A large air-filled 0.221-kg plastic ball is thrown up into the air with an initial speed of 11.9 m/s. At a height of 2.77 m, the
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Answer:
A)3.8196 * Newtons B) 2.153 *
Newtons
Explanation:
Force = Pressure * Area.
Pressure is given as 120GPa
Area = Cross Sectional Area of Rod = Area Of A circle = π.
Area Expansivity β =
Area Expansivity β = 2α
α = 16.9x10-6/ ° C.
Radius of Rod AB = Half 0f Diameter, 200mm = = 0.1 m
Radius of Rod BC = Half Of Diameter, 150mm = = 0.075 m.
Note: To convert from millimeter to meter you divide by 1000.
Area of Rod AB = π * = 0.0314
Area of Rod BC = π * = 0.0177
.
Change in Area = 2α * Original Area * Temperature Rise. Therefore for
Rod AB = 2 * 16.9* * 0.0314 * 30 = 3.183 *
.
For Rod BC we have = 2 * 16.9* * 0.0177 * 30 = 1.794∈-5
.
The forces on each rods will be given by.
Force AB = Pressure * Area of Rod AB
= 120GPa * 3.183 * gives 3.8196 *
Newtons.
Force BC = Pressure * Area of Rod BC
= 120GPa * 1.794 * gives 2.153 *
Newtons
Answer:
A. α = - 1.047 rad/s²
B. θ = 14.1 rad
C. θ = 2.24 rev
Explanation:
A.
We can use the first equation of motion to find the acceleration:
where,
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s
t = time = 3 s
α = angular acceleration = ?
Therefore,
<u>α = - 1.047 rad/s²</u>
B.
We can use the second equation of motion to find the angular distance:
<u>θ = 14.1 rad</u>
C.
θ = (14.1 rad)(1 rev/2π rad)
<u>θ = 2.24 rev</u>