Answer:
d = 69 .57 meter
Explanation:
First case
Speed of car ( v ) = 20.5 mi/h = 9.164 M/S
distance ( d ) = 11.6 meter ( m = mass of the car )
Work done = 0.5 m v² = 0.5 * 9.164² * m J = 41.99 m J
Force = ( workdone /distance ) = ( 41.99 m / 11.6 ) = 3.619 m N
Second case
v = 50.2 mi/h = 22.44135 m/s
d = ?
Work done = 0.5 * 22.44² * m J = 251.7768 * m J
Since the braking force remains the same .
3.619 m = ( 251.7768 m / d )
d = 69 .57 meter
If the applied force is in the same direction as the object's displacement, the work done on the object is:
W = Fd
W = work, F = force, d = displacement
Given values:
F = 45N
d = 12m
Plug in and solve for W:
W = 45(12)
W = 540J
A beta particle. Hoped I help. Sorry if it wrong.
300 000 0 squared = 2 x 9.8 distance
KINEMATICS
Uniform or constant motion in a straight line (rectilinear). Speed or velocity constant and/or acceleration constant. If motion is up and down and/or has an up and down component then acceleration omn earth will be g. g is about 10m/s/s.
speed = distance/time
velocity = displacement/time
s=distance ... u=initial speed ... v = final speed ... a = acceleration ... t = time
v=u+at
v^2=u^2+2as
s=ut+1/2at^2
The mode in this case would be 125 because it occurs the most in the sequence of numbers.
