I think this is false
hope this helps...
Answer:
The correct option is A.
Explanation:
Following the equation of continuum, AV remains constant.
Case a
(3A)(V0) = AV1 + AV1 + AV1
3AV0 = 3AV1
V1 = V0
Case b
(A)(V0) = (A/3)V2 + (A/3)V2 + (A/3)V2 + (A/3)V2
AV0 = 4V2/3
V2 = 3/4V0
Case c
(A/2)(V0) = AV3 + AV3 + AV3
AV0/2 = 3AV3
V3 = V0/6
Case d
(3A)(V0) = 2AV4 + 2AV4
3AV0 = 4AV4
V4 = 3V0/4
Comparing all the flow speeds, V1 is the largest.
Thus, the correct option is A.
Answer:
(a) 1.2 rad/s
(b) 1.8 rad
Explanation:
Applying,
(a) α = (ω-ω')/t................ Equation 1
Where α = angular acceleration, ω = final angular velocity, ω' = initial angular velocity, t = time.
From the question,
Given: α = 0.40 rad/s², t = 3 seconds, ω' = 0 rad/s (from rest)
Substitute these values into equation 1
0.40 = (ω-0)/3
ω = 0.4×3
ω = 1.2 rad/s
(b) Using,
∅ = ω't+αt²/2.................. Equation 2
Where ∅ = angle turned.
Substitutting the values above into equation 2
∅ = (0×3)+(0.4×3²)/2
∅ = 1.8 rad.
Answer:
Answered
Explanation:
A) The work done by gravity is zero because displacement and the gravitational force are perpendicular to each other.
W= FS cosθ
θ= 90 ⇒cos90 = 0 ⇒W= 0
B) work done by tension
W= Tcosθ×S= 5cos30×2.30= 10J
C) Work done by friction force
W= f×s=1×2.30= 2.30 J
D) Work done by normal force is Zero because the displacement and the normal force are perpendicular to each other.
E) The net work done= Work done by tension in the rope - frictional work
=10-2.30= 7.7 J
