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3 years ago

Why do you need to breath faster at the top of a mountain

1 answer:

6 0

Because of the higher elevation. As much as we go away from earth by increasing altitude, there will occur lacoccuren. due to this, we need to breathe fast to get suffient oxygen by time at top of the mountain. We breathe faster at the top of a tall mountain due to a natural reflex. As we go higher up a mountain, the density of air and oxygen decreases, i.e. the amount of oxygen in the air decreases along with other gases. Therefore, to maintain the regular supply of oxygen, the body breathes faster. If we go too high without oxygen tanks, we could fall unconscious and die within 8 minutes due to oxygen deprivation.

Hope this helps! let me know if you need anything else!!

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The answer can't be D because I got it wrong when I took my test so it has to be A,B,or C

Cerrena [4.2K]

To what question please put the question ill be glad to help

5 0

3 years ago

Ben walks 500 meters from his house to the corner store. He then walks back toward his house but stops to talk to a neighbor whe

spin [16.1K]

Average velocity =

(displacement) / (time for the displacement)
and
(direction of the displacement) .

Displacement =

(distance from the start-point to the end-point)
and
    (direction from the start-point to the end-point) .

When Ben is 200 meters from the corner store,
he is (500 - 200) = 300 meters from his house.

His displacement is

   300 meters in the direction
   from his house to the neighbor .

His average velocity is

   (300/910) = 0.33 meters per second, in the
   direction from his house to the neighbor .

7 0

3 years ago

Read 2 more answers

A small piece of Styrofoam packing material is dropped from a height of 2.60 m above the ground. Until it reaches terminal speed

Vadim26 [7]

Answer:

Explanation:

a ) After the attainment of terminal speed , object takes 4.5 s to cover a distance of 2 m

So terminal speed V = 2 / 4.5

= .444 m /s

When it attains terminal speed , acceleration becomes zero

0 = g - B x .444

B = 22.25 s⁻¹

b ) At t = 0 , v = 0

a = g - B v

a = g at t = 0

c ) When v = .15

a = g - 22.25 x .15

= 9.8 - 3.31

= 6.5 m /s²

7 0

3 years ago

A classmate is attempting to draw a free body diagram for a box being pushed across the floor at a constant speed. The image bel

aivan3 [116]

Answer:

A. The applied force should be the same size as the friction force

Explanation:

Whenever we apply a force to an object it moves if the force applied to that object is unbalanced and there is no force or a lesser force to counter it. According to Newton's Second Law of motion, when an unbalanced force is applied to an object it produces an acceleration in the object in its own direction. So, the two forces acting on this box are the frictional force and the applied force in horizontal direction. In order to move the box at constant speed, the applied force must first, overcome the frictional force, so the object can start its motion. Since, the motion has constant velocity, it means no acceleration. So, the force must be balanced in order to avoid acceleration as a consequence of Newton's Second Law of motion. Therefore, the correction in this case will be:

<u>A. The applied force should be the same size as the friction force</u>

7 0

2 years ago

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu

Alekssandra [29.7K]

Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰, force = 4.07 N

(b) When the angle, θ = 90 ⁰, force = 4.7 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;

$F = BIl \ sin(\theta)$

(a) When the angle, θ = 60 ⁰

$F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N$

(b) When the angle, θ = 90 ⁰

$F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N$

$F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N$

6 0

2 years ago