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ella [17]
3 years ago
5

What is the force the when does when Gravity pushes you​

Physics
2 answers:
BARSIC [14]3 years ago
6 0

Answer:

Image result for what is the force the when does when Gravity pushes you​

The important thing to remember is that gravity is neither a push nor a pull; what we interpret as a “force” or the acceleration due to gravity is actually the curvature of space and time — the path itself stoops downward.

Explanation:

Image result for what is the force the when does when Gravity pushes you​

The important thing to remember is that gravity is neither a push nor a pull; what we interpret as a “force” or the acceleration due to gravity is actually the curvature of space and time — the path itself stoops downward.

Nezavi [6.7K]3 years ago
5 0

Answer:

It is called gravitational force.

Explanation:

{ \boxed{ \rm{F =  \frac{GM_{a} m _{a}}{ {r}^{2} } }}}

  • F is the gravitational force
  • Ma is the mass of a planetary body
  • ma is mass lf body being attracted
  • r is the separation distance
  • G is the universal gravitational constant

→ Assume you are in space and you have a mass of ma, and you are r metres from the Earth of mass Ma. The force F will be experienced and you will tend to be pulled towards the Earth when you are in its gravitational field.

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The radius of the aorta is «10 mm and the blood flowing through it has a speed of about 300 mm/s. A capillary has a radius of ab
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Answer:

(I). The effective cross sectional area of the capillaries is 0.188 m².

(II). The approximate number of capillaries is 3.74\times10^{9}

Explanation:

Given that,

Radius of aorta = 10 mm

Speed = 300 mm/s

Radius of capillary r=4\times10^{-3}\ mm

Speed of blood v=5\times10^{-4}\ m/s

(I). We need to calculate the effective cross sectional area of the capillaries

Using continuity equation

A_{1}v_{1}=A_{2}v_{2}

Where. v₁ = speed of blood in capillarity

A₂ = area of cross section of aorta

v₂ =speed of blood in aorta

Put the value into the formula

A_{1}=A_{2}\times\dfrac{v_{2}}{v_{1}}

A_{1}=\pi\times(10\times10^{-3})^2\times\dfrac{300\times10^{-3}}{5\times10^{-4}}

A_{1}=0.188\ m^2

(II). We need to calculate the approximate number of capillaries

Using formula of area of cross section

A_{1}=N\pi r_{c}^2

N=\dfrac{A_{1}}{\pi\times r_{c}^2}

Put the value into the formula

N=\dfrac{0.188}{\pi\times(4\times10^{-6})^2}

N=3.74\times10^{9}

Hence, (I). The effective cross sectional area of the capillaries is 0.188 m².

(II). The approximate number of capillaries is 3.74\times10^{9}

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3 years ago
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