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Inessa [10]
2 years ago
5

What inflammatory molecule aids with muscle building and is released when muscles experience microscopic damage?

Physics
1 answer:
Reika [66]2 years ago
3 0

Answer:

Cytokines

Explanation:

Cytokines are known as inflammatory molecules which are also proteinous and aid signaling of certain processes and conditions in the body.

They are also normally involved in aiding muscle building and are released when muscles experience microscopic damage which may lead to the muscles being sore.

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A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and
irina [24]

The eccentricity of its orbit is $$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

<h3>What is mass?</h3>
  • Mass is a physical body's total amount of matter. It also serves as a gauge for the body's inertia or resistance to acceleration (change in velocity) in the presence of a net force. The strength of an object's gravitational pull to other bodies is also influenced by its mass.
  • The kilogram is the SI unit of mass (kg). In science and technology, a body's weight in a given reference frame is the force that causes it to accelerate at a rate equal to the local acceleration of free fall in that frame.
  • For instance, a kilogram mass weighs around 2.2 pounds at the surface of the planet. However, the same kilogram mass would weigh just about 0.8 pounds on Mars and about 5.5 pounds on Jupiter.
  • An object's mass is a crucial indicator of how much stuff it contains. Weight is a measurement of an object's gravitational pull. It is influenced by the object's location in addition to its mass. As a result, weight is a measurement of force.

The length of the semi-major axis is calculated as follows:

where, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$

$M=1.99 \times 10^{30} \mathrm{~kg}=$mass of sur

$m=1.20 \times 10^{10} \mathrm{~kg}$ - a mass of the comet

$$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

$$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

To learn more about mass, refer to:

brainly.com/question/3187640

#SPJ4

4 0
1 year ago
Which of the following is an example if harmonic motion?
german

Answer:

"A pendulum swinging back and forth" is an example of harmonic motion

X = Xo cos ω t

Explains the back and forth motion of the pendulum

3 0
2 years ago
While standing on a balcony a child drops a penny. The penny lands on the ground floor 1.5 s later. How fast was the penny trave
sveticcg [70]

Answer:

14.7 m/s.

Explanation:

From the question given above, the following data were obtained:

Time (t) = 1.5 s

Acceleration due to gravity (g) = 9.8 m/s².

Height = 11.025 m

Final velocity (v) = 0 m/s

Initial velocity (u) =?

We, can obtain the initial velocity of the penny as follow:

H = ½(v + u) t

11.025 = ½ (0 + u) × 1.5

11.025 = ½ × u × 1.5

11.025 = u × 0.75

Divide both side by 0.75

u = 11.025/0.75

u = 14.7 m/s

Therefore, the penny was travelling at 14.7 m/s before hitting the ground.

8 0
3 years ago
An airplane engine starts from rest; and 2 seconds later, it is rotating with an angular speed of 300 rev/min. if the angular ac
elena-s [515]

First of all, we need to convert the angular speed from rev/min into rev/s:

\omega_f=300 rev/min=5 rev/s

The angular acceleration is the variation of angular speed divided by the time:

\alpha=\frac{\omega_f-\omega_i}{t}=\frac{5 rev/s-0}{2 s}=2.5 rev/s^2

And this is constant, so we can use the following equation to calculate the angle through which the engine has rotated:

\theta(t)=\frac{1}{2}\alpha t^2 =\frac{1}{2}(2.5 rev/s^2)(2 s)^2=5 rev

so, 5 revolutions.

3 0
2 years ago
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm . The explorer finds that
larisa86 [58]

Answer:

12.4 m/s²

Explanation:

L = length of the simple pendulum = 53 cm = 0.53 m

n = Number of full swing cycles = 99.0

t = Total time taken = 128 s

T = Time period of the pendulum

g = magnitude of gravitational acceleration on the planet

Time period of the pendulum is given as

T = \frac{t}{n}

T = \frac{128}{99}

T = 1.3 sec

Time period of the pendulum is also given as

T = 2\pi \sqrt{\frac{L}{g}}

1.3 = 2(3.14) \sqrt{\frac{0.53}{g}}

g = 12.4 m/s²

4 0
3 years ago
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