All categories

3 years ago

Helpppppppppppppppppp

Biology

1 answer:

dangina [55]3 years ago

4 0

Being cute can help species because they would attract people,environmentalists, who would want to save them from extinction.

You might be interested in

The pKa for the side chain of histidine is 6.0. What is the ratio of the deprotonated imidazole side chain to the protonated sid

Solnce55 [7]

Answer:

When the pKa is 6.0, we can determine the fraction of protonated H is by:

pH = pKa + log [A]/[HA]

Where

A = Deprotonated imidazole side

HA = Protonated side

Given, pH = 5.0

5 = 6 + log [A]/[HA]

log [A]/[HA] = -1 (take antilog of both side)

[A]/[HA] = 0.1

The ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 = 0.1

Given, pH = 7.5

7.5 = 6 + log [A]/[HA]

log [A]/[HA] = 1.5 (take antilog of both sides)

[A]/[HA] = 31.62

The ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 = 31.62

7 0

2 years ago

What is the disadvantage of a shallow shoulder socket

Mila [183]

The shallow shoulder socket makes the joint unstable, which makes the bone easy to dislocate.

4 0

3 years ago

Transcription produces Translation takes place in

kodGreya [7K]

Answer:

Transcription produces ✔ mRNA .

Translation takes place in the ✔ ribosome .

Explanation:

7 0

2 years ago

Read 2 more answers

Anyone know the answer

inessss [21]

Answer:

b.nuclear energy to heat energy to electrical energy

4 0

2 years ago

Mars had an orbital period of 1.88 years. In two or more complete sentences,explain how to calculate the average distance from m

Vanyuwa [196]

This is an interesting problem involving astronomy, in fact, simple physics.
Let r=distance of sun to mars, in metres

Mars had an orbital period of 1.88 years.
=>
tangential velocity, v, of the planet, in m/s is
$v=\frac{2\pi{r}}{T}$
 $=\frac{2\pi{r}}{1.88*365.25*86400}$ m/s, accounting for leap years
$=3.371*10^{-8}\pi{r}$ m/s

The centripetal force, Fc, generated is
$Fc=\frac{mv^2}{r}$
where m=mass of mars = 6.39*10^(24) kg
$=\frac{mv^2}{r}$
$=\frac{6.39*10^{24}v^2}{r}$
$=7.26168*10^9\pi^2r$

The gravitation pull from the sun, Fg, is given by
$Fg=\frac{GMm}{r^2}$
where G=grav. const., =6.67408*10^(-11) m^3 kg^(-1) s^(-2)
M=mass of sun=1.989*10^(30) kg
$=\frac{6.67408*10^{-11}1.989*10^{30}6.39*10^{24}}{r^2}$
$=\frac{8.4826*10^44}{r^2}$

Since the radial distance is in equilibrium, the average distance, r can be found by equation Fc=Fg and solving for r:
Fc=Fg
=>
$7.26168*10^9\pi^2r=\frac{8.4826*10^44}{r^2}$
Solving for the real root:
$r^3=\frac{8.48256*10^44}{7.26168*10^9*%pi^2}$
$=\frac{1.1681263*10^{35}}{\pi^2}$
=2.279*10^11 m

5 0

3 years ago