Answer:


Explanation:
m = Mass of proton = 
v = Speed of proton = 0.5c = 
Circumference of the colider is 7 km


Centripetal acceleration is 

Force on protons is 
Answer:
1) Transition states are short-lived
Explanation:
Transition state theory explains the rates of elementary chemical reactions. It assumes a quasi-equilibrium between reactants and activated transition state complexes.
The following are the characteristics of transition states
- Instability
- Ill-defined
- High energy
- short-lived
The species that must collide for the reaction to occur are shown by the mechanism of reaction and not the balanced reaction itself
Intermediates are consumed in each step of the overall reaction, they are not short lived
Total distance = 36500 m
The average velocity = 19.73 m/s
<h3>Further explanation</h3>
Given
vo=initial velocity=0(from rest)
a=acceleration= 1 m/s²
t₁ = 20 s
t₂ = 0.5 hr = 1800 s
t₃= 30 s
Required
Total distance
Solution
State 1 : acceleration


State 2 : constant speed

State 3 : deceleration


Total distance : state 1+ state 2+state 3

the average velocity = total distance : total time

<span>0.52%
First, let's convert that speed into m/s.
150 km/h * 1000 m/km / 3600 s/h = 41.667 m/s
Now let's see how much time gravity has to work on the ball. Divide the distance by the speed.
18 m / 41.667 m/s = 0.431996544 s
Now multiply that time by the gravitational acceleration to see what the vertical component to the ball's speed that gravity adds.
0.431996544 s * 9.8 m/s^2 = 4.233566131 m/s
Use the pythagorean theorem to get the new velocity of the ball.
sqrt(41.667^2 + 4.234^2) = 41.882 m/s
Finally, let's see what the difference is
(41.882 - 41.667)/41.667 = 0.005159959 = 0.5159959%
Rounding to 2 figures, gives 0.52%</span>
As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards
So here we know that

now from the above equation


so both of the charges will apply 0.288 N force on q3 charge along the line joining them
now the net force due to vector sum is given by

here we know that angle is

now we have


so net force on q3 is 0.46 N vertically upwards along +Y axis