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2 years ago

What is required for both the light-dependent and light-independent reactions to proceed?

1 answer:

3 0

<span>ATP is required for both light-dependent and light-independent reactions.
ATP stands for </span> adenosine triphosphate.
Hope this helps ;)

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The MAC is 58 inches, The CG limits are from 26% to 43% MAC. If the CG is found to be

eimsori [14]

By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.

<h3>What are the limits?</h3>

First, we need to find the limits.

We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.

So if 58 in is the 100%, the 26% and 43% of that are:

26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.

But we know that the CG is found to be 45.5% MAC, then it measures:

(45.5%/100%)*58in = 0.455*58in = 26.39 in

We need to compare it with the largest limit, so we get:

26.39 in - 24.94 in = 1.45 in

This means that the CG is 1.45 inches out of limits.

If you want to learn more about percentages, you can read:

brainly.com/question/14345924

6 0

2 years ago

Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic

VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

$m_Agh=\frac{1}{2}m_Av_A^2$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

$m_A=0.5 kg$ is the mass of the block

$v_A$ is the speed of the block A just before touching block B

Solving for the speed,

$v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s$

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

$e=\frac{v'_B-v'_A}{v_A-v_B}$

where:

e = 0.7 is the coefficient of restitution in this case

$v_B'$ is the final velocity of block B

$v_A'$ is the final velocity of block A

$v_A=3.83 m/s$

$v_B=0$ is the initial velocity of block B

Solving,

$v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s$

Re-arranging it,

$v_A'=v_B'-2.68$ (1)

Also, the total momentum must be conserved, so we can write:

$m_A v_A + m_B v_B = m_A v'_A + m_B v'_B$

where

$m_B=2 kg$

And substituting (1) and all the other values,

$m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s$

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

$\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2$

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m$

5 0

3 years ago

An empty plate capacitor is connected between the terminals ofa 9.0 V battery and charged up. The capacitor is then disconnected

777dan777 [17]

Answer:

The new voltage between the parallel plates of the capacitor is 18V, because for a constant electric field, doubling the space between the parallel capacitor plates, will also double the potential difference (voltage) between the plates.

Explanation:

ΔV = E*Δd

Where;

ΔV is the change in potential difference

Δd is the change in the distance between the parallel plates

E is the electric field potential.

Assuming a constant electric field; $E = \frac{v}{d} , then; \frac{v_1}{d_1} =\frac{v_2}{d_2}$

when the spacing between the capacitor plates is doubled, d₂ = 2d₁

v₂ = (v₁*d₂)/(d₁)

v₂ = (v₁*2d₁)/(d₁)

v₂ = 2v₁

v₂ = 2(9) = 18 V

Therefore, for a constant electric field, doubling the space between the parallel capacitor plates, will also double the potential difference (voltage).

5 0

3 years ago

A dust particle with mass of 5.4×10−2 g and a charge of 2.3×10−6 C is in a region of space where the potential is given by V(x)=

user100 [1]

Answer:

1.6 m/s^2

Explanation:

Hello!

To calculate the acceleration we must know the electric field. The electric field and the potential are related by:

$E = -\frac{dV}{dx} =- 2.6(\frac{V}{m^{2}})x + 8.1(\frac{V}{m^{3}})x^{2}$

If the particle starts at 2.3m, the electric field is:

E = 36.869 V/m = 36.869 N/C

So, the force on the particle is:

F = q E = 2.3×10^−6 C * 36.869 N/C = 8.48 x 10^-5 N

And its acceleration is :

a = F/m = 8.48 x 10^-5 N / 5.4×10−5 kg = 1.57 m/s^2

Rounded to two significant figures:

1.6 m/s^2

6 0

3 years ago

A man walks 30m to the west, then 5m to the east in 45 seconds. What is the average velocity?

solong [7]

1.28m/s the velocity is found by distance/time.

8 0

3 years ago

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