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3 years ago

Mutations are avoided during replication because DNA polymerase is able to _____?

Physics

2 answers:

Ulleksa [173]3 years ago

3 0

D. add more codes
hope this helped

Igoryamba3 years ago

3 0

<h3>Answer</h3><h2>Repair the DNA</h2>

Meanwhile DNA construction, maximum DNA polymerases "review their performance," getting the bulk of mis-paired bases in a method called proofreading. Shortly after DNA synthesis, any left mis-paired bases can be identified and substituted in a procedure called mismatch repair. If DNA gets destroyed, it can be fixed by different mechanisms, including chemical repeal, excision restoration, and double-stranded break repair.

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What happens to a circuit's resistance (R), voltage (V), and current (1) when

hammer [34]

Answer:

R increases

V is constant

I decreases

Explanation:

The resistance of a wire is given by the following formula:

$R = \frac{(Resistivity)(L)}{A}$

It is clear from this formula that resistance is directly proportional to the length of wire. So, when length of wire is increased, the resistance of circuit increases.

The voltage in the circuit will be constant as the voltage source remains same and it is not changed.

Now, we can use Ohm Law:

V = IR

at constant V:

I ∝ 1/R

it means that current is inversely proportional to resistance. Hence, the increase of resistance causes the current in circuit to decrease.

Therefore, the correct option will be:

D.

R increases

V is constant

I decreases

6 0

2 years ago

The properties of metals depend mainly on the number and arrangement of neutrons True False

Sergeeva-Olga [200]

Answer:

False

Explanation:

Electrons

7 0

2 years ago

How do you determine the specific heat of a substance?

svetlana [45]

Answer:

I think it's A

Explanation:

3 0

2 years ago

An object is most likely to become electrically charged by gaining or losing which type of particles?

elena-14-01-66 [18.8K]

Answer:

negative particles

Explanation:

An atom can be defined as the smallest unit comprising of matter that forms all chemical elements. Thus, atoms are basically the building blocks of matters and as such defines the structure of a chemical element.

Generally, these atoms are typically made up of three distinct particles and these are protons, neutrons and electrons.

In Chemistry, electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.

Valence electrons can be defined as the number of electrons present in the outermost shell of an atom. Valence electrons are used to determine whether an atom or group of elements found in a periodic table can bond with others. Thus, this property is typically used to determine the chemical properties of elements.

Hence, an object is most likely to become electrically charged by gaining or losing negative particles.

5 0

3 years ago

A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a

Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

$Z_{R}$ = R

R = resistance

$Z_{L}$ = jωL

ω = voltage source angular frequency, L = inductance

$Z_{C}$ = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

$Z_{R}$ = 217Ω

$Z_{L}$ = j(220)(0.875) = j192.5Ω

$Z_{C}$ = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = $Z_{R}$ + $Z_{L}$ + $Z_{C}$

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

$v_{R}$ = I× $Z_{R}$

$v_{L}$ = I× $Z_{L}$

$v_{C}$ = I× $Z_{C}$

Given values:

I = (0.0569∠1.147+220t)A

$Z_{R}$ = 217Ω = (217∠0)Ω

$Z_{L}$ = j192.5Ω = (192.5∠π/2)Ω

$Z_{C}$ = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

$v_{R}$ = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

$v_{L}$ = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

$v_{C}$ = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

$v_{R}$ = (12.35∠1.147+220t)V

$v_{L}$ = (10.95∠2.718+220t)V

$v_{C}$ = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

$v_{R}$ = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

$v_{L}$ = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

$v_{C}$ = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0

3 years ago