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Aleksandr [31]
3 years ago
9

How many steps in glycolysis have atp as a substrate or product?

Physics
1 answer:
Lana71 [14]3 years ago
5 0

Answer:

4

Explanation:

1. Phosphorylation of Glucose

2. Production of Fructose-6 Phosphate

3. Production of Fructose 1, 6-Diphosphate

4. Splitting of Fructose 1, 6-Diphosphate

5. Interconversion of the Two Sugars

6. Formation of NADH and 1,3-Diphoshoglyceric acid

7. Production of ATP and 3-Phosphoglyceric Acid

8. Relocation of Phosphorus Atom

9. Removal of Water

10. Creation of Pyruvic Acid and ATP

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THIS MARCIN
nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

f

1

=

v

1

−

u

1

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

h

h

′

=

u

v

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

h

h

′

=

u

v

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

f

1

=

v

1

−

(−25)

1

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

v

1

=

f

1

+

(−25)

1

=

10

1

−

25

1

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

v

1

=

250

25−10

=

250

15

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

15

250

=

3

50

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

5

h

′

=−

25

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

25

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

5 0
2 years ago
A 47.2 kg girl is standing on a 177 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,
Softa [21]

Answer:

v_g,i = 1.208 m/s

Explanation:

We are given;

Mass of girl; m_g = 47.2 kg

Mass of plank; m_p = 177 kg

Let the velocity of girl to ice be v_g,i

Let the velocity of plank to ice be v_p,i

Since the velocity of the girl is 1.53 m/s relative to the plank, then;

v_g,i + v_p,i = 1.53

From conservation of momentum;

m_g × v_g,i = m_p × v_p,i

Thus;

47.2(v_g,i) = 177(v_p,i)

Dividing both sides by 47.2 gives;

v_g,i = 3.75(v_p,i)

v_pi = (v_g,i)/3.75

Thus, from v_g,i + v_p,i = 1.53, we have;

v_g,i + ((v_g,i)/3.75) = 1.53

v_g,i(1 + 1/3.75) = 1.53

1.267v_g,i = 1.53

v_g,i = 1.53/1.267

v_g,i = 1.208 m/s

5 0
3 years ago
ILL MARK BRAINIEST IF YOU DO THIS CORRECTLY!!!
IRINA_888 [86]

Explanation:

1)5.8m/s

2)5.15m/s^2

3)12.69m/s

4)

8 0
3 years ago
Which of the following best describes a person with social health
Anuta_ua [19.1K]
D. you develop positive interactions with your peers
7 0
3 years ago
Read 2 more answers
A car is traveling at 15m/s on a horizontal road. the brakes are applied and the car skids to a stop in 4.0s . the coefficient o
iren2701 [21]

Answer:

the coefficient of Kinetic friction between the tires and road is 0.38

Option A) .38 is the correct answer

Explanation:

Given that;

final velocity v = 0

initial velocity u = 15m/s

time taken t = 4 s

acceleration  a = ?

from the equation of motion        

v   =   u   +   at

we substitute

0 = 15 + a × 4

acceleration a = -15/4 =  - 3.75 m/s²    

the negative sign tells us that its a  deacceleration so the sign can be ignored.

Deacceleration due to friction a = μ × g

we substitute

3.75 = μ × 9.8    

μ = 3.75 / 9.8 = 0.3826 ≈ 0.38

Therefore the coefficient of Kinetic friction between the tires and road is 0.38

Option A) .38 is the correct answer

8 0
3 years ago
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