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3 years ago

What is the substance that is broken down in cellular respiration

Physics

1 answer:

Harman [31]3 years ago

4 0

Glucose.... have a nice day

You might be interested in

Why are the stars that appear in the night sky not visible during the day?

xxTIMURxx [149]

Because the sun is not hitting the light on them and there not as bright like when we see them when it's dark.

6 0

3 years ago

The granulation pattern that astronomers have observed on the surface of the Sun tells us that: a. the Sun is a lot cooler on th

Norma-Jean [14]

Answer:

c. hot material must be rising from the Sun's hotter interior

Explanation:

Granulation is the grainy appearance of the solar photosphere produced by the top of the convection cells in the sun.

The grainy appearance are produced by granules on the photosphere of the sun and granules are caused by convection currents of plasma within the sun's convection zone.

The interior of these granules are brighter (and thus hotter) than the exterior of the granules which are darker.

<u>So, the granulation pattern that astronomers have observed on the surface of the Sun tells us that hot material must be rising from the Sun's hotter interior.</u>

4 0

3 years ago

100 g of Ice at -10°C is added into a

Andrei [34K]

Answer:

The mass of the juice responsible for melting the ice is 949.043 grams.

Explanation:

By the First Law of Thermodynamics, we understand that juice releases heat to the ice, which turns into water under the assumption that interactions between the ice-juice system and surroundings are negligible and energy processes are done in steady-state. Since juice is done with water, its specific heat will be taken as of the water. The process is described by the following formula:

$m_{i} \cdot [c_{i}\cdot (T_{1}-T_{2}) - L_{f} + c_{w}\cdot (T_{2}-T_{3})] + m_{w} \cdot c_{w}\cdot (T_{4}-T_{3}) = 0$ (1)

Where:

$m_{i}$ - Mass of ice, in grams.

$m_{w}$ - Mass of the juice, in grams.

$c_{i}$ - Specific heat of ice, in joules per gram-degree Celsius.

$c_{w}$ - Specific heat of water, in joules per gram-degree Celsius.

$L_{f}$ - Latent heat of fusion, in joules per gram.

$T_{1}$ - Initial temperature of ice, in degrees Celsius.

$T_{2}$ - Melting point of water, in degrees Celsius.

$T_{3}$ - Final temperature of the ice-juice system, in degrees Celsius.

$T_{4}$ - Initial temperature of the juice, in degrees Celsius.

If we know that $m_{i} = 100\,g$ , $c_{i} = 2.090\,\frac{J}{g\cdot ^{\circ}C}$ , $c_{w} = 4.18\,\frac{J}{g\cdot ^{\circ}C}$ , $L_{f} = 334\,\frac{J}{g}$ , $T_{1} = -10\,^{\circ}C$ , $T_{2} = 0\,^{\circ}C$ , $T_{3} = 10\,^{\circ}C$ and $T_{4} = 20\,^{\circ}C$ , then the mass of the juice is:

$m_{w} = \frac{m_{i}\cdot [c_{i}\cdot (T_{1}-T_{2}) - L_{f} + c_{w}\cdot (T_{2}-T_{3})]}{c_{w} \cdot (T_{3}-T_{4})}$

$m_{w} = \frac{(100\,g)\cdot \left[\left(2.090\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C) - 334\,\frac{J}{g} +\left(4.18\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C) \right]}{\left(4.180\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C)}$

$m_{w} = 949.043\,g$

The mass of the juice responsible for melting the ice is 949.043 grams.

5 0

3 years ago

Important result of french revolution

qaws [65]

The french revolution led to many deaths and impacted history because of the amount of life lost

8 0

3 years ago

So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

tia_tia [17]

Answer:

a) $a=33.73mm/s^{2}$

b) mg>N

c) $\%_{change}=0.343\%$

d) $a=24.07mm/s^{2}$

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

$a_{c}=\omega ^{2}r$

where:

$\omega=\frac{2\pi}{T}$

we know the period of rotation of the earth is about 24 hours, so:

$T=24hr*\frac{3600s}{1hr}=86400s$

so we can now find the angular speed:

$\omega=\frac{2\pi}{86400s}$

$\omega=72.72x10^{-6} rad/s^{2}$

So the centripetal acceleration will be:

$a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)$

which yields:

$a_{c}=33.73mm/s^{2}$

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

$\sum F=0$

so we get that:

$N-mg+ma_{c} = 0$

and solve for the normal force:

$N=mg-ma_{c}$

In this case, we can clearly see that:

$mg>mg-ma_{c}$

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

$mg=(60kg)(9.81m/s^{2})=588.6N$