Explanation:
the impact cull ball stop,bit transfer all of it
momentum and kinetic energy to the another ball resunting in the hit ball rolling with the initial speed of the cue ball .
Answer:
The object´s displacement vector is Δr = 8i - 6 j
Explanation:
Hi there!
The position vector is given by the following function:
r = t²i - (3t + 3) j
Let´s find the position of the object at time t1 and t2:
At t1 = 1 s:
r1 = (1)² i - (3 · (1) + 3 )j
r1 = 1 i - 6 j
At t2 = 3 s:
r2 = (3)² i - (3 · (3) + 3) j
r2 = 9 i - 12 j
The displacement is calculated as follows:
displacement = Δr = final position - initial position = r2 - r1
r2 - r1 = 9 i - 12 j - (1 i - 6 j)
r2 - r1 = 9 i - 12 j - 1 i + 6 j
r2 - r1 = 8 i - 6 j
The object´s displacement vector is Δr = 8i - 6 j
Answer:The cannons total flight time is 2.23seconds
Explanation:
A ball fired diagonally is fired at an angle of 45° to the horizontal. The motion of the fired ball is a projectile motion. A projectile is a motion in which an object fired into space with an initial velocity U is allowed to fall freely under the influence of gravitational force.
To total time of flight T of the cannon ball can be expressed as;
T = 2Usin(theta)/g where;
U is the initial velocity or speed of the ball = 31m/s
theta is the angle that the ball make with the horizontal = 45°
g is the acceleration due to gravity = 9.81m/s²
Substituting the given datas into the formula we have;
T = 31sin45°/9.81
T = 31×0.7071/9.81
T = 21.92/9.81
T = 2.23seconds
Answer:
(a) x=ASin(ωt+Ф₀)=±(√3)A/2
(b) x=±(√2)A/2
Explanation:
For part (a)
V=AωCos(ωt+Ф₀)⇒±0.5Aω=AωCos(ωt+Ф₀)
Cos(ωt+Ф₀)=±0.5⇒ωt+Ф₀=π/3,2π/3,4π/3,5π/3
x=ASin(ωt+Ф₀)=±(√3)A/2
For part(b)
U=0.5E and U+K=E→K=0.5E
E=K(Max)
(1/2)mv²=(0.5)(1/2)m(Vmax)²
V=±(√2)Vmax/2→ωt+Ф₀=π/4,3π/4,7π/4
x=±(√2)A/2
Answer:
Part a)
A = 66.2 m
Part b)
Angle = 38.35 degree
Explanation:
Part a)
Length of the vector is the magnitude of the vector
here we know that
now we have
Part b)
Angle made by the vector is given as
