The magnitude of their resultant vector is 4.6 meters/seconds
Since we are to add the velocity vectors in order to find the magnitude of their resultant vector.
Hence:
Resultant vector magnitude=5.8 meters/seconds + (1.2 meters/seconds)
Resultant vector magnitude=5.8 meters/seconds-1.2 meters/seconds
Resultant vector magnitude 4.6 meters/seconds
Inconclusion The magnitude of their resultant vector is 4.6 meters/seconds
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S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)
S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m
Distance double 720m*2=1440m
V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places
Answer:
v=u+at is the first equation of motion. In this v=u+at equation, u is initial velocity.
Answer:
The angular velocity is
5.64rad/s
Explanation:
This problem bothers on curvilinear motion
The angular velocity is defined as the rate of change of angular displacement it is expressed in rad/s
We know that the velocity v is given as
v= ωr
Where ω is the angular velocity
r is 300mm to meter = 0.3m
the radius of the circle
described by the level
v=1.64m/s
Making ω subject of the formula and solving we have
ω=v/r
ω=1.64/0.3
ω=5.46 rad/s
The total momentum of the system has to be conserved to satisfy the principle of conservation of momentum. Before the ball hits the bottle, the momentum of the system is 0.4 x 18 = 7.2 kg m/s
The momentum of the bottle after being hit is 0.2 x 25 = 5 kg m/s
So the momentum of the ball now is 7.2 - 5 = 2.2 kg m/s
Hence its velocity is 2.2/0.4 = 5.5 m/s