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3 years ago

7

An argument against the validity of evolutionary theory says that there are too few transitional forms representing the gap betw

een any two living species that supposedly share a common ancestor. How do the human evolutionary record and this criticism relate?

1 answer:

ICE Princess25 [194]3 years ago

6 0

Answer:They both relate in that almost every specie in the pre-mordern human fossil record can be deregarded as a transitional specie between the African Ape and the modern man.

Explanation:

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A child on a swing sweeps out a distance of 45 ft on the first pass. If she is allowed to continue swinging until she stops, an

earnstyle [38]

Answer:

d = 90 ft

Explanation:

Here in each swing the distance sweeps by the swing is half of the initial distance that it will move

So here we can say that total distance in whole motion is given as

$d = 45 + \frac{45}{2} + \frac{45}{4} + \frac{45}{8}..........$

since it is half of the distance that it will move in each swing so it would be a geometric progression with common ratio of 1/2

so sum of such GP is given by the formula

$S = \frac{a}{1 - r}$

$d = \frac{45}{1 - \frac{1}{2}}$

$d = 90 ft$

6 0

4 years ago

On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo

ludmilkaskok [199]

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

$\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i$

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

$\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i$

42 m/s, negative direction (to the west).

5 0

3 years ago

What scientific discovery did Daniel Bernoulli work on with Leonhard Euler?

Dmitriy789 [7]

He worked with Euler on elasticity and the development of the Euler-Bernoulli beam equation.

8 0

3 years ago

What is the vertical acceleration of a paper airplane that is launched horizontally with an initial velocity of 2.3 m/s?

mr_godi [17]

Answer:

<em>The vertical acceleration is -9.81 m/s^2</em>

Explanation:

When a body is projected, the body experiences an acceleration in the vertical axis that is proportional to the acceleration due to gravity of the earth, which is equal to 9.81 m/s^2. In this case, the acceleration acts to stop the vertical motion of the paper plane, and hence is a deceleration, which explains the negative vertical acceleration.

8 0

4 years ago

A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t

Arisa [49]

A) $2.4\cdot 10^{-16}kg$

The radius of the oil droplet is half of its diameter:

$r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m$

Assuming the droplet is spherical, its volume is given by

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3$

The density of the droplet is

$\rho=885 kg/m^3$

Therefore, the mass of the droplet is equal to the product between volume and density:

$m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg$

B) $1.5\cdot 10^{-18}C$

The potential difference across the electrodes is

$V=17.8 V$

and the distance between the plates is

$d=11 mm=0.011 m$

So the electric field between the electrodes is

$E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m$

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

$qE=mg$

So, from this equation, we can find the charge of the droplet:

$q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C$

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

$q=-1.5\cdot 10^{-18}C$

we can think this charge has made of N excess electrons, so the net charge is given by

$q=Ne$

where

$e=-1.6\cdot 10^{-19}C$ is the charge of each electron

Re-arranging the equation for N, we find:

$N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9$

so, a surplus of 9 electrons.

3 0

3 years ago