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Novosadov [1.4K]
3 years ago
5

Why is it correct to say that the linear electron flow of the light reactions directly captures energy in nadph but only indirec

tly captures energy in atp?
Physics
1 answer:
denis-greek [22]3 years ago
5 0

Answer:

he electron is directly transferred to NADP+ to NADPH, but electron flow is used to generate a proton gradient for ATP synthesis. Electron is not directly transferred to ATP

NADP= Nicotin amide adenine dinucleotide phosphate

NADPH =Nicotinamide adenine dinucleotide phosphate is the reduced form of  NADP

therefore electron is not directly transferred to ATP

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How much heat is needed to melt 13.74 kg of silver that is initially at 20°C?
salantis [7]

Answer:

Q = 4.63 \times 10^6 J

Explanation:

As we know that melting point of silver is

T = 961.8 degree C

Latent heat of fusion of silver is given as

L = 111 kJ/kg

specific heat capacity of silver is given as

s = 240 J/kg C

now we will have

Q = ms\Delta T + mL

\Delta T = 961.8 - 20

\Delta T = 941.8 degree C

now from above equation

Q = (13.74)(240)(941.8) + (13.74)(111 \times 10^3)

Q = 3.1 \times 10^6 + 1.53 \times 10^6

Q = 4.63 \times 10^6 J

3 0
2 years ago
Why the unit of power is called derived unit? <br>​
Lilit [14]
A :-) it was given the name Newton (N). from this, the derived unit of energy (or work) is defined ,as the work produced when the unit of force causes a displacement equal to the unit of length of its point of application along its direction . It was given the name Joule (J).
7 0
2 years ago
A disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular a
stepan [7]

Answer:

a) α = 1.875 \frac{rad}{s^{2} }

b) t = 8 s

Explanation:

Given:

ω1 = 0 \frac{rad}{s}

ω2 = 15 \frac{rad}{s}

theta (angular displacement) = 60 rad

*side note: you can replace regular, linear variables in kinematic equations with angular variables (must entirely replace equations with angular variables)*

a) α = ?

(ω2)^2 = (ω1)^2 + 2α(theta)

15^{2} = 0^{2} + 2(α)(60)

225 = 120α

α = 1.875 \frac{rad}{s^{2} }

b)

α = (ω2-ω1)/t

t = (ω2-ω1)/α = (15-0)/1.875 = 8

t = 8 s

4 0
2 years ago
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
IRINA_888 [86]

Answer:

Time : <u>7.96 s</u>

Distance Traveled : <u>357.8 m</u>  

Explanation:

In order to solve this problem, we first consider the accelerated motion of rocket. We will be using the subscript 1 for accelerated motion.

So, for accelerated motion, we have:

Acceleration = a₁ = 14.5 m/s²

Time Period = t₁ = 3.1 s

Initial Velocity = Vi₁ = 0 m/s    (Since, it starts from rest)

Final Velocity = Vf₁

Distance covered by sled during acceleration motion = s₁

Now, using 1st equation of motion:

Vf₁ = Vi₁ + (a₁)(t₁)

Vf₁ = 0 m/s + (14.5 m/s²)(3.1 s)

Vf₁ = 44.95 m/s

Now, using 2nd equation of motion:

s₁ = (Vi₁)(t) + (0.5)(a₁)(t₁)

s₁ = (0 m/s)(3.1 s) + (0.5)(14.5 m/s²)(3.1 s)

s₁ = 22.5 m

Now, we first consider the decelerated motion of rocket. We will be using the subscript 2 for decelerated motion.

So, for accelerated motion, we have:

Deceleration = a₂ = - 5.65 m/s²

Time Period = t₂ = ?

Initial Velocity = Vi₂ = Vf₁ = 44.95 m/s    (Since, decelerate motion starts, where accelerated motion ends)

Final Velocity = Vf₂ = 0 m/s    (Since, rocket will eventually stop)

Distance covered by sled during deceleration motion = s₂

Now, using 1st equation of motion:

Vf₂ = Vi₂ + (a₂)(t₂)

0 m/s = 44.95 m/s + (- 5.65 m/s²)(t₂)

t₂ = (44.95 m/s)/(5.65 m/s²)

<u>t₂ = 7.96 s</u>

Now, using 2nd equation of motion:

s₂ = (Vi₂)(t₂) + (0.5)(a₂)(t₂)

s₂ = (44.95 m/s)(7.96 s) + (0.5)(- 5.65 m/s²)(7.96 s)

s₂ = 357.8 m - 22.5 m

s₂ = 335.3 m

Thus, the total distance covered by sled will be:

Total Dustance = S = s₁ + s₂

S = 22.5 m + 335.3 m

<u>S = 357.8 m</u>

7 0
3 years ago
Two cars A and B are moving with velocities 20 m/s and 15 m/s in the direction east and west respectively. If
Sphinxa [80]

Answer:

<u>Distance</u><u> </u><u>between</u><u> </u><u>them</u><u> </u><u>is</u><u> </u><u>4</u><u>,</u><u>2</u><u>0</u><u>0</u><u> </u><u>meters</u><u>.</u>

Explanation:

Consinder car A:

{ \bf{distance =  speed \times time }}

substitute:

distance = 20 \times (2 \times 60) \\  = 2400 \: m

Consider car B:

distance = 15 \times (2 \times 60) \\  = 1800 \: m

since these cars move in opposite directions, distance between them is their summation:

distance \: between = { \sum(distance \: of \: each \: car)} \\  = 2400 + 1800 \\  = 4200 \: m

3 0
3 years ago
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