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3 years ago

What is the final electron acceptor in noncyclic electron flow?

Physics

1 answer:

Phoenix [80]3 years ago

5 0

Answer:

final electron acceptor in noncyclic electron flow is NADP+

Explanation:

solution

in the plant final electron acceptor in noncyclic electron flow is NADP+

because we know here that NADP+ is oxidized by nicotinamide adenine di nucleotide phosphates

and coenzyme that is act as electron transported agent

so as a oxidized substance that act as electron acceptor

so that as non cyclic electron flow final electron acceptor is NADP +

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Un objeto tiene una masa de 120 kg y un volumen de 5 m3

melisa1 [442]

Answer: dont ever worry mate

Explanation:

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3 years ago

Kenetic energy increasing because velocity increased

Ymorist [56]

Answer:

Explanation:

The formula for kinetic energy Ke=1/2mv^2 where Ke is kinetic energy, m=mass, and v= velocity.

If the kinetic energy is increased that means that there must be an increase in mass or velocity. However, since the velocity is squared that means that velocity affects the kinetic energy of the object more than the mass does.

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3 years ago

A cheetah spotted a Gazelle. The cheetah runs at its top speed of 30 m/s for 15 seconds. Durning this time , a gazelle ,160 m fr

Xelga [282]

Answer:

A) 450 m

B) 27 m/s

C) 81 m, 243 m

D) Gazelle

Explanation:

Since, the Cheetah is running at constant speed. Therefore, we use the equation:

s₁ = v₁t

where,

s₁ = distance covered by Cheetah = ?

v₁ = speed of Cheetah = 30 m/s

t = time taken = 15 s

Therefore,

s₁ = (30 m/s)(15 s)

s₁ = 450 m

For final speed of Gazelle at the end of 6 s acceleration time we use 1st equation of motion:

Vf = Vi + at

where,

Vf = Final Speed of Gazelle at the end of 6 s = ?

Vi = Initial Speed of Gazelle = 0 m/s (Since, Gazelle is initially at rest)

t = time taken = 6 s

a = acceleration = 4.5 m/s²

Therefore,

Vf = (0 m/s) + (4.5 m/s²)(6 s)

Vf = 27 m/s

For the distance covered by Gazelle at the end of 6 s acceleration time we use 2nd equation of motion:

s₂ = Vi t + (0.5)at²

where,

Vi = Initial Speed of Gazelle = 0 m/s (Since, Gazelle is initially at rest)

t = time taken = 6 s

a = acceleration = 4.5 m/s²

Therefore,

s₂ = (0 m/s)(6 s) + (0.5)(4.5 m/s²)(6 s)²

s₂ = 81 m

Now, for the distance covered during the last 9 s at constant velocity Vf, we use equation:

s₃ = Vf t

where,

s₃ = distance covered by Gazelle in last 9 s = ?

t = time = 9 s

Therefore:

s₃ = (27 m/s)(9 s)

s₃ = 243 m

We know that, at the end of 15 seconds:

Distance covered by Cheetah = s₁ = 450 m

Distance Covered by Gazelle = s₂ + s₃ = 81 m + 243 m = 324 m

If we take the initial position of Cheetah as origin. Then the positions of both Gazelle and Cheetah with respect to origin will be:

Position of Cheetah = 450 m ahead of origin

Position of Gazelle = 324 m + 160 m = 484 m (Since, Gazelle was initially 160 m ahead of Cheetah)

Hence, it is clear that Gazelle is ahead at the end of 15 s.

5 0

3 years ago

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

$F_{No}= 7771.125 \ N$

$F_p = 2.2*10^{6} N$

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

4 years ago

Sean, after being so happy for two full days that he reported he "never needed much sleep," now is stating he is so sad that he

Kaylis [27]

D or A i'm not that sure

5 0

3 years ago

Read 2 more answers