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Mrrafil [7]
3 years ago
7

What is the significance of the similar number and arrangement of bones in human arm and a bat wing

Physics
1 answer:
IRISSAK [1]3 years ago
6 0
The significance of the arrangement of bones in a human arm and a bat wing is that humans and bats share a common ancestor.
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I love buying physics toys. I recently broke out my new electromagnetic field meter and started playing with it. After turning i
harkovskaia [24]

Answer:

ionized particles from the sun.

* interactions in radiation belts.

* the friction of the planet in the solar wind

q = +9 10⁵ C

Explanation:

Due to being made up of matter, the planet Earth has a series of positive and negative charges, in general these charges should be balanced and the net charge of the planet should be zero, but there are several phenomena that introduce unbalanced charges, for example:

* ionized particles from the sun.

* interactions in radiation belts.

* the friction of the planet in the solar wind

This creates that the planet has a net electrical load

         

We can roughly calculate the charge of the planet

             E = k q / r²

             q = E r² / k

let's calculate

             q = 200 (6.37 10⁶)²/9 10⁹

              q = +9 10⁵ C

3 0
3 years ago
Calculate the hydrostatic difference in blood pressure between the brain and the foot in a person of height 1.93 m. The density
Slav-nsk [51]

Answer:

Explanation:

Given: Density of blood = 1.03 × 10³ Kg/m³, Height =  1.93 m g = 9.8 m/s²

pressure at the brain is equal to atmospheric pressure. = Hydro-static

pressure(ρ₀)

∴ pressure of the foot = pressure of the brain(ρ₀) + ( density of blood × acceleration due to gravity × height)(ρgh)

Hydro-static pressure = pressure at the feet- pressure at the brain(ρ₀)

Hydro-static pressure (Δp) = (ρgh + ρ₀) - ρ₀ = ρgh

Hydro-static pressure = 1.03 × 10³ × 9.8 × 1.93 = 1.948 × 10⁴ Pa

∴  Hydro-static pressure ≈ 1.95 × 10⁴ Pa

3 0
3 years ago
Tomorrow i have a science test, could someone please fill me in on the water cycle?
Lorico [155]

the cycle of processes by which water circulates between the earth's oceans, atmosphere and land involving precipitation as rain and snow drainage in streams and rivers and return to the atmosphere by evaporation and transpiration.

Hope this gives you a little bit more information!

5 0
4 years ago
what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame
Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

      riders.

      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

5 0
3 years ago
A weightlifter liftsa 1,250-N barbell 2 m in 3 s. how much power was used to lift the barbell?
STatiana [176]
Power = Force * Distance/ time
P = 1,250 * 2/3
P = 2,500/3
P = 833.33 Watts

So, your final answer is 833.33 Watts
5 0
4 years ago
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