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eduard
3 years ago
6

The part of a cell division that causes the seperation of the cytoplasm

Physics
1 answer:
KatRina [158]3 years ago
4 0
Cytokinesis is the part of a cell division that causes the separation of the cytoplasm. 

Good luck :)
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Visible light falls into wavelength ranges of 400-700 nm, for which 1 m = 1 × 10 9 nm . The energy and wavelength of light are r
Paul [167]

Answer:

E = 3.54 x 10⁻¹⁹ J

Explanation:

The energy of the photon can be given in terms of its wavelength by the use of the following formula:

E = \frac{hc}{\lambda}

where,

E = energy = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ Js

c = speed of light = 2.998 x 10⁸ m/s

λ =  wavelength of light = 560.6 nm = 5.606 x 10⁻⁷ m

Therefore,

E = \frac{(6.626\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{5.606\ x\ 10^{-7}\ m}\\

<u>E = 3.54 x 10⁻¹⁹ J</u>

7 0
3 years ago
How are types of energy in the electromagnetic spectrum defined?
Drupady [299]
It is defined by their wavelength. Different colors have different wavelengths. For example, radio waves have a really long wavelength, whereas gamma-rays have a very short wavelength.
6 0
3 years ago
Read 2 more answers
Compare and contrast the terms vaporizing and condensation.
JulsSmile [24]
The terms are both about changing states. Vaporizing is when you heat something up into a vapor; condensation is when you lower a vapors temperature to make it become a liquid state.
5 0
3 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
The cube with 2.00m wide and 2.00m long and 2.00m high has a weight of 960.00 N what pressure does it exert
andre [41]
2m X 2m = 4m ^2
960n / 4 = 240pa
3 0
3 years ago
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