Answer:
the velocity of car when it passes the truck is u = 16.33 m/s
Explanation:
given,
constant speed of truck = 28 m/s
acceleration of car = 1.2 m/s²
passes the truck in 545 m
speed of the car when it just pass the truck = ?
time taken by the truck to travel 545 m
time =
time =
time =19.46 s
velocity of the car when it crosses the truck
u = 16.33 m/s
the velocity of car when it passes the truck is u = 16.33 m/s
Answer:
C) W = - 190 J
Explanation:
Notation
Wf = work done by the friction force (unknown)
Ff = force of the friction
d = distance travelled by the box = (2 pi 1.82 m) = 11.435 m
Finding acceleration= final velocity-initial velocity/ time taken (or A= V-U/T)
Final speed= 2m
Initial speed= 0m
Time taken= 2 seconds
2-0/2 so it’ll be 1m/s
2-0=0
2/2=
Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal
Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s
The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s
The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.
Answer:
t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37
Answer:
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