What tool is used to measure air pressure?

What structures can be found in the axial region of the body?

Softa [21]

Answer:

The axial region of the body consists of the bones of the head, trunk of a vertebrate, skull, vertebral column, and thoracic cage. The human skeleton consists of 80 bones.

Explanation:

The axial region of the body consists of the bones of the head, trunk of a vertebrate, skull, vertebral column, and thoracic cage. The human skeleton consists of 80 bones.

It is composed of the following six parts:

1. Skull (22 bones)

2. Ossicles of the middle ear

3. Hyoid bone

4. Rib cage

5. Sternum

6. Vertebral column

The axial region of the body forms the vertical axis of the body as the axial skeleton supports the head, neck, back, and chest.

3 0

3 years ago

What are the principals of electric arc welding

Wewaii [24]

Principles of arc welding. Arc welding is a welding process, in which heat is generated by an electric arc struck between an electrode and the work piece. Electric arc is luminouselectrical discharge between two electrodes through ionized gas.

7 0

3 years ago

A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 35.0 min

Lubov Fominskaja [6]

Explanation:

Solution:

Let the time be

t1=35min = 0.58min

t2=10min=0.166min

t3=45min= 0.75min

t4=35min= 0.58min

let the velocities be

v1=100km/h

v2=55km/h

v3=35km/h

a. Determine the average speed for the trip. km/h

first we have to solve for the distance

S=s1+s2+s3

S= v1t1+v2t2+v3t3

S= 100*0.58+55*0.166+35*0.75

S=58+9.13+26.25

S=93.38km

V=S/t1+t2+t3+t4

V=93.38/0.58+0.166+0.75+0.58

V=93.38/2.076

V=44.98km/h

b. the distance is 93.38km

6 0

3 years ago

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie

leva [86]

To solve the problem it is necessary to apply conservation of the moment and conservation of energy.

By conservation of the moment we know that

$MV=mv$

Where

M=Heavier mass

V = Velocity of heavier mass

m = lighter mass

v = velocity of lighter mass

That equation in function of the velocity of heavier mass is

$V = \frac{mv}{M}$

Also we have that $m/M = 1/7 times$

On the other hand we have from law of conservation of energy that

$W_f = KE$

Where,

W_f = Work made by friction

KE = Kinetic Force

Applying this equation in heavier object.

$F_f*S = \frac{1}{2}MV^2$

$\mu M*g*S = \frac{1}{2}MV^2$

$\mu g*S = \frac{1}{2}( \frac{mv}{M})^2$

$\mu = \frac{1}{2} (\frac{1}{7}v)^2$

$\mu = \frac{1}{98}v^2$

$\mu = \frac{1}{g(98)(5.1)}v^2$

Here we can apply the law of conservation of energy for light mass, then

$\mu mgs = \frac{1}{2} mv^2$

Replacing the value of $\mu$

$\frac{1}{g(98)(5.1)}v^2 mgs = \frac{1}{2}mv^2$

Deleting constants,

$s= \frac{(98*5.1)}{2}$

$s = 249.9m$

7 0

3 years ago

A mosquito can fly with a speed of 1.10 m/s with respect to the air. Suppose a mosquito flies east at this speed across a swamp.

sineoko [7]

Answer:

The velocity of Mosquito with respect to earth will be 0.302m/s

Explanation:

V(ma) = 1.10 m/s, east Velocity of mosquito with respect to air

V(ae) = 1.4 m/s at 35° Velocity of air with respect to Earth in west of south direction.

Velocity of Mosquito with respect to earth will be

V(me) = V(ma) + V(ae)

We need to find the mosquito’s speed with respect to Earth in the x direction.

V(x, me) = V(x, ma) + V(x, ae) = V(ma) + V(ae)(cos theta(ae) )

Angle (ae) = –90.0° − 35°=−125°

V(x, me) = 1.10 + (1.4)Cos(-125)

= 1.10 + 1.4(-0.57)

= 1.10 -0.798

= 0.302

So the velocity of Mosquito with respect to earth will be 0.302m/s

7 0

2 years ago