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:
The cross-sectional area of the larger piston is 392cm ^{2}[/tex]
Explanation:
To solve this problem we apply the following formula:
Pascal principle: F=P*A Formula (1)
F=Force applied to the piston
P: Pressure
A= Piston area
Nomenclature:
Fp= Force on the primary piston= 500N
W= car weight =m*g=2000kg*9.8m/s2= 19600N
Fs= Force on the secondary piston= W = 19600N

As= Secondary piston area=?
The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.
In the equation (1)
P=F/A
Pp=Ps





Answer: Change in ball's momentum is 1.5 kg-m/s.
Explanation: It is given that,
Mass of the ball, m = 0.15 kg
Speed before the impact, u = 6.5 m/s
Speed after the impact, v = -3.5 m/s (as it will rebound)
We need to find the change in the magnitude of the ball's momentum. It is given by :
So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.
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Answer:
(a) k =
(b) τ =
∝
Explanation:
The moment of parallel pipe rotating about it's axis is given by the formula;
I =
---------------------------------1
(a) The kinetic energy of a parallel pipe is also given as;
k =
--------------------------------2
Putting equation 1 into equation 2, we have;
k = 
k =
(b) The angular momentum is given by the formula;
τ = Iw -----------------------3
Putting equation 1 into equation 3, we have
τ = 
But
τ = dτ/dt =
------------------4
where
dw/dt = angular acceleration =∝
Equation 4 becomes;
τ =
∝