Something that claims to be science but is not is called

Which of the following is an example of an even population distribution?

icang [17]

Even population distribution refers to a type of population distribution in which the arrangement is done in such a way that the distance between neighboring individual is maximized and uniform. This type of population distribution is usually found on the farm land, where the space in which the crops are planted have been carefully measured out. Thus an example of even population distribution is corn planted in a field.

8 0

4 years ago

car company wants to build a wind- powered car that converts 100 percent of the mechanical energy in the wind to the mechanical

navik [9.2K]

EVEN IF they can build such a machine, it's not too useful.

-- If the wind starts and stops, your car would do the same thing.

-- If the wind isn't blowing at all, your car is going nowhere. (and fast)

-- You could never move along the road faster than the wind is moving along the road.

-- You could never move in the direction towards where the wind is coming from. This has been proven before, with the technological marvel known as the "sailboat".

Call your broker immediately. Tell him you do NOT want to buy any stock in CarCompany.

3 0

3 years ago

Read 2 more answers

A bicycle rider travels 50.0 km in 2 and a half hours.what is the cyclists average speed

kakasveta [241]

25 is the answer enjoy 25

4 0

3 years ago

Read 2 more answers

John runs 120 meters in 10 seconds and then runs back to where he started in another 10 seconds. Which statement is true? Rememb

atroni [7]

Answer:

B

Explanation:

A. His speed is 0 m/s

B. His velocity is 12 m/s

C. His velocity is 0 m/s

D. His acceleration is 12 m/s

Total distance traveled by John = 120 + 120 = 240 meters

Total time taken by John to cover the distance = 10 + 10 = 20 s

Average speed of John = total distance traveled/total time taken

= 240/20 = 12 m/s

Hence, the average speed/velocity of John throughout the journey is 12 m/s.

The correct option is B.

4 0

3 years ago

a particle is moving with shm of period 8.0s and amplitude 5.0cm. find (a) the speed of particle when it is 3.0m from the centre

Fudgin [204]

Answer:

a) $speed=\pi cm/s$

b) $v_{max}=\frac{5\pi}{4} cm/s$

c) $a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

Explanation:

The very first thing we must do in order to solve this problem is to find an equation for the simple harmonic motion of the given particle. Simple harmonic motion can be modeled with the following formula:

$y=Asin(\omega t)$

where:

A=amplitude

$\omega$ = angular frequency

t=time

we know the amplitude is:

A=5.0cm

and the angular frequency can be found by using the following formula:

$\omega=\frac{2\pi}{T}$

so our angular frequency is:

$\omega=\frac{2\pi}{8s}$

$\omega=\frac{\pi}{4}$

so now we can build our equation:

$y=5sin(\frac{\pi}{4} t)$

we need to find the speed of the particle when it is 3m from the centre of its motion, so we need to find the time t when this will happen. We can use the equation we just found to get this value:

$y=5sin(\frac{\pi}{4} t)$

$3=5sin(\frac{\pi}{4} t)$

so we solve for t:

$sin(\frac{\pi}{4} t)=\frac{3}{5}$

$\frac{\pi}{4} t=sin^{-1}(\frac{3}{5})$

$t=\frac{4}{\pi}sin^{-1}(\frac{3}{5})$

you can directly use this expression as the time or its decimal representation:

t=0.81933

since we need to find the speed of the particle at that time, we will need to get the derivative of the equation that represents the particle's position, so we get:

$y=5sin(\frac{\pi}{4} t)$

$y'=5cos(\frac{\pi}{4} t)*\frac{\pi}{4}$

which simplifies to:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

and we can now substitute the t-value we found previously, so we get:

$y'=\frac{5\pi}{4}cos(\frac{\pi}{4} (0.81933))$

$y'=\pi$

so its velocity at that point is $\pi$ cm/s

b) In order to find the maximum velocity we just need to take a look at the velocity equation we just found:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

its amplitude will always give us the maximum velocity of the particle, so in this case the amplitude is:

$A=\frac{5\pi}{4}$

so:

$v_{max}=\frac{5\pi}{4} cm/s$

c) we can use a similar procedure to find the maximum acceleration of the particle, we just need to find the derivative of the velocity equation and determine its amplitude. So we get:

$y'= \frac{5\pi}{4}cos(\frac{\pi}{4} t)$