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2 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!!

You might be interested in

Two protons are released from rest, with only the electrostatic force acting. Which of the following statements must be true abo

ycow [4]

Answer:

(A)

Explanation:

We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.

Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.

7 0

3 years ago

How many hydrogen atoms are in a water molecule?

earnstyle [38]

The subscript after the element indicates the number of atoms of that element in the molecule. So, in H20, the subscript after the H, which stands for hydrogen, is 2. This means that there are 2 hydrogen atoms in a water molecule.

Hope this helps! :)

4 0

3 years ago

Two electric charges, held a distance, dd, apart experience an electric force of magnitude, FF, between them. If one of the char

lorasvet [3.4K]

Answer:

$F'=2F$

Explanation:

The Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:

$F=\frac{kq_1q_2}{d^2}$

In this case, we have $q_1'=2q_1$ :

$F'=\frac{kq'_1q_2}{d^2}\\F'=\frac{k(2q_1)q_2}{d^2}\\F'=2\frac{kq_1q_2}{d^2}\\F'=2F$

3 0

3 years ago

What type of acceleration does an object moving with constant speed in a circular path experience?Select one:A) constant acceler

olga_2 [115]

ANSWER:

D) centripetal acceleration.

STEP-BY-STEP EXPLANATION:

When a body performs a uniform circular motion, the direction of the velocity vector changes at every instant. This variation is experienced by the linear vector, due to a force called centripetal, directed towards the center of the circumference that gives rise to the centripetal acceleration.

Therefore, the answer is centripetal acceleration.

3 0

1 year ago

A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.

kow [346]

Answer:

The maximum height the pebble reaches is approximately;

A. 6.4 m

Explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile, $h_{max}$ , is given by the following equation;

$h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}$

Therefore, substituting the known values for the pebble, we have;

$h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439$

Therefore, the maximum height of the pebble projectile, $h_{max}$ ≈ 6.4 m.

3 0

3 years ago