Answer:
D
Explanation:
as one burns then all stop simple
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Answer: A surface wave is a wave that moves along the interface of two different materials, like air and water.
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
<u />
Answer:
114.44 J
Explanation:
From Hook's Law,
F = ke................. Equation 1
Where F = Force required to stretch the spring, k = spring constant, e = extension.
make k the subject of the equation
k = F/e.............. Equation 2
Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.
Substitute into equation 2
k = 44.5/1.016
k = 43.799 N/m
Work done in stretching the 9 in beyond its natural length
W = 1/2ke²................. Equation 3
Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m
Substitute into equation 3
W = 1/2×43.799×2.286²
W = 114.44 J
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