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brilliants [131]
3 years ago
8

In which century was the DNA database established?

Physics
1 answer:
krok68 [10]3 years ago
8 0
DNA database was established in 1995
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the answer is B: earth takes to rotate once on its axis

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How far apart would you have to place the poles of a 1. 5 v battery to achieve the same electric field?
Zarrin [17]

To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m

The potential difference is related to the electric field by:

∆V=Ed

where,

∆V is the potential difference

E is the electric field

d is the distance

what is potential difference?

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.

We want to know the distance the detectors have to be placed in order to achieve an electric field of

E=1v/cm=100v/cm

when connected to a battery with potential difference

∆v=1.5v

Solving the equation,we find

d =  \frac{ \:Δv}{e}

=  \frac{1.5v}{100v/m}

= 1.5 \times 10 {}^{ - 2} m

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10 months ago
Which type of mirror produces images that are always upright and at the same distance from the mirror as the object is?
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3 years ago
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Two balls, one twice as massive as the other, are dropped from the roof of a building (freefall). Just before hitting the ground
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KE= 1/2 mv^2

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2 years ago
A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the
Maurinko [17]

Answer:

(a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity R_{0}=10\ mCi

Time t_{1}=4\ hours

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

R=R_{0}e^{-\lambda t}

\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})

Put the value into the formula

\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})

\lambda=0.0000154\ s^{-1}

\lambda=1.55\times10^{-5}\ s^{-1}

We need to calculate the half life

Using formula of half life

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}

Put the value into the formula

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}

T_{\dfrac{1}{2}}=44.719\times10^{3}\ s

T_{\dfrac{1}{2}}=11.3\ hr

(b). We need to calculate the value of N₀

Using formula of N_{0}

N_{0}=\dfrac{3.70\times10^{6}}{\lambda}

Put the value into the formula

N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}

N_{0}=2.38\times10^{11}\ nuclei

(c). We need to calculate the sample's activity

Using formula of activity

R=R_{0}e^{-\lambda\times t}

Put the value intyo the formula

R=10e^{-(1.55\times10^{-5}\times30\times3600)}

R=1.87\ mCi

Hence, (a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

4 0
3 years ago
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