Answer:
Mass of bullet is m=0.01kg
Mass of the block is M=4kg
Coefficient=0.25,distance=20m
So, let the speed of the block just after the bullet embedded in it be V and v be the speed of bullet before striking the block,
By applying conservation of momentum,
mv=(m+M)V
V=
M+m
mv
Explanation:
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Answer:
A. Vx = 3.63 m/s
B. Vy = -45.73 m/s
C. |V| = 45.87 m/s
D. θ = -85.46°
Explanation:
Given that position, r, is given as:
r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk
Velocity is the derivative of position, r:
V = dr/dt = 3.63 - 11.46t^j
A. x component of velocity, Vx = 3.63 m/s
B. y component of velocity, Vy = -11.46t
t = 3.99 secs,
Vy = - 11.46 * 3.99 = -45.73 m/s
C. Magnitude of velocity, |V| = √[(-45.73)² + 3.63²]
|V| = √(2091.2329 + 13.1769)
|V| = √(2104.4098)
|V| = 45.87 m/s
D. Angle of the velocity relative to the x axis, θ is given as:
tanθ = Vy/Vx
tanθ = -45.73/3.63
tanθ = -12.6
θ = -85.46°
Answer with Explanation:
We are given that
Diameter of fighter plane=2.3 m
Radius=
a.We have to find the angular velocity in radians per second if it spins=1200 rev/min
Frequency=
1 minute=60 seconds
Angular velocity=
Angular velocity=
b.We have to find the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac.
c.Centripetal acceleration=
Centripetal acceleration==
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