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VladimirAG [237]
3 years ago
8

Match the nitrogenous base of RNA with its complement. TTACCGG TAACGCA CGCATGT TGACCTA GCCAATT ACTGGAT arrowRight GCGTACA arrowR

ight CGGTTAA arrowRight AATGGCC arrowRight ATTGCGT arrowRight

Physics
thunderdus
2 years ago
Thanks this is extremely helpful
3 answers:
Ksju [112]3 years ago
4 0

Answer: The nitrogenous bases are complementary to each other which means the complementary bases will bind with each other. Example: Adenine will bind with Thymine and Guanine will bind to Cytosine.

Explanation:

ACTGGAT→    TGACCTA

GCGTACA→   CGCATGT

CGGTTAA→   GCCAATT

AATGGCC→   TTACCGG

ATTGCGT→   TAACGCA

This is how the nitrogenous bases are paired in DNA(deoxyribonucleic acid) which makes one individual different from other.

egoroff_w [7]3 years ago
3 0

I took the test and this is the answer key showing I got it right..

Hope this helps!

thunderdus2 years ago
0 0

this is extremly helpfull thanks

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Determine the inductive reactance for a 50 mH inductor that is across a 15 volt, 400 Hz source.
tresset_1 [31]

Answer:

Inductive reactance is 125.7 Ω

Explanation:

It is given that,

Inductance, L=50\ mH=0.05\ H

Voltage source, V = 15 volt

Frequency, f = 400 Hz

The inductive reactance of the circuit is equivalent to the impedance. It opposes the flow of electric current throughout the circuit. It is given by :

X_L=2\pi fL

X_L=2\pi \times 400\ Hz\times 0.05\ H

X_L=125.66\ \Omega

X_L=125.7\ \Omega

So, the inductive reactance is 125.7 Ω. Hence, this is the required solution.

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3 years ago
What motion makes the acceleration unchanged?
Rashid [163]

Answer:

The acceleration is equal to the net force divided by the mass. If the net force acting on an object doubles, its acceleration is doubled. If the mass is doubled, then acceleration will be halved. If both the net force and the mass are doubled, the acceleration will be unchanged.

7 0
2 years ago
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The LR5 is the specialist submarine for underwater rescue. The average density of seawater is 1028 kg/ m3.
sladkih [1.3K]

Answer:

P = 7196 [kPa]

Explanation:

We can solve this problem using the expression that defines the pressure depending on the height of water column.

P = dens*g*h

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Answer:

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A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
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