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3 years ago

13

Why don't gorilla evolve to human if scientist was about to said that human evolve from gorilla to human?

1 answer:

Zanzabum3 years ago

7 0

Answer:

"Man's mind is far too advanced for his survival needs."

See also "Wallace", Darwin's junior partner who couldn't explain the gap in intelligence.

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Describe how a battery works

Whitepunk [10]

Magic, Nah im just kidding. A battery has two parts, the anode and the cathode. Which anode is positive and cathode is negative, which they are connected to the electrolyte. Once they are connected to a device they once start working from separate ends. Which is the flow of energy

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3 years ago

A procedure called______is used to find the volume of a rock

KengaRu [80]

Displacement is usually how the messure the rock

8 0

3 years ago

Read 2 more answers

I need help with this

lisabon 2012 [21]

A. Condensation

Hope this helps!!!

5 0

2 years ago

The midpoint M of a guitar string is pulled a distance d = 1.7 mm from equilibrium and released. Point M is observed to undergo

nlexa [21]

Answer:

ω = 380π rad/s

Explanation:

The formula for the angular frequency is the oscillation frequency f (hertz) multiplied by 2π

ω = 2πf

then

ω = 2π(190)

ω = 380π rad/s

8 0

3 years ago

A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude

Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$ .

Now, let us call $\lambda$ the charge per unit length, then we know that

$dQ = \lambda Rd\theta$ ;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda }{R}.$

Now, we know that

$\lambda = 3.0*10^{-9}C/m,$

$k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},$

and the radius of the semicircle is

$\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };$

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9}) }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0

3 years ago