The velocity of the ball when it reaches the ground is equal to B. 68.6 m/s. This value was obtained from the formula Vf = Vi + at. Vf is the final velocity. Vi is the initial velocity. The acceleration is "a", while the time of travel is "t". The solution is:
<span>Vf = Vi + at
</span>Vf = 0 + (-9.8 m/s^2) (7 s)
Vf = -68.6 m/s
The negative sign denotes the direction of the ball.
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Answer:
a= 23.65 ft/s²
Explanation:
given
r= 14.34m
ω=3.65rad/s
Ф=Ф₀ + ωt
t = Ф - Ф₀/ω
= (98-0)×
/3.65
98°= 1.71042 rad
1.7104/3.65
t= 0.47 s
r₁(not given)
assuming r₁ =20 in
r₁ = r₀ + ut(uniform motion)
u = r₁ - r₀/t
r₀ = 14.34 in= 1.195 ft
r₁ = 20 in = 1.67 ft
= (1.667 - 1.195)/0.47
0.472/0.47
u= 1.00ft/s
acceleration at collar p
a=rω²
= 1.67 × 3.65²
a = 22.25ft/s²
acceleration of collar p related to the rod = 0
coriolis acceleration = 2ωu
= 2× 3.65×1 = 7.3 ft/s²
acceleration of collar p
= 22.5j + 0 + 7.3i
√(22.5² + 7.3²)
the magnitude of the acceleration of the collar P just as it reaches B in ft/s²
a= 23.65 ft/s²
** Missing information: The vertical distance from surface of liquid to bottom of the object is sought in this question, with the condition that the object is at equilibrium **
Ans: The vertical distance = y = M/(ρA)
Explanation:Support the vertical distance = y
Object's density = M/(A*h) (since A*h = volume)
By applying the condition,
(M/(Ah))/ρ = y/h
M/(ρAh) = y/h
y = M/(ρA)
