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3 years ago

13

The theory of light has been expounded differently by scientists in different periods of time. What is an important reason for t

hese different theories?

1 answer:

wlad13 [49]3 years ago

8 0

Technological advances clearly.

Theories change due to other peoples view on the matter and how they experiment with it, but Technology advancements are the greatest factor of all. I mean seriously, we wouldn't know of cells before the microscope and when they first invented it, it advanced further and now we can see things on a whole deeper lever.

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I will give brainliest if right

e-lub [12.9K]

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Prokaryotic Cells are cells that lack Cell Nucleus so the answer is bacteria

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Read 2 more answers

Which has the same meaning as air pressure?

SVETLANKA909090 [29]

B) atmosphere pressure, i believe

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3 years ago

17. (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106

sleet_krkn [62]

Answer:

Explanation:

Distance between plates d = 2 x 10⁻³m

Potential diff applied = 5 x 10³ V

Electric field = Potential diff applied / d

= 5 x 10³ / 2 x 10⁻³

= 2.5 x 10⁶ V/m

This is less than breakdown strength for air 3.0×10⁶ V/m

b ) Let the plates be at a separation of d .so

5 x 10³ / d = 3.0×10⁶ ( break down voltage )

d = 5 x 10³ / 3.0×10⁶

= 1.67 x 10⁻³ m

= 1.67 mm.

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3 years ago

Which positions experience low tide

MrMuchimi

Answer:

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3 years ago

The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

irina [24]

Part a)

At t = 0 the position of the object is given as

$x = 0$

At t = 2

$x = 2 sin(\pi/2) = 2cm$

so displacement of the object is given as

$d = 2 - 0 = 2cm$

so average speed is given as

$v_{avg} = \frac{2}{2} = 1 cm/s$

Part b)

instantaneous speed is given by

$v = \frac{dy}{dt}$

$v = 2cos(\pi t/4 ) * \frac{\pi}{4}$

now at t= 0

$v = \frac{\pi}{2} cm/s$

at t = 1

$v = 2 cos(\pi/4) * \frac{\pi}{4}$

$v = \frac{\pi}{2\sqrt2}$

at t = 2

$v = 0$

Part c)

Average acceleration is given as

$a_{avg} = \frac{v_f - v_i}{t}$

$a_{avg} = \frac{0 - \frac{\pi}{2}}{2}$

$a = -\frac{\pi}{4} cm/s^2$

Part d)

Now for instantaneous acceleration

As we know that

$a =- \omega^2 y$

at t = 0

$a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2$

at t = 1

$y = \sqrt2 cm$

now we have

$a = -\frac{\pi^2}{16}*\sqrt2$

At t = 2 we have

$y = 2 cm$

$a = -\frac{\pi^2}{16}*2$

$a = -\frac{\pi^2}{8}$

<em>so above is the instantaneous accelerations</em>

7 0

3 years ago