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3 years ago

14

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

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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Why are some consolation visible to New York State observers at midnight during April , but not visible at midnight during Octob

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"Midnight" means looking away from the Sun. But in 6 months from April to October the earth goes halfway around the Sun. So midnight in April and midnight in October are exactly opposite directions.

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3 years ago

1. A 1.5 kg ball moves in a circle that is 0.5 m in radius at a speed of 5.1 m/s.

kolezko [41]

Answer: a= 52.02 m/s²

Fc= 78.03 N

Explanation: Solution attached:

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3 years ago

How was dance used in primitive cultures

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3 years ago

Which statement correctly lists the type of electromagnetic waves from highest to lowest energy? gamma rays, ultraviolet waves,

ahrayia [7]

Answer:

Thank you for posting your question on Brainly.

Explanation:

The waves from highest to lowest are :

* Gamma

* X- Rays

* UV

* Visible

* Infrared

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I really hope this was helpful. Have a great day : )

6 0

3 years ago

The acceleration of a particle is defined by the relation a 5 28 m/s2. Knowing that x 5 20 m when t 5 4 s and that x 5 4 m when

mamaluj [8]

Explanation:

It is given that,

The acceleration of a particle, $a=-8\ m/s^2$ (negative as the particle is decelerating)

Initial distance, x₁ = 20 m

Initial time, t₁ = 4 s

New distance x₂ = 4 m

Velocity, v = 10 m/s

(A) Calculating initial distance using second equation of motion as :

$x_1=ut_1+\dfrac{1}{2}at^2$

$20=4u+\dfrac{1}{2}(-8)\times 4^2$

u = 21 m/s

When velocity of the particle is zero, time taken is t (say). Using first equation of motion as :

$v=u+at$

$0=21+(-8)t$

t = 2.62 seconds

So, the velocity of the particle is zero at t = 2.62 seconds.

(B) Velocity at t = 11 s

$v=21+(-8)\times 11$

v = 13 m/s

Total distance covered at t = 11 s. The overall path travelled by the particle during its entire journey is called total distance covered.

$d=ut+\dfrac{1}{2}at^2+|ut+\dfrac{1}{2}at^2|$

$d=21\times 2.62+\dfrac{1}{2}\times (-8)(2.62)^2+|21\times 8.38+\dfrac{1}{2}\times (-8)(8.38)^2|$

d = 132.48 m

So, the distance travelled by the particle at t = 11 seconds is 132.48 meters.

5 0

3 years ago