Ask question

Ask question

All categories

3 years ago

6

In a client with lower crossed syndrome, which of the following muscles is lengthened?

1 answer:

jek_recluse [69]3 years ago

6 0

Answer:

Internal Oblique.

Explanation:

Lower crossed syndrome is a condition in which there are strong and weak muscles. So there is an imbalance of muscle strengths. It occurs when some muscles constanly get shortened or lengthened just like in this case internal oblique muscle got lengthened.

You might be interested in

I need help on question 5

11Alexandr11 [23.1K]

Weathering, erosion, impact, pressure

3 0

3 years ago

I have a pen<br> I have a apple<br> what do I have now?

sammy [17]

Answer:

You have an apple pen. :)

7 0

3 years ago

Read 2 more answers

How do you spell hard water in three letters.

Kay [80]

Ice, which is the solid state of water, the liquid

5 0

3 years ago

Read 2 more answers

Supposed an object is weighted with a spring balance, first in air then while totally immersed in water.The readings on the bala

Sedbober [7]

Is An object completely submerged in a fluid displaces its own volume of fluid ". This is: D

8 0

2 years ago

The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I

erma4kov [3.2K]

Answer:

Part a)

$I_{end} = \frac{mL^2}{3}$

Part b)

$I_{edge} = \frac{2ma^2}{3}$

Explanation:

As we know that by parallel axis theorem we will have

$I_p = I_{cm} + Md^2$

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

$I = \frac{mL^2}{12}$

now if we need to find the inertia from its end then we will have

$I_{end} = I_{cm} + Md^2$

$I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2$

$I_{end} = \frac{mL^2}{3}$

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

$I = \frac{ma^2}{6}$

now if we need to find the inertia about an axis passing through its edge

$I_{edge} = I_{cm} + Md^2$

$I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2$

$I_{edge} = \frac{2ma^2}{3}$

7 0

3 years ago