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3 years ago

1. Define cellular respiration, and state where it takes place.

Physics

1 answer:

larisa [96]3 years ago

7 0

Answer:

1. Cellular respiration is the process in which cells break down glucose, release the stored energy, and use it to make ATP. The process begins in the cytoplasm and is completed in a mitochondrion. Cellular respiration occurs in three stages: glycolysis, the Krebs cycle, and electron transport.

2. How much ATP is produced in all three stages combined? Glycolysis produces 2 ATP molecules, and the Krebs cycle produces 2 more. Electron transport from the molecules of NADH and FADH2 made from glycolysis, the transformation of pyruvate, and the Krebs cycle creates as many as 32 more ATP molecules.

3. Photosynthesis converts carbon dioxide and water into oxygen and glucose. ... Cellular respiration converts oxygen and glucose into water and carbon dioxide. Water and carbon dioxide are by- products and ATP is energy that is transformed from the process.

4. Aerobic respiration takes place in presence of oxygen; whereas anaerobic respiration takes place in absence of oxygen. Carbon dioxide and water are the end products of aerobic respiration, while alcohol is the end product of anaerobic respiration. Aerobic respiration releases more energy than anaerobic respiration.

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The electric motor of a model train accelerates the train from rest to 0.540 m/s in 27.0 ms. The total mass of the train is 610

Gnesinka [82]

Answer:

The value is $P =3.294 \ W$

Explanation:

From the question we are that

The velocity $v = 0.540 \ m/s$

The time taken is $t = 27.0 ms = 27.0 *10^{-3} \ s$

The total mass of the train is $m = 610 \ g = 0.610 \ kg$

Generally the average power delivered is mathematically represented as

$P =\frac{KE }{t}$

$P =\frac{ \frac{1}{2} * m * v^2 }{t}$

=> $P =\frac{ \frac{1}{2} * 0.610 * 0.540 ^2 }{ 27.0 *10^{-3}}$

=> $P =3.294 \ W$

5 0

3 years ago

Read 2 more answers

A person is using a rope to lower a 5.0-n bucket into a well with a constant speed of 2.0 m/s. What is the magnitude of the forc

OLga [1]

Answer:

5 N

Explanation:

The bucket is moving at a constant speed of 2m/s Therefore F=ma is 0 N for this to be correct the magnitude of the force exerted by the rope must be equal to the weight of the bucket which is 5 N

7 0

3 years ago

Read 2 more answers

An astronaut drops a rock on the surface of an asteroid.The rock is released from rest at a height of 0.86 m above the ground, a

SCORPION-xisa [38]

Answer:

$a_y=0.92m/s^2$

Explanation:

To solve this problem we use the formula for accelerated motion:

$y=y_0+v_{y0}t+\frac{a_yt^2}{2}$

We will take the initial position as our reference ( $y_0=0m$ ) and the downward direction as positive. Since the rock departs from rest we have:

$y=\frac{a_yt^2}{2}$

Which means our acceleration would be:

$a_y=\frac{2y}{t^2}$

Using our values:

$a_y=\frac{2(0.86m)}{(1.37s)^2}=0.92m/s^2$

5 0

3 years ago

A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

photoshop1234 [79]

Answer:

The height is $h_c = 42.857$

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

From the question we are told that

The height is $h_s = 30 \ cm$

The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

$mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2$

Where I is the moment of inertia which is mathematically represented as this for a sphere

$I = \frac{2}{5} mr^2$

The angular velocity $w$ is mathematically represented as

$w = \frac{v}{r}$

So the equation for conservation of energy becomes

$mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

$mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

$v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

$h_c = \frac{v^2}{g}$

Recall that $v^2 = \frac{10gh_s}{7}$

$h_c= \frac{\frac{10gh_s}{7} }{g}$

$= \frac{10h_s}{7}$

Substituting values

$h_c = \frac{10(30)}{7}$

$h_c = 42.86 \ cm$

8 0

3 years ago

An electron (q=-1.602×10-19C) is placed .03m away from spherical object with a net charge of -7.2 C.

vovangra [49]

Answer:

Explanation:

electric field at the location of electron

= 9 x 10⁹ x 7.2 / .03²

= 72 x 10¹² N/C

force on electron = electric field x charge on electron

= 72 x 10¹² x 1.6 x 10⁻¹⁹

= 115.2 x 10⁻⁷ N .

C )

work done = charge on electron x potential difference at two points

potential at .03 m

= 9 x 10⁹ x 7.2 / .03

= 2.16 x 10¹² V

potential at .001 m

= 9 x 10⁹ x 7.2 / .001

= 64.8 x 10¹² V

potential difference = (64.8 - 2.16 )x 10¹² V

= 62.64 x 10¹² V .

work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹

= 100.224 x 10⁻⁷ J .

D )

There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction .

Work done in case of electron will be positive and work done in case of positron will be negative .

electric field due to charge will be same in both the cases .

8 0

3 years ago