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serious [3.7K]
3 years ago
6

The cells lie odjacent to the sieve tubes​

Physics
1 answer:
nignag [31]3 years ago
4 0

Answer:

<h2><em>Almost</em><em> </em><em>always</em><em> </em><em>adjacent</em><em> </em><em>to</em><em> </em><em>nucleus</em><em> </em><em>containing</em><em> </em><em>companion</em><em> </em><em>cells</em><em>,</em><em> </em><em>which</em><em> </em><em>have</em><em> </em><em>been</em><em> </em><em>produced</em><em> </em><em>as</em><em> </em><em>sister</em><em> </em><em>cells</em><em> </em><em>with</em><em> </em><em>the</em><em> </em><em>sieve</em><em> </em><em>elements</em><em> </em><em>from</em><em> </em><em>the</em><em> </em><em>same</em><em> </em><em>mother</em><em> </em><em>cell</em><em>.</em><em> </em></h2>
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How many grams of CO-60 result in 1 Millicuire of activity? How many years until the activity decays to 1 microcuire tl/2 =5.3 Y
vredina [299]

Answer:

m =8.81*10^{-6}grams

time t = 52.8 year

Explanation:

GIVEN DATA:

the half life of the CO-60 is, T_1/2 = 5.27 years = 1.663 e+8 s

activity dN/dt = 1 mCi = 3.7 X 10^7 decay/s

activity ,  dN/dt = N* \lambda

              \frac{dN}{dt} = N* \frac{ln2}{T_1/2}

                   N = \frac{(dN/dt )(T1/2)}{ln2}

                      = ( 3.7 X 10^7 )(1.663*10^8 ) / ln2

                      = 8.877*10^{16}

Number of moles:

     n = N/NA = 8.877*10^{16} / 6.022X10^23 = 1.474*10^{-7} mol

mass of the CO-60 is,

   m = n*M = [1.474*10^{-7} mol]*[59.93 grams /mol] = 8.81*10^{-6}grams

-----------------------------------------------------------------------------------------

    time t = -[T1/2 / ln2]*ln[N/N0]

             = - [5.3 years / ln2]*ln[1x10-6/1x10-3]

             = 52.8 year

8 0
4 years ago
The toy car is about 3 inches long is an example of a ?
balandron [24]

Answer:

a quantitative observation because it includes numerical data

8 0
3 years ago
A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, q=3\times 10^{-6}\ C

Mass of charge particle, m=2\times 10^{-6}\ C

Speed, v=5\times 10^{6}\ m/s

Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

ma=qvB\ sin\theta

For minimum magnetic field,

ma=qvB

B=\dfrac{ma}{qv}

B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0
3 years ago
The low-pressure area near Earth’s equator is filled by cool air moving in from ________. Btw this is science
bonufazy [111]

Answer:

the north and south pole

Explanation:

this should be the correct answer

5 0
2 years ago
Determine the weight of an average physical science textbook whose mass is 3.1 kilograms. The acceleration due to gravity is 9.8
zvonat [6]
Weight equals mass*gravity
W = mg

Given m = 3.1 kg, g = 9.8 m/s^2

W = (3.1)(9.8)
W = 30.38
6 0
3 years ago
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