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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

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a ball slows down as it rolls up hill this is an example of? neutral motion, relative velocity, positive acceleratio, negative a
Zarrin [17]

Answer: negative acceleration

Explanation:

Acceleration is speeding up, the ball is slowing down making it negative acceleration

8 0
4 years ago
Read 2 more answers
What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?
Stolb23 [73]

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

8 0
3 years ago
Đổi 20độC = ? độ F =? Độ k
san4es73 [151]

Answer:

IDC

Explanation:

I DON'T UNDERSTAND........

7 0
2 years ago
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

than the equation of motion becomes

s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0
3 years ago
Which spacecraft did john glenn pilot when he became the first man to orbit the earth in 1962?
Makovka662 [10]
Spacecraft used is "Friendship 7". Hope it helps.
7 0
3 years ago
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