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-Dominant- [34]

3 years ago

11

Why is Sean Carrol studying the fruit fly?

1 answer:

andrew11 [14]3 years ago

3 0

Answer:

Carroll has spent his career studying how evolution occurs at the genetic level, pinpointing many of the molecular changes that lead to new traits and new species. Most of Carroll's discoveries have come to light by studying the common fruit fly Drosophila melanogaster, the workhorse of modern molecular genetics.

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Two objects moving at the same speed can have different velocities. <br> True or False

Rom4ik [11]

Answer:

True

Explanation:

Objects have the same velocity only if they are moving at the same speed and in the same direction.

6 0

3 years ago

The table below shows a sample wavelength of each type of electromagnetic radiation, along with its frequency and energy.

trasher [3.6K]

Answer:

D lower energy waves is most likely the safest if one is exposed to.

5 0

3 years ago

One way to probe the nucleus is to bombard a sample with high-energy electrons. To learn about the nuclear structures in a sampl

Radda [10]

Answer:

E = 1.38 x 10⁸ eV = 138 MeV

Explanation:

The energy associated with the given wavelength can be found from the following formula:

$E = \frac{hc}{\lambda}$

where,

E = Energy of electron = ?

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

c = Speed of Light = 3 x 10⁸ m/s

λ = wavelength = 9 fm = 9 x 10⁻¹⁵ m

Therefore,

$E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{9\ x\ 10^{-15}\ m}\\\\E = (2.21\ x\ 10^{-11}\ J)(\frac{1\ eV}{1.6\ x\ 10^{-19}\ J})$

<u>E = 1.38 x 10⁸ eV = 138 MeV</u>

5 0

3 years ago

Which chart places the elements of plot in the correct order?

jekas [21]

Is there a picture ?

5 0

3 years ago

Read 2 more answers

A capacitor is connected to an ac generator that has a frequency of 3.4 kHz and produces a voltage of 2.0 V. The current in the

garik1379 [7]

Answer:

$V_G=3.302857143V$

Explanation:

Let´s find the capacitor impedance:

$V=IZ_C$

Using the data provided:

$Z_C=\frac{V}{I}=\frac{2}{0.035}=57.14285714$ Ω

Now let´s find the capacitance:

$Z_C=\frac{1}{j2\pi fC }$

$C=\frac{1}{j2\pi fZ_C} =\frac{1}{j2\pi (3400)(57.14285714)}= 0.8191798542$ μF

We can calculate the capacitor impedance for the second case:

$Z_C=\frac{1}{j2\pi fC}=\frac{1}{j2\pi (5000)(0.8191798542X10^{-6}) }=38.85714285$ Ω

Finally we can calculate the voltage of the second generator:

$V_G=IZ_C=(0.085)(38.85714285)=3.302857143V$

7 0

4 years ago