Answer:
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We know that
g = LcosΘ
<span>where g, L and Θ are centripetal gravity length, and angle of object
</span><span>ω² = g/LcosΘ </span>
<span>ω = √(g / LcosΘ) </span>
Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
Answer:
a) 
b) 
Explanation:
a)
Given:
amount of heat transfer occurred,
initial temperature of car, 
final temperature of car, 
We know that the change in entropy is given by:
(heat is transferred into the system of car)
b)
amount of heat transfer form the system of house, 
initial temperature of house, 
final temperature of house, 
Answer:
The magnitude of the tension in the cable, T is 1,064.315 N
Explanation:
Here we have
Length of beam = 4.0 m
Weight = 200 N
Center of mass of uniform beam = mid-span = 2.0 m
Point of attachment of cable = Beam end = 4.0 m
Angle of cable = 53° with the horizontal
Tension in cable = T
Point at which person stands = 1.50 m from wall
Weight of person = 350 N
Therefore,
Taking moment about the wall, we have
∑Clockwise moments = ∑Anticlockwise moments
T×sin(53) = 350×1.5 + 200×2
T = 850/sin(53) = 1,064.315 N.
