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3 years ago

Your diagram is a muscle that helps you breathe. do you think this muscle is voluntary or involuntary? Explain.

Physics

1 answer:

PIT_PIT [208]3 years ago

3 0

The diaphragm is primarily an involuntary muscle. Although voluntary means can be achieved, such as when you hold your breath while swimming, this is temporary as the diaphragm will eventually work on its own to supply your body with oxygen.

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The rotational inertia of a collapsing spinning star changes to 2/5 its initial value. what is the ratio of the new rotational k

Alexandra [31]

Let K.E $_{i}$ be the intial K.E and K.E $_{new}$ be new K.E,

I $_{new}$ = $\frac{2}{5} I _{i}$

K.E $_{i} = \frac{I_{i} w^{2} }{2}$ --- (i)
K.E $_{new} = \frac{I_{new}w^{2} }{2}$

Therefore,
K.E $_{new} = \frac{2I_{i}w^{2} }{5*2}$ ----(ii)

Dividing i and ii,

$\frac{K.E_{new} }{K.E_{i} } = \frac{2}{5}$

4 0

3 years ago

A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 8.1 m from this surface, the potenti

igomit [66]

Answer:

The radius of the sphere is 4.05 m

Explanation:

Given;

potential at surface, $V_s$ = 450 V

potential at radial distance, $V_r$ = 150

radial distance, l = 8.1 m

Apply Coulomb's law of electrostatic force;

$V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}$

$450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r} \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m$

Therefore, the radius of the sphere is 4.05 m

4 0

2 years ago

List the color of the stars from hottest to coldest

masha68 [24]

Blue, blue-white, Yellow-white, yellow, yellow-orange, and red. Hope this helps.

8 0

3 years ago

1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of

laiz [17]

(a) The maximum height reached by the ball from the ground level is 75.87m

(b) The time taken for the ball to return to the elevator floor is 2.21 s

<u>The given parameters include:</u>

constant velocity of the elevator, u₁ = 10 m/s

initial velocity of the ball, u₂ = 20 m/s

height of the boy above the elevator floor, h₁ = 2 m

height of the elevator above the ground, h₂ = 28 m

To calculate:

(a) the maximum height of the projectile

total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)

at maximum height the final velocity of the projectile (ball), v = 0

Apply the following kinematic equation to determine the maximum height of the projectile.

$v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m$

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection

h = h₁ + h₂ + h₃

h = 28 m + 2 m + 45.87 m

h = 75.87 m

(b) The time taken for the ball to return to the elevator floor

Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m

Apply the following kinematic equation to determine the time to return to the elevator floor.

$h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s$

To learn more about projectile calculations please visit: brainly.com/question/14083704

6 0

3 years ago

In his novel From the Earth to the Moon (1866), Jules Verne describes a spaceship that is blasted out of 12,000 yards/s. the Col

saw5 [17]

Answer:

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

Explanation:

Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration ( $a$ ), measured in meters per square second, is estimated by this kinematic formula:

$a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s }$ (1)

Where:

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ , $v$ - Initial and final speeds of the spaceship, measured in meters.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v = 10968\,\frac{m}{s}$ and $\Delta s = 212.8\,m$ , then the acceleration experimented by the spaceship is:

$a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}$

$a = 282652.782\,\frac{m}{s^{2}}$

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

5 0

3 years ago