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3 years ago

15

The chemical formula for glucose is C6H12O6. Therefore, four molecules of glucose will have carbon atoms, hydrogen atoms, and ox

ygen atoms.

2 answers:

bezimeni [28]3 years ago

7 0

<h3><u>Answer;</u></h3>

<em>- 24 carbon atoms, 48 hydrogen atoms and 24 oxygen atoms</em>

<h3><u>Explanation;</u></h3>

<em><u>Glucose </u></em>is a simple carbohydrate and a compound that is made of elements carbon, hydrogen and oxygen.
One molecules of glucose<em><u> contains six carbon atoms, 12 hydrogen atoms, and six oxygen atoms.</u></em>
Therefore, four molecules of glucose will have <em><u>24 carbon atoms, 48 hydrogen atoms and 24 oxygen atoms.</u></em>
That is; 4(C₆H₁₂O₆), thus, C = 4×6 =24, H = 4×12 =48, and O = 4×6 =24.

Gekata [30.6K]3 years ago

5 0

If there are 4 molecules of glucose, there will be 24 carbon, 48 hydrogen, and 24 oxygen.

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3 years ago

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The tape in a videotape cassette has a total

Softa [21]

Answer:

1.37 rad/s

Explanation:

Given:

Total length of the tape is, $d= 297$ m

Total time of run is, $t = 2.1$ hours

We know, 1 hour = 3600 s

So, 2.1 hours = 2.1 × 3600 = 7560 s

So, total time of run is, $t= 7560$ s

Inner radius is, $r = 10\ mm = 0.01\ m
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Outer radius is, $R = 47\ mm = 0.047\ m
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Now, linear speed of the tape is, $v =\frac{d}{t}=\frac{297}{7560}=0.039\ m/s
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Let the same angular speed be $\omega$ .

Now, average radius of the reel is given as the sum of the two radii divided by 2.

So, average radius is, $R_{avg}=\frac{R+r}{2}=\frac{0.047+0.01}{2}=\frac{0.057}{2}=0.0285\ m
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Now, common angular speed is given as the ratio of linear speed and average radius of the tape. So,

$\omega=\dfrac{v}{R_{avg}}\\\\\\\omega=\dfrac{0.039}{0.0285}\\\\\\\omega=1.37\ rad/s
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5 0

3 years ago

Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?

Alexandra [31]

11) $1.04\cdot 10^7 J$

12) $1.04\cdot 10^7 J$

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

$g=9.8 m/s^2$ (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

$PE=(8325)(9.8)(127)=1.04\cdot 10^7 J$

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

$KE_t + PE_t = KE_b + PE_b$

where

$KE_t$ is the kinetic energy at the top

$PE_t$ is the potential energy at the top

$KE_b$ is the kinetic energy at the bottom

$PE_b$ is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

$KE_t=0$

Also, at the bottom the height is zero, so the potential energy is zero

$PE_b=0$

Therefore, we find:

$KE_b=PE_t=1.04\cdot 10^7 J$

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

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

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

$KE=1.04\cdot 10^7 J$

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s$

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

$KE_1 + PE_1 = KE_2 + PE_2$

where

$KE_1$ is the kinetic energy at the top of the 1st hill

$PE_1$ is the potential energy at the top of the 1st hill

$KE_2$ is the kinetic energy at the top of the 2nd hill

$PE_2$ is the potential energy at the top of the 2nd hill

From the previous questions, we know that

$KE_1=0$

and

$PE_1=1.04\cdot 10^7 J$

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

$PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J$

So, the kinetic energy at the second hill is

$KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J$

And so, the speed is

$v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s$

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3 years ago

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