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4 years ago

Which type of molecule is added to this genetic sequece might create a insertion mutation? A-G-A-T-G-C-A-G-A-G-T-T-A-C-G-G

1 answer:

3 0

D. Adding another base pair will re-arrange your DNA sequences and cause an insertion mutation. This will make your codons group of differently, and possibly give you a BAD mutation. However, sometimes the codons still make the same proteins as its supposed to, the mutation will NOT affect you.

Example:

THE BIG FAT CAT ATE THE RAT

Now, if i were to add a letter (X) to this and make the letters group up in three aka the codons:

THE XBI GFA TCA TAT ETH ERA T

As you can you can see, adding a base pair in a DNA insertion will usually have a negative effect, specifically a insertion mutation.

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A compound machine is a combination of two or more blank

Arturiano [62]

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3 years ago

If a hiker that weighs 600 newtons climbs a 50 meter hill, how much gravitational potential energy has the hiker gained?

Illusion [34]

The gravitational potential energy U is defined as the product of mass m, the acceleration of gravity g and the height of object h.

$U = mgh$

We do not have the mass of the hiker. But we know that its W weight is:

$W = mg$

Where

$g = 9.8 m/s^2$

So:

$m = \frac{W}{g}\\\\m = \frac{600}{g}$ .

So:

$U = (\frac{600}{g})gh\\\\U = (600N)(50m)$

$U = 30000$ J

The hiker has gained 30,000 J of energy

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3 years ago

Which of the following best describes Gravitational Potential Energy?

kupik [55]

Answer:

Stored motion due to position

Explanation:

5 0

3 years ago

Read 2 more answers

The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic fie

telo118 [61]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The uncertainty in inverse frequency is $\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s$

Explanation:

From the question we are told that

The value of the proportionality constant is $k = 5 \frac{Hz }{T}$

The strength of the magnetic field is $B = 20 \ T$

The change in this strength of magnetic field is $\Delta B = 3 \ T$

The magnetic field is given as

$B = \frac{k}{\frac{1}{w} }$

Where $w$ is frequency

The uncertainty or error of the field is given as

$\Delta B = \frac{k }{[\frac{1}{w}^]^2 } \Delta [\frac{1}{w} ]$

The uncertainty in inverse frequency is given as

$\Delta [\frac{1}{w} ] = \frac{\Delta B}{k [\frac{1}{w^2} ]}$

$\Delta [\frac{1}{w} ]= \frac{\Delta B}{k (B)^2 }$

substituting values

$\Delta [\frac{1}{w} ]= \frac{3}{5 (20)^2 }$

$\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s$

6 0

3 years ago

A simple series circuit consists of a 130 ? resistor, a 30.0V battery, a switch, and a 2.10 pF parallel-plate capacitor (initial

V125BC [204]

Answer with Explanation:

We are given that

Resistance,R=130 ohm

Potential difference, V=30 V

Capacitor,C=2.1pF= $2.1\times 10^{-12} F$

$1pF=10^{-12} F$

$d=5.0 mm=5\times 10^{-3} m$

$1mm=10^{-3} m$

a.Maximum flux

$\phi=EA=\frac{V}{d}\times \frac{Cd}\epsilon_0}=\frac{CV}{\epsilon_0}$

$\epsilon_0=8.85\times 10^{-12}$

$\phi=\frac{2.1\times 10^{-12}\times 30}{8.85\times 10^{-12}}=7.12Vm$

b.Maximum displacement current, $I=\frac{V}{R}=\frac{30}{130}=0.23 A$

c.We have to find electric flux at t=0.5 ns

$t=0.5ns=0.5\times 10^{-9} s$

$1ns=10^{-9}s$

$q=CV(1-e^{-\frac{t}{RC}})$

$q=30\times 2.1\times 10^{-12}(1-e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-9}})$

$q=52.9\times 10^{-12} C$

$\phi=\frac{q}{\epsilon_0}=\frac{52.9\times 10^{-12}}{8.85\times 10^{-12}}=5.98Vm$

d.Displacement current at t=0.5ns

$I=(\frac{V}{R})e^{-\frac{t}{RC}}=\frac{30}{130}e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-12}}}$

$I=0.037 A$

5 0

3 years ago