All categories

3 years ago

contrast the location of genetic material in bacteria cells to its location in plant and animal cells.

1 answer:

6 0

First bacterial cells are usually much smaller than plant or animal cells. a human skin cells , for example is about times as larger as an average bacterial cell.

You might be interested in

Sunburn is caused by

Snowcat [4.5K]

Answer:

too much exposure to the sun's rays

5 0

3 years ago

Read 2 more answers

The net effect of each cycle of the sodium-potassium pump is to ____.? ?decrease the number of positively charged ions within th

stealth61 [152]

The correct answer may be A

3 0

3 years ago

If 25.0 g of water at 21c is mixed with 45.0 g of water at 75c, what is the final temperature of the mixture?

mrs_skeptik [129]

56C

Hope that helps, Good luck! (:

5 0

3 years ago

What is the change in potential energy if the distance separating the electron and proton is increased to 1.0 nm?

Vlada [557]

Answer:

$Ep=-2.3*10^{-19}J$

Explanation:

The change in potential energy can be expressed as:

$Ep=K.\frac{q1.q2}{r}$

where K is a constant with a value of $9*10^{9}\frac{N.m^{2}}{C^{2}}$ , q1 and q2 are the charges of the proton and the electron and r is the distance between them.

The charge for the proton is $+1.6*10^{-19}C$ and the charge for the electron is $-1.6*10^{-19}C$ .

Converting r=1.0nm to m:

$1.0nm*\frac{1*10^{-9}m}{1.0nm}=1*10^{-9}m$

Replacing values:

$Ep=9*10^{9}\frac{N.m^{2}}{C^{2}}.\frac{(+1.6*10^{-19}C).(-1.6*10^{-19}C)}{1*10^{-9}m}$

$Ep=-2.3*10^{-19}J$

5 0

2 years ago

Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the ce

jasenka [17]

Answer:

$\lambda= 506.25 nm$

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

$y=\frac{m \lambda D}{a}$

Where:

$y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1$

Solving for λ:

$\lambda=\frac{y*a}{mD}$

Replacing the data provided by the problem:

$\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm$

7 0

2 years ago