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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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Un móvil se desplaza con una rapidez inicial de 100 km/h, se le aplican los frenos con una

KIM [24]

Answer:

in English please I am quite puzzled

4 0

3 years ago

An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at

andrew-mc [135]

Answer:

9.3m/s

Explanation:

Based on the law of conservation of momentum

Sum of momentum before collision = sum of momentum after collision

m1u1 +m2u2 = m1v1+m2v2

m1 = 8kg

u1 = 15.4m/s

m2 = 10kg

u2 = 0m/s(at rest)

v1 = 3.9m/s

Required

v2.

Substitute

8(15.4)+10(0) = 8(3.9)+10v2

123.2=31.2+10v2

123.2-31.2 = 10v2

92 = 10v2

v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

6 0

3 years ago

How many dots should be drawn in the lewis dot structure for an atom of sodium A)19 B)8 C)4 D)1

Stells [14]

Na is in the first column on the periodic table so therefore it would have 1 valence electron
D 1

6 0

3 years ago

By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 12.1 kg and

puteri [66]

Answer:

14.8 kg

Explanation:

We are given that

$m_1=43.7 kg$

$m_2=12.1 kg$

$g=9.8 m/s^2$

$a=\frac{1}{2}(9.8)=4.9 m/s^2$

We have to find the mass of the pulley.

According to question

$T_2-m_2 g=m_2 a$

$T_2=m_2a+m_2g=m_2(a+g)=12.1(9.8+4.9)=177.87 N$

$T_1=m_1(g-a)=43.7(9.8-4.9)=214.13 N$

Moment of inertia of pulley= $I=\frac{1}{2}Mr^2$

$(T_2-T_1)r=I(-\alpha)=\frac{1}{2}Mr^2(\frac{-a}{r})=\frac{1}{2}Mr(-4.9)$

Where $\alpha=\frac{a}{r}$

$(177.87-214.13)=-\frac{1}{2}(4.9)M$

$-36.26=-\frac{1}{2}(4.9)M$

$M=\frac{36.26\times 2}{4.9}=14.8 kg$

Hence, the mass of the pulley=14.8 kg

6 0

3 years ago

? What is the magnitude of force required to accelerate a car of mass 1.7 × 103 kilograms by 4.75 meters/second2

KIM [24]

f=ma

f=175.1*4.75m/s

f= 831.725N

4 0

3 years ago

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