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fenix001 [56]
2 years ago
8

Lucille is finding it difficult to play soccer after school. Her doctor thinks that her cells might not be getting enough oxygen

. When Lucille talked about it with her friend, her friend said there might be a problem with Lucille‘s respiratory system or circulatory system.
What do Lucille cells need an order for her body to function properly? How might a problem with Lucille‘s respiratory system and circulatory system make it difficult for Lucille to play soccer?
Physics
1 answer:
lana66690 [7]2 years ago
4 0

Answer:

See the explanation below.

Explanation:

Circulation of blood and oxygen is possible in body when circulatory system work along with respiratory system. Through tranche air moves in and out from lungs, whereas, through pulmonary arteries and veins (both connected to heart) blood moves in and out from the lungs. As Lucille is facing problem in his respiratory and circulatory system hence, it is difficult for him to play soccer because under normal circumstances when there is increase in physical activity then muscle cell respire more as compare to when the body is on rest. So, with increase of physical activity there is also increase in the rate of breathing which result in more absorption of oxygen and more removal of carbon dioxide but if there is problem in respiratory and circulatory system, for example, infection in throat due to pollution,etc. then this normal process of breathing gets affected which sometime may prove fatal to the person.

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2 years ago
Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of
vredina [299]

Answer:

The angular speed after 6s  is \omega = 1466.67s^{-1}.

Explanation:

The equation

I\alpha  = Fd

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In our case, the force applied is F = 22N, at a distance d = 6cm =0.06m, to a ring with the moment of inertia of I =mr^2; therefore, the angular acceleration is

$\alpha =\frac{Fd}{I} $

$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2} $

\alpha  = 244.44\: s^{-2}

Therefore, the angular speed \omega which is

\omega  = \alpha t

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\boxed{\omega = 1466.67s^{-1}}

7 0
3 years ago
A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st
Maru [420]

Answer:

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Explanation:

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\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r

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r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

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Applying values we get

\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa

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2 years ago
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valkas [14]

Answer:

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For planet 1, g = 26 m/s^2.

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Since MG/R^2 = 26 m/s^2, then:

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