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maria [59]
2 years ago
7

DON'T ADD LINKS!

Physics
1 answer:
Amanda [17]2 years ago
6 0

Two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.

<h3>How many cells are produced? </h3>

When the body cell passes through cell division process, it divides into two daughter cells. In the second cell division, two cells divides into four cells. In the third cell division, the four cells divides into eight cells. In the fourth cell division, the eight cells divides and sixteen cells are produced.

So we can conclude that two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.

Learn more about cell division here: brainly.com/question/796780

#SPJ1

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Neutrons incident on a heavy nucleus with spin J 0 show a resonance at an incident energy ER = 250 eV in the total cross-section
ivolga24 [154]

Answer:

elastic partial width is 2.49 eV

Explanation:

given data

ER  E = 250 eV

spin J = 0

cross-section magnitude σ = 1300 barns

peak P = 20ev

to find out

elastic partial width W

solution

we know here that

σ = λ²× W /  ( E × π × P )     ...................1

put here all value

σ = (0.286)² × W  × 10^{-16}/  ( 250 × π × 20 )

1300 × 10^{-24} = (0.286)² × W  × 10^{-16}/  ( 250 × π × 20 )

solve it and we get W

W = 249.56 × 10^{-2}

so elastic partial width is 2.49 eV

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4 years ago
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PLEASE HELP, THANK YOU!.. :)
Firlakuza [10]

Answer:

<u><em>First Reaction:</em></u>

^{234}U  =>  ^{230} Th + ^{4}He

<u><em>Second Reaction:</em></u>

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<u><em>Combined Reaction:</em></u>

^{234} U  => ^{226}Ra + 2( ^{4} He)

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Los anillos de Saturno están compuestos de trozos de hielo que están en órbita alrededor del planeta. El radio interior de los a
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Explanation:

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3 0
3 years ago
A circular wire loop of radius LaTeX: RR lies in the xy-plane with the z-axis running through its center. There is initially no
frosja888 [35]

Answer:

Explanation:

Given a circular loop of radius R

r = R

Note: the radius lies in the xy plane

Area is given as

A = πr² = πR²

At t = 0, no magnetic field B=0

The magnetic field is given as a function of time

B = C•exp(t) •i + D•t² •k

Where C and D are constant

We want to find the magnitude of EMF in the circular loop.

EMF is given as

ε = - N•dΦ/dt

Where,

N is number of turn and in this case we will assume N = 1.

Φ is magnetic flux and it is given as

Φ = BA

ε = - N•d(BA)/dt

Where A is a constant, then we have

ε = - N•A•dB/dt

B = C•exp(t) •i + D•t² •k

dB/dt = C•exp(t) •i + 2D•t •k

Then,

ε = - N•A•dB/dt

ε = - 1•πR²•(C•exp(t) •i + 2D•t •k)

ε = -πR²•(C•exp(t) •i + 2D•t •k)

So, let find the magnitude of EMF

Generally finding magnitude of two vectors R = a•i + b•j

Then, |R| = √a² + b²

So, applying this we have,

ε = πR² (√(C²•exp(2t) + 4D²t²))

From the given magnetic field, we are given that,

B = 0 at t = 0

B = C•exp(t) •i + D•t² •k

B = 0 = C•exp(0) •i + D•0² •k

0 = C

Then, C = 0.

So, substituting this into the EMF.

ε = πR² (√(0²•exp(2t) + 4D²t²))

ε = πR² (√4D²t²)

ε = πR² × 2Dt

ε = 2πDR²t

So, the EMF is also a function of time

ε = 2πDR²t

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