Answer:
For Xenon fluoride, the average bond energy is 132kj/mol
For tetraflouride,the average bond energy is 150.5kj/mol.
For hexaflouride, the average bond energy is 146.5 kj/mol
Explanation:
For xenon fluoride
105/2 = 52.5
For F-F
159/2 = 79.5
Average bond energy of Xe-F = 79.5 + 52.5 = 132kj/mole
For tetraflouride
284/4 = 71
For F-F
159/2 = 79.5
Average bond energy = 79.5 + 71 = 150.5kj/mol
For hexaflouride
402/6 = 67
F-F = 159/2 = 79.5
Average bond energy = 67 + 79.5 = 146.5kj/ mol
Answer:
the final speed of object A changed by a factor of 
= 0.58
the final speed of object B changed by a factor of 
= 1.29
Explanation:
Given;
kinetic energy of object A, = 27 J
let the mass of object A = 
then, the mass of object B = 
work done on object A = -18 J
work done on object B = -18 J
let 
be the initial speed
let 
be the final speed
For object A;
Thus, the final speed of object A changed by a factor of = 0.58
To obtain the change in the final speed of object B, apply the following equations.
Thus, the final speed of object B changed by a factor of = 1.29
The force of the racket affects the ball's motion because it changes the momentum of the ball.
<h3>Impulse received by the ball</h3>
The impulse received by the ball through the racket affects the motion because it changes the momentum of the ball.
The ball which is initially at rest, will gain momentum after been hit with the racket.
J = ΔP = Ft
where;
- J is the impulse received by the ball
- ΔP is change in momentum of the ball
- F is the applied force
- t is the time of action
Thus, the force of the racket affects the ball's motion because it changes the momentum of the ball.
Learn more about impulse here: brainly.com/question/25700778
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
Answer:
It always acurs after a 1st quarter
do you have photo?
Explanation:
