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4 years ago

6

Some life forms move so slowly that their movement cannot be detected by normal observation. The best way to determine any movem

ent is by placing a mark at some point, and observing any change in position from it. Why is this true?

2 answers:

romanna [79]4 years ago

5 0

Not sure the precise concept of "normal observation", but I assume that is observed by "eyes".

Eye observation is basically macroscopic, but when you use a mark, which can be regarded as a point of mass, then it goes to microscopic.

Mark is a reference point which you can compare the relative position change, but with your eyes, first you cannot notice microscopic changes, second the eyes cannot precisely set a stable reference point.

kirza4 [7]4 years ago

3 0

The mark is a reference point.

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3 years ago

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Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of

Ratling [72]

Answer:

6.97 E 16

Explanation:

Frequency is a function of velocity of light to it's wavelength.

Mathematically written as

F = Velocity / wavelength

Velocity of light = 3 x 10^8

Wavelength =430 nm =430 x 10^-9 m

converting wavelength from nanaometer to meter we divide by 10^9

Frequency = (3 x 10^8)/(430 x10^-9) =6.97 E 16

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3 years ago

A tennis ball is dropped from a height of 10.0 m. It rebounds off the floor and comes up to a height of only 4.00 m on its first

Dominik [7]

Answer:

a) $V=14.01 m/s$

b) $V=8.86 \, m/s$

c) $t = 2.33s$

Explanation:

Our most valuable tool in solving this problem will be the conservation of mechanical energy:

$E_m = E_k +E_p$

That is, mechanical energy is equal to the sum of potential and kinetic energy, and the value of this $E_m$ mechanical energy will remain constant. (as long as there is no dissipation)

For a point particle, we have that kinetic energy is:

$E_k = \frac{1}{2} m \, V^2$

Where m is the mass, and V is the particle's velocity,

Potential energy on the other hand is:

$E_p= m\, g\, h$

where g is the acceleration due to gravity ( $g=9.81 \, m/s^2$ ) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which $h=0$ , a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.

Let's calculate the mechanical energy just before the ball is dropped:

As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:

$E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m$

That's it, the actual value of m is not important now, as we will see.

Now, what's the potential energy at the bottom? Let's see:

At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set $h=0$

$E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2$

So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:

$E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\$

And so we have found the velocity of the ball as it hits the floor.

$V = \sqrt[]{2g\cdot 10m}=14.01\, m/s$

Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:

$V = \sqrt[]{2g\cdot h}$

where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:

$V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s$

Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.

As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average? It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:

$V_{avg}\cdot t=h\\t=h/V_{avg}$

what's the average speed when the ball is descending?

$V_{avg}=\frac{1}{2} (14.01\, m/s+0)=7 \, m/s$

so the time it takes the ball to go down is:

$t=h/V_{avg}=\frac{10m}{7m/s} =1.43s\\$

Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:

$V_{avg}=\frac{1}{2} (8.86\, m/s+0)=4.43 \, m/s$

and the time it takes to go up is: $t=h/V_{avg}=\frac{4m}{4.43m/s} =0.90s$

When we add both times , we get:

$t_{total}=t_{down}+t_{up}=1.43s+0.90s = 2.33s$

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3 years ago

What happens to the velocity of an object when balanced forces act on it

mixas84 [53]

Nothing happens to velocity at all. Speed and direction remain constant.

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3 years ago

You are in a contest with your friends to see who can drive a golf ball the farthest. should you hit a "line drive" (low to the

WITCHER [35]

Line drive would be more farthest from the very high angle shot because when we increase the angle of flight the range started to increase and at some point it becomes maximum and that angle is 45° after that as you go on increasing the angle it won't be covering more distance as compared to the max at (45°) .

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3 years ago