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Vedmedyk [2.9K]
2 years ago
13

Write a simple rule that will tell a person how many water molecule will be lost while putting monosaccharides together to form

polysaccharide ?
Physics
1 answer:
belka [17]2 years ago
3 0
<span>For hydrolysis to monosaccharides, one molecule of a disaccharide needs only one molecule of water. C12H22O11 (sucrose) + H2O = C6H12O6 (glucose) + C6H12O6 (fructose) Structurally, a disaccharide molecule may be viewed as a product formed by the condensation of two molecules of monosaccharides with the elimination of a water molecule. So, only one H2O molecule is needed for the reverse process.</span>
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A ball is thrown vertically upward (assumed to be the positive direction) with a speed of 24.0 m/s from a height of 3.0 m. (a) H
SashulF [63]

Answer:

a) 29.36 m

b) 2.44 s

c) 2.57 s

d) 25.117 m/s

Explanation:

t = Time taken

u = Initial velocity = 24 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

b)

v=u+at\\\Rightarrow 0=24-9.81\times t\\\Rightarrow \frac{-24}{-9.81}=t\\\Rightarrow t=2.44 \s

Time taken by the ball to reach the highest point is 2.44 seconds

a)

s=ut+\frac{1}{2}at^2\\\Rightarrow s=24\times 2.44+\frac{1}{2}\times -9.81\times 2.44^2\\\Rightarrow s=29.35\ m

The highest point reached by the ball above its release point is 29.36 m

c) Total height is 3+29.35 = 32.35 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 32.35=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{32.35\times 2}{9.81}}\\\Rightarrow t=2.57\ s

The ball reaches the ground 2.57 seconds after reaching the highest point

d)

v=u+at\\\Rightarrow v=0+9.81\times 2.57\\\Rightarrow v=25.2117\ m/s

The ball will hit the ground at 25.2117 m/s

8 0
3 years ago
A light year is approximately 9.5 million km long. 'Barnard's Star' is 6 light years away from Earth. Calculate how many million
hammer [34]

Answer:

6 light years = 57 million km

Explanation:

Given;

A light year = 9.5 million km

To calculate how far is 6 light years;

6 light years = 6 × 1 light year = 6 × 9.5 million km

6 light years = 57 million km

7 0
3 years ago
Read 2 more answers
An electromagnetic flowmeter is useful when it is desirable not to interrupt the system in which the fluid is flowing (e.g. for
AURORKA [14]

Answer:

<em>2 m/s</em>

<em></em>

Explanation:

The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as

E = Blv

where E is the induced voltage = 2.88 mV = 2.88 x 10^-3 V

l is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m

v is the velocity of the fluid through the field = ?

B is the magnetic field = 0.120 T

substituting, we have

2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x v

2.88 x 10^-3 = 1.44 x 10^-3 x v

v = 2.88/1.44 = <em>2 m/s</em>

8 0
3 years ago
The drawing shows a wire composed of three segments, AB, BC, and CD. There is a current of I = 2.0 A in the wire. There is also
alexdok [17]

Answer:

The magnitude of the magnetic force acting on the wire is zero, because the magnetic field is parallel to the wire.

In fact, the magnetic force exerted by the magnetic field on the wire is

where I is the current in the wire, L the length of the wire, B the magnetic field intensity and  the angle between the direction of B and the wire. In our problem, B and the wire are parallel, so the angle is  and so , therefore the magnetic force is zero: F=0.

7 0
3 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

f = f_0 (\frac{v_0}{v_0-v})

Here,

f_0 = Frequency of Source

v_s = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v  = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

f = f_0 (\frac{v_0}{v_0-v})

300 = f_0 (\frac{343}{343-v})

(300*343) - 300v = 343f_0

Now the second expression will be,

f' = f_0 (\frac{v_0}{v_0-v/2})

290 = (343)(\frac{v_0}{343-v/2})

290*343-145v = 343f_0

Dividing the two expression we have,

\frac{(300*343) - 300v}{290*343-145v} = 1

Solving for v, we have,

v = 22.12m/s

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0
2 years ago
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