1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
suter [353]
3 years ago
10

Which statement describes the redox reaction involved in photosynthesis?

Physics
2 answers:
vodka [1.7K]3 years ago
6 0
The answer will be d
Hopefully it’s right :)
NeTakaya3 years ago
3 0

Answer:

D. O2 is removed from the atmosphere, and CO2 is released

Explanation:

Hope it helps you

You might be interested in
In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that
erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m

This means that the relation between the wavelength and the length of the string is

3\lambda/2 = L

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)

a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0

For this equation to be equal to zero, sin(59.94t) = 0. So,

59.94t = \pi\\t = \pi/59.94 = 0.0524~s

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s

4 0
3 years ago
A woman flies 300 kilometers north, 200 kilometers west, and 500 kilometers south. How far is she from her starting point?
maria [59]
She is
\sqrt{200^2+200^2}=  ~282.84 kilometers away from her starting point
5 0
4 years ago
you decide to work part time at a local supermarket. The job pays eight dollars and 60 per hour and you work 20 hours per week.
aleksley [76]

Answer:

I guess that we want to find how much money you get each week.

We know that the job pays $8.60 per hour.

We know that you work 20 hours per week.

Then the gross pay (the total money that you earn) in a week is 20 times $8.60, or:

20*$8.60 = $172.

Now we know that your employer witholds:

10% + 7.65% + 5% = 22.65%

Then your employer withholds 22.65% of your gross pay.

if the 100% of your gross pay is $172

Then the 22.65% will be:

(22.65%/100%)*$172 = 0.2265*$172 = $38.96

This means that your employer withholds $38.96 of your weekly gross pay.

Then each week you get:

$172 - $38.96 = $133.04

4 0
3 years ago
What would a series circuit be used for?
igor_vitrenko [27]

Answer:

C

Explanation:

a series circuit would be an odd choice to power a battery or light a lamp when a direct would be much more efficient, and it's not converting types of energy, so C is the best possible answer

6 0
3 years ago
A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  T_R  = 11.8 \  days  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  a_R = 45.0\  AU

   The semi - major axis of  Planet D is  a_D  = 60 \  AU

    The orbital  period of planet D is  T_D = 18.164 \  days

Generally from Kepler third law

          T \  \ \alpha \ \ a^{\frac{3}{2} }

Here T is the  orbital period  while a is the semi major axis

So  

        \frac{T_D}{T_R}  =  \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}

=>     T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }  

=>     T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }

=>      T_R  = 11.8 \  days  

   

7 0
3 years ago
Other questions:
  • A physics book slides off a horizontal table top with a speed of 1.10m/s. It strikes the floor after a time of 0.400s. Find the
    14·1 answer
  • Juan is standing on the street. An ambulance moves toward him and then passes by. What best describes the pitch that Juan heats?
    8·1 answer
  • A boy is inning with a kinetic energy of 810j. If the boy has a mass of 80kg. What is his speed ?
    5·1 answer
  • Please help 100 points if right and if its not an awnser then will be reported and deleted
    15·2 answers
  • Introduction to conductors and insulators and electrification processes. Please
    12·1 answer
  • A roller coaster car moving at a velocity of 30 meters/second has a momentum of 2.5 _ 104 kilogram meters/second. What is its ma
    15·1 answer
  • the distance between any two bodies is 10 M and the gravitational force between them is 3.2×10-⁹m. if the mass of one object is
    15·1 answer
  • A heat engine has three quarters the thermal efficiency of a carnot engine operating between temperatures of 65°C and 435°C if
    15·1 answer
  • Why this app says im in high school.?. im in middle and im 13
    7·1 answer
  • Why aren't there lunar and solar eclipses every month?
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!