Answer:
hello the diagram relating to this question is attached below
a) angular accelerations : B1 = 180 rad/sec, B2 = 1080 rad/sec
b) Force exerted on B2 at P = 39.2 N
Explanation:
Given data:
Co = 150 N-m ,
<u>a) Determine the angular accelerations of B1 and B2 when couple is applied</u>
at point P ; Co = I* ∝B2'
150 = ( (2*0.5^2) / 3 ) * ∝B2
∴ ∝B2' = 900 rad/sec
hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec
at point 0 ; Co = Inet * ∝B1
150 = [ (2*0.5^2) / 3 + (2*0.5^2) / 3 + (2*0.5^2) ] * ∝B1
∴ ∝B1 = 180 rad/sec
hence angular acceleration of B1 = 180 rad/sec
<u>b) Determine the force exerted on B2 at P</u>
T2 = mB1g + T1 -------- ( 1 )
where ; T1 = mB2g ( at point p )
= 2 * 9.81 = 19.6 N
back to equation 1
T2 = (2 * 9.8 ) + 19.6 = 39.2 N
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Answer:
80 meters high
Explanation:
The velocity of the balloon would be g*t (I won't calculate, but will us this later)
We know that the kinetic energy at the bottom equals the potential at the top.
KE = PE
1/2 * m * v^2 = m * g * h
1/2 * m * (g * t)^2 = m * g * h (substitution)
1/2 * m * g^2 * t^2 = m * g * h
1/2 * g * t^2 = h (simplification by dividing the commons between both sides)
h = 1/2 * 9.81 * 4^2
h = 78.48 m (roughly 80 m)
Given:
Dy= 20 m
Vi = 5.0 m/s horizontally
A=9.81 m/s^2
Find:
Horizontal displacement
Solution:
D=ViT+(1/2)AT^2
Dy=(1/2)AT^2
T^2=Dy/(1/2)A
T=sqrt(Dy/(1/2)A)
T=sqrt(20/4.905)
T=2.0s
Dx=ViT
Dx=(5.0)(2.0)
Dx=10. meters
Answer:
Law of conservation of momentum states that. For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed.
Explanation:
Hope it helps
