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Describe the proton gradient generated by the electron transport chain

Oksana_A [137]

Answer:

The proton gradient produced by proton pumping during the electron transport chain is used to synthesize ATP. Protons flow down their concentration gradient into the matrix through the membrane protein ATP synthase, causing it to spin (like a water wheel) and catalyze conversion of ADP to ATP.

Explanation:

4 0

3 years ago

Fill in the blank.<br> You don’t ____ me thass fine _ ___<br><br> -Man Tiffany

Jet001 [13]

Answer:

like, k bye

Hope this helps!

5 0

3 years ago

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You are 12 miles north of your base camp when you begin walking north at a speed of 2mph. what is your location, relative to you

Leona [35]

Http://www.calculator.net/pace-calculator.html?ctype=distance&ctime=05%3A00%3A00&cdistance=5&cdistanceunit=Miles&cpace=02%3A00%3A00&cpaceunit=tpm&printit=0&x=87&y=24 a pace calculator

4 0

3 years ago

Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder o

Verdich [7]

Answer:

V = 20 miles /sec

Explanation:

We have remaining distance = d = 96 miles

Lets call Pascal velocity V in miles per hour

Now if he increases his velocity by 50 % (equivalent to multiply by 1.5 ) he will need a time t₁ to arrive then as V = d/t

1.5* V = d/ t₁ ⇒ 1.5 * V = 96 /t₁

And in the case of reducing his velocity

(V / 4) = d/ (t₁ + 16 ) ⇒ V * (t₁ + 16 ) = 4*d ⇒ V*t₁ + 16*V = 384

So we a 2 equation system with two uknown variables

1.5*V = 96/t₁ (1)

V*t₁ + 16*V = 384 (2)

We solve from equation (1) t₁ = 64/V

And by substitution in equation (2)

V * (64/V) + 16* V = 384

64 + 16 *V = 384 ⇒ 16*V = 320 ⇒ V= 320/16

V = 20 miles /sec

6 0

3 years ago

One object (m1 = 0.220 kg) is moving to the right with a speed of 2.10 m/s when it is struck from behind by another object (m2 =

blagie [28]

Answer:

vf₁ = 6.86 m/s , to the right

vf₂ = 2.96 m/s, to the right

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:

p=m*v

where

p:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀ = Pf Formula (1)

P₀ :Initial linear momentum quantity

Pf : Final linear momentum quantity

Data

m₁= 0.220 kg : mass of object₁

m₂= 0.345 kg : mass of object₂

v₀₁ = 2.1 m/s ₁ , to the right : initial velocity of m₁

v₀₂= 6 m/s, to the right i :initial velocity of m₂

Problem development

We appy the formula (1):

P₀ = Pf

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂

We assume that the two objects move to the right at the end of the collision, so, the sign of the final speeds is positive:

(0.22)*(2.1) + (0.345)*(6) = (0.22)*vf₁ +(0.345)*vf₂

2.532 = (0.22)*vf₁ +(0.345)*vf₂ Equation (1)

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.

$e= \frac{v_{f2}-v_{f1} }{v_{o1} -v_{o2} }$

1*(v₀₁ - v₀₂ ) = (vf₂ -vf₁)

(2.1 - 6 ) = (vf₂ -vf₁)

-3.9 = (vf₂ -vf₁)

vf₂ = vf₁ - 3.9

vf₂ = vf₁ - 3.9 Equation (2)

We replace Equation (2) in the Equation (1)

2.532 = (0.22)*vf₁ +(0.345)*( vf₁ - 3.9)

2.532 = (0.22)*vf₁ +(0.345)* (vf₁) -(0.345)( 3.9)

2.532 + 1.3455 = (0.565)*vf₁

3.8775 = (0.565)*vf₁

vf₁ = (3.8775) / (0.565)

vf₁ = 6.86 m/s, to the right

We replace vf₁ = 6.86 m/s in the Equation (2)

vf₂ = 6.86 - 3.9

vf₂ = 2.96 m/s, to the right

8 0

3 years ago