Ask question

Ask question

All categories

3 years ago

5

A tiger is running across a field. What would happen if the hydrogen ion pumps into the tigers mitochondria stopped functioning

1 answer:

galben [10]3 years ago

3 0

Answer:

The tiger would not be able to produce glucose causing it to stop running

Explanation:

Since the mitochondria is in charge of producing ATP the tiger would not be able to use any glucose causing it to not be able to run.

You might be interested in

What are the characteristics and ph level of acids?

12345 [234]

General Characteristics of Acids & Bases. Chemists measure the strength of an acid or base by its pH, which is a term that refers to the "power of hydrogen.". The midpoint of the pH scale is neutral. Compounds with a pH lower than the midpoint value are acidic while those with a higher value are basic or alkaline.

7 0

3 years ago

Does someone know how to do math with that equation

mestny [16]

Answer:

f = 5 cm

Explanation:

using the thin lens equation, given as follows:

$\frac{1}{f} = \frac{1}{d_{o}}+\frac{1}{d_{i}}\\$

where,

f = focal length = ?

do = the distance of object from lens = 20 cm

di = the distance of image from lens = 6.6667 cm

Therefore,

$\frac{1}{f} = \frac{1}{20\ cm}+\frac{1}{6.6667\ cm}\\\\\frac{1}{f} = 0.199999\ cm^{-1}\\\\f = \frac{1}{0.199999\ cm^{-1}}\\\\$

<u>f = 5 cm</u>

8 0

3 years ago

Please Help!! When do ionic bonds occur?

iren2701 [21]

A is the answer. I think

8 0

3 years ago

In what scientific study did scientists discover that when atoms split they _________.

Shtirlitz [24]

Answer:OOF

Explanation:

6 0

3 years ago

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0

3 years ago