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
hoa [83]
3 years ago
14

Which of the following is most closely related to an activated complex?

Physics
1 answer:
Anon25 [30]3 years ago
5 0

Among the choices above, the one that is most closely related to an activated complex is the transition state. The answer is letter D. This formation forms quickly and does not stay in a way compound is. It usually forms during the enzyme – substrate reaction.

You might be interested in
Pube Goldberg machine is a complicated contraption designed to do a very simple task, like the one shown above
Elanso [62]

Answer:

I think is 2

Explanation:

6 0
2 years ago
The first person to map the Milky Way Galaxy using radio waves was
Art [367]

Answer:B

Explanation:Grote reber was the first scientist to map the milky way galaxy using radio waves.

7 0
2 years ago
Read 2 more answers
A gyre is a set of currents that form:
tangare [24]
A gyre is a set of currents that form b. a loop. The circulation of gyres are affected by global wind patterns, landmasses, and the planet's rotation. The circulation is also affected by temperature, as warm water goes up and cold water sinks. There are five major gyres in the world: <span>North Atlantic, South Atlantic, Indian, North Pacific, and South Pacific.</span>
7 0
3 years ago
Read 2 more answers
While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)
ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration=2.40m/s^{2}, c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = 2.40m/s^{2}

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}

after cancelling the pertinent units, I get that  my answer will be given in meters. So I get:

x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}

which solves to:

x=172.8m

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

a=\frac{V_{f}-V_{0} }{t}

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

at=V_{f}-V_{0}

and finally I can add V_{0} to both sides so I get:

V_{f}=at+V_{0}

and now I can proceed and substitute the known values:

V_{f}=at+V_{0}

V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0

which solves to:

V_{f}=28.8m/s

8 0
3 years ago
Read 2 more answers
The magnitude of the electric current is directly proportional to the _____________ of the electric field.
MatroZZZ [7]

Answer;

the potential difference

The magnitude of the electric current is directly proportional to the potential difference of the electric field

Explanation;

An electric current results from the collective movement of free charges under the effect of an electric field. An electric field exists and can be observed in the space around a single charge or a number of charges.

Electric fields cause charges to move. It stands to reason that an electric field applied to some material will cause currents to flow in that material. In other words, the current density is directly proportional to the electric field. The constant of proportionality σ is called the material’s conductivity.

8 0
3 years ago
Read 2 more answers
Other questions:
  • A book 15.6 cm tall is 28.7 cm to the left of a plane mirror. What is the height of the image formed?
    8·1 answer
  • WILL MARK BRAINLIEST!
    12·1 answer
  • A very elastic rubber ball is dropped from a certain height and hits the floor with a (2pts) downward speed v. Since it is so el
    12·1 answer
  • A car in front of the school goes from rest to 5.6 m/s in 3 seconds. What is its acceleration?
    12·1 answer
  • A researcher measured the absorbance and percent transmittance of a blue dye at wavelengths between 425 and 700 nmnm. The measur
    7·1 answer
  • Why isn't solar energy usually the only power source for a region​
    13·1 answer
  • A baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby has a mass of 4kg. The carriage ha
    10·1 answer
  • It took 3 seconds for an object that was thrown up with velocity vo from
    6·1 answer
  • A charged comb contains 1000 electrons. Calculate the charge on the comb.
    9·1 answer
  • What do i do for #17
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!