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Darya [45]
3 years ago
8

two strands of DNA produced during replication are copies of each other because each strand in the double helix is

Physics
2 answers:
Luden [163]3 years ago
6 0
The correct answer you should be looking for is complementary. :)
natulia [17]3 years ago
3 0
The answer is complementary
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A perfectly elastic collision occurs between a 15.0-kg mass traveling at 3.50 m/s and a 9.00-kg mass traveling at 2.35 m/s. if t
BaLLatris [955]
Momentum is conserved in a collision. Momentum is mass*velocity, so you can find your answer by calculating initial and final momentums and setting them equal to each other.

15kg * 3.50 m/s + 9kg * 2.35 m/s = 73.65 kg m/s

73.65 = 9 * 2.8 + 15x

solve for x
x= 3.23

The final velocity is 3.23 m/s
5 0
3 years ago
If the pressure in a gas is doubled while its volume is held constant, by what factor do vrms change
Nat2105 [25]

Answer is given below

Explanation:

given data

pressure = double

volume = constant

solution

As we know that an Average velocity and rms velocity is directly proportional to square root of PV ..................1

so if we take P is doubled while keeping V constant

than Velocity increases by a factor \sqrt{2}  

so that Factor = 1.414 for both the cases

8 0
3 years ago
Officials in the House of Representatives serve __ terms.
Salsk061 [2.6K]
Officials in the House of Representives have are able to serve a 2 year term and there are no limits as to how many terms.
7 0
3 years ago
Read 2 more answers
A bus contains a 1440 kg flywheel (a disk that has a 0.63 m radius) and has a total mass of 10200 kg. Calculate the angular velo
CaHeK987 [17]

Answer:\omega =93.51 rad/s

Explanation:

Given

mass of Flywheel m_1=1440 kg

mass of bus m_b=10200 kg

radius of Flywheel r=0.63 m

final speed of bus v=21 m/s

Conserving Energy i.e.

0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus

Let \omegabe the angular velocity of Flywheel

0.9\cdot \frac{I\omega ^2}{2}=\frac{m_bv^2}{2}

I=moment\ of\ Inertia =mr^2=1440\cdot 0.63^2=571.536 kg-m^2

0.9\cdot \frac{571.536\cdot \omega ^2}{2}=\frac{10200\cdot 21^2}{2}

\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}

\omega =21\times 4.45=93.51 rad/s

8 0
3 years ago
What is the volume V of a sample of 4.00 mol of copper? The atomic mass of copper (Cu) is 63.5 g/mol, and the density of copper
rusak2 [61]

Answer : The volume of a sample of 4.00 mol of copper is 28.5cm^3

Explanation :

First we have to calculate the mass of copper.

\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}

\text{ Mass of copper}=(4.00moles)\times (63.5g/mole)=254g

Now we have to calculate the volume of copper.

Formula used :

Density=\frac{Mass}{Volume}

Now put all the given values in this formula, we get:

8.92\times 10^3kg/m^3=\frac{254g}{Volume}

Volume=\frac{254g}{8.92\times 10^3kg/m^3}=2.85\times 10^{-2}L=2.85\times 10^{-2}\times 10^3cm^3=28.5cm^3

Conversion used :

1kg/m^3=1g/L\\\\1L=10^3cm^3

Therefore, the volume of a sample of 4.00 mol of copper is 28.5cm^3

7 0
3 years ago
Read 2 more answers
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