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3 years ago

Know the monosaccharides, the disaccharides and the polysaccharides. Know the hormones involved in glucose regulation.

Physics

1 answer:

BabaBlast [244]3 years ago

7 0

Answer:

Explanation:

Monosaccharides are the simplest form of carbohydrates which are called as single sugars. These are the building blocks of bigger carbohydrates.

Disaccharides are the sugars that are formed when two monosaccharides combine together by glycosidic bonds.

Polysaccharides are the long chains of carbohydrate molecules. These are formed by the monosaccharide units bonded by the glycosidic linkages.

The insulin and glucagon are the two hormones secreted by the pancreas that regulate the blood glucose levels. Insulin is secreted by the beta cells of pancreas. It is secreted when the blood glucose level is high. Glucagon is secreted by the beta cells of pancreas when the blood glucose level is low.

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uranmaximum [27]

Answer:

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Explanation:

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2 years ago

A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h

hram777 [196]

Answer:

$v_{2}=3.5 m/s$

Explanation:

Using the conservation of energy we have:

$\frac{1}{2}mv^{2}=mgh$

Let's solve it for v:

$v=\sqrt{2gh}$

So the speed at the lowest point is $v=7 m/s$

Now, using the conservation of momentum we have:

$m_{1}v_{1}=m_{2}v_{2}$

$v_{2}=\frac{1*7}{2}$

Therefore the speed of the block after the collision is $v_{2}=3.5 m/s$

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3 years ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=8.7\cdot 10^{13}kg$ is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s$

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4 years ago

How far will you travel if you run for 10 minutes at 2 m/s? SHOW WORK PLZS

ella [17]

10 minutes are the same as 600 seconds.

If you run 2 meters in 1 second then you run 2 * 600 meters in 600 seconds.

7 0

3 years ago

A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the

AlekseyPX

Answer:

73.5 m/s

Explanation:

The position of the first ball is:

y = y₀ + v₀ t + ½ at²

y = h + (0)(18) + ½ (-9.8)(18)²

y = h − 1587.6

The position of the second ball is:

y = y₀ + v₀ t + ½ at²

y = h + (-v) (18−6) + ½ (-9.8)(18−6)²

y = h − 12v − 705.6

Setting the positions equal:

h − 1587.6 = h − 12v − 705.6

-1587.6 = -12v − 705.6

1587.6 = 12v + 705.6

882 = 12v

v = 73.5

The second ball is thrown downwards with a speed of 73.5 m/s

8 0

3 years ago