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galina1969 [7]
3 years ago
6

The hormone glucagon is released by number of different tissues in the body to stabilize blood glucose levels. Which of the foll

owing pathways is least ikely to be activated by glucagon in hepatocytes? cells in the pancreas when blood sugar levels are low. It acts on a gluconeogenesis O glycogenolysis O ? oxidation glycolysis
Physics
1 answer:
Citrus2011 [14]3 years ago
4 0

Answer:

Glycolysis

Explanation:

In human body, glucose levels are regulated by hormones  insulin and glucagon, secreted from pancreas. Glucagon from alpha cells and insulin from beta cells of the pancreas. Glucagon are regulated along depending upon the blood sugar levels. During fasting when blood sugar levels are decreased the glucagon levels are increased. Glucagon increases hepatic glucose through glycogenelysis.

So, Glycolysis of glucagon is least likely to be activated by glucagon in hepatocytes

You might be interested in
Is it proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinde
Gelneren [198K]

Answer:

No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.

Explanation:

A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.

Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:

* When the diameter and length are comparable (i.e have the same measurement)

When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.

Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.

8 0
2 years ago
What is the speaker’s power output if the sound intensity level is 102 dBdB at a distance of 25 mm ? Express your answer to two
lesya [120]

Answer:

Power  = 124.50 W

Explanation:

Given that:

The Sound intensity of a speaker output is 102 dB

and the distance r = 25 m

For the intensity of sound,

\beta (dB)= 10 \  log_{10 } (\dfrac{I}{I_o})

where;

the threshold of hearing   I_o = 10^{-12} (W/m^2)

\dfrac{102 }{10}= log_{10}( \dfrac{I}{10^{-12}})

10^{10.2} =  \dfrac{I}{10^{-12}}

I = 10^{10.2} \times 10^{-12}

I = 0.01585 W/m²

If we recall, we know remember that ;

Power = Intensity × A rea

Power = 0.01585 W/m² × 4 × 3.142 × (25 m)²

Power  = 124.50 W

7 0
3 years ago
We are running late for school and we want to make our 0.5 kg tea (it’s at 90 C) colder. Let’s assume we can drink tea when it’s
PSYCHO15rus [73]

Answer:

x=0.154kg

Explanation:

(x*L)+(0.5kg*4200*50)+(x*4200*(-50)=0

(x*333 000J/kg*c)+(0.5kg*4200J/kg*C*(-40C))+(x*4200J/kg*C*50C)=0

6 0
2 years ago
If the wind or current is pushing your boat away from the dock as you prepare to dock, which line should you secure first?
Akimi4 [234]

Answer:

Bow Line

Explanation:

If the wind or current is pushing your boat away from the dock, bow line should be secured first.

1- We should cast off the bow and stern lines.

2-With the help of an oar or boat hook, keep the boat clear of the dock.

3-Leave the boat on its own for sometime and let the wind or current carry the boat away from the dock.

4 - As you see there is sufficient clearance, shift into forward gear and slowly leave the area.

7 0
2 years ago
0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary for
Paladinen [302]

Assuming the accleration applied was constant, we have

v=v_0+at\implies v_0+20.0\,\dfrac{\mathrm m}{\mathrm s}=v_0+a(0.10\,\mathrm s)

\implies20.0\,\dfrac{\mathrm m}{\mathrm s}=a(0.10\,\mathrm s)

\implies a=200\,\dfrac{\mathrm m}{\mathrm s^2}

Then the force applied to the ball is given by

F=ma=(0.45\,\mathrm{kg})\left(200\,\dfrac{\mathrm m}{\mathrm s^2}\right)

\implies F=90\,\dfrac{\mathrm{kg}\,\mathrm m}{\mathrm s^2}=90.\,\mathrm N

8 0
2 years ago
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