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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
Question is missing. Found on google:
<em>"Part A What is the acceleration of the ball? Express your answer to two significant figures and include the appropriate units. </em>
<em>Part B </em>
<em>What is the net force on the ball during the hit? </em>
<em>Express your answer to two significant figures and include the appropriate units."</em>
Solution:
A)
The acceleration of the ball is given by
where
v = 12 m/s is the final velocity
u = 0 is the initial velocity (the ball is stationary)
t = 2.0 ms = 0.002 s is the time of contact
Substituting,
B)
The force on the ball can be found by using Newton's second law:
where
m = 140 g = 0.14 kg is the mass of the ball
is the acceleration
Substituting,