Physics help please..............
1 answer:
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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:
It must be 4 times high.
Explanation:
- Assuming that the car can be treated as a point mass, and that the ramp is frictionless, the total mechanical energy must be conserved.
- This means, that at any time, the following must be true:
- ΔK (change in kinetic energy) = ΔU (change in gravitational potential energy)
⇒
- Let's call v₁, to the final speed of the car, and h₁ to the height of the ramp.
So, at the bottom of the ramp, all the gravitational potential energy
must be equal to the kinetic energy of the car (Defining the bottom of
the ramp as our zero reference for the gravitational potential energy):
(1)
- Now, let's do v₂ = 2* v₁
- Replacing in (1) we get:
(2)
- Dividing (2) by (1), and rearranging terms, we get:
- h₂ = 4* h₁