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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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The ground state energy of an oscillating electron is 1.23 eV. How much energy must be added to the electron to move it to the t
Vikki [24]

Answer:

  • The energy that must be added to the electron to move it to the third excited state is  -1.153 eV
  • The energy that must be added to the electron to move it to the fourth excited state is  -1.181 eV

Explanation:

Given;

Energy of electron in ground state (n = 1 ) = 1.23 eV

E₁ = 1.23 eV

Eₙ = E₁ /n²

where;

E₁ is the energy of the electron in ground state

n is the energy level,

For third excited state, n = 4

E₄ = E₁ /4²

E₄ = (1.23 eV) / 16

E₄ = 0.077 eV

Change in energy level, = E₄ - E₁ = 0.077 eV - 1.23 eV = -1.153 eV

The energy that must be added to the electron to move it to the third excited state is  -1.153 eV

For fourth excited state, n = 5

E₅ = E₁ /5²

E₄ = (1.23 eV) / 25

E₄ = 0.049 eV

Change in energy level, = E₅ - E₁ = 0.049 eV - 1.23 eV = -1.181 eV

The energy that must be added to the electron to move it to the fourth excited state is  -1.181 eV

5 0
3 years ago
I need help, please answer
Burka [1]

This being a perfect collision means no energy is lost during the collision. Because this question asks for speed and not velocity, the speed will be the same because the final energy is the same. The speed after the collision would therefore be 1.27 m/s.

7 0
3 years ago
Caleb rides his horse with a constant speed of 12 miles per hour. How far can she travel in 3 1/2 hours?
baherus [9]
12 × 3= 36
12 × 1/2 = 6
36+ 6= 42
6 0
3 years ago
Three packing crates of masses, M1 = 6 kg, M2 = 2 kg and M3 = 8 kg are connected by a light string of negligible mass that passe
never [62]

Answer:

39.81 N

Explanation:

I attached an image of the free body diagrams I drew of crate #1 and #2.  

Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.

∑Fₓ = maₓ

∑Fᵧ = maᵧ

Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:

  • ∑Fᵧ = maᵧ  
  • T₁ - m₂g = m₂aᵧ

Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.

Let's solve for T in the equation...

  • T₁ = m₂aᵧ + m₂g
  • T₁ = m₂(a + g)

We'll come back to this equation later. Now let's go to the free body diagram for crate #1.

We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.

  • ∑Fₓ = maₓ
  • F_f - F_g sinΘ = maₓ

The normal force is equal to the x-component of the force of gravity.

  • (F_n · μ_k) - m₁g sinΘ = m₁aₓ
  • (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
  • [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
  • [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
  • [2.539595871] - [-58.0962595] = 6aₓ
  • 60.63585537 = 6aₓ
  • aₓ = 10.1059759 m/s²

Now let's go back to this equation:

  • T₁ = m₂(a + g)  

We have 3 known variables and we can solve for the tension force.

  • T = 2(10.1059759 + 9.8)
  • T = 2(19.9059759)
  • T = 39.8119518 N

The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.

3 0
2 years ago
A block is released from rest, at a height h, and allowed to slide down an inclined plane. There is friction on the plane. At th
GREYUIT [131]

Here the block has two work done on it

1. Work done by gravity

2. Work done by friction force

So here it start from height "h" and then again raise to height hA after compressing the spring

So work done by the gravity is given as

W_g = m_A g(h - h_A)

Now work done by the friction force is to be calculated by finding total path length because friction force is a non conservative force and its work depends on total path

W_f = -(\mu m_A g cos\theta)(\frac{h}{sin\theta} + \frac{h_A}{sin\theta})

W_f = -\mu m_A g cot\theta(h + h_A)

Total work done on it

W = m_A g(h - h_A) - \mu m_A g cot\theta(h + h_A)

So answer will be

None of these

7 0
3 years ago
Read 2 more answers
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