

Answer:

superscript

Explanation:

When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.

An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.

6 0

3 years ago

Is a mirrors surface opaque or transparent or translucent?

dangina [55]

A mirror is opaque you can not see through it but you can see a reflection within it

5 0

3 years ago

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axi

Marrrta [24]

Answer:

h = 2.64 meters

Explanation:

It is given that,

Mass of one ball, $m_1=3\ kg$

Speed of the first ball, $v_1=20\ m/s$ (upward)

Mass of the other ball, $m_2=2\ kg$

Speed of the other ball, $v_2=-12\ m/s$ (downward)

We know that in an inelastic collision, after the collision, both objects move with one common speed. Let it is given by V. Using the conservation of momentum to find it as :

$V=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}$

$V=\dfrac{3\times 20+2\times (-12)}{3+2}$

V = 7.2 m/s

Let h is the height reached by the combined balls of putty rise above the collision point. Using the conservation of energy as :

$mgh=\dfrac{1}{2}mV^2$

$h=\dfrac{V^2}{2g}$

$h=\dfrac{7.2^2}{2\times 9.8}$

h = 2.64 meters

So, the height reached by the combined mass is 2.64 meters. Hence, this is the required solution.

5 0

3 years ago

A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di

siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

$E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\$

part a)

Electric Field strength at point P: a = 0.75 m

$E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C$

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

$E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o$

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

6 0

3 years ago

.1 An 8-ft 3 tank contains air at an initial temperature of 808F and initial pressure of 100 lbf/in. 2 The tank develops a small

Alina [70]

Correct temperature is 80°F

Answer:

T_f = 38.83°F

Explanation:

We are given;

Volume; V = 8 ft³

Initial Pressure; P_i = 100 lbf/in² = 100 × 12² lbf/ft²

Initial temperature; T_i = 80°F = 539.67 °R

Time for outlet flow; t_o = 90 s

Mass flow rate at outlet; m'_o = 0.03 lb/s

Final pressure; P_f = 30 lbf/in² = 30 × 12² lbf/ft²

Now, from ideal gas equation,

Pv = RT

Where v is initial specific volume

R is ideal gas constant = 53.33 ft.lbf/°R

Thus;

v = RT/P

v_i = 53.33 × 539.67/(100 × 12²)

v_i = 2 ft³/lb

Formula for initial mass is;

m_i = V/v_i

m_i = 8/2

m_i = 4 lb

Now change in mass is given as;

Δm = m'_o × t_o

Δm = 0.03 × 90

Δm = 2.7 lb

Now,

m_f = m_i - Δm

Thus; m_f = 4 - 2.7

m_f = 1.3 lb

Similarly in above;

v_f = V/m_f

v_f = 8/1.3

v_f = 6.154 ft³/lb

Again;

Pv = RT

Thus;

T_f = P_f•v_f/R

T_f = (30 × 12² × 6.154)/53.33

T_f = 498.5°R

Converting to °F gives;

T_f = 38.83°F

7 0

3 years ago