The 3003 aluminum alloy is made up of 1.25% Magnesium and 0.1% Copper. This combination is designed to increase the strength of the material over other types of alloys such as those of the 1000 series. This alloy provides a medium strength and can be educated by cold work.
The alloy is not heat treatable and generally has good formability, corrosion resistance and weldability.
However, being a material that hardens by cold work, welding a 3003 Aluminum structure will cause the body to undergo recrystallization which will generate a loss in the 'resistance' of the material and the force capable of withstanding. If this aluminum will be used for structural purposes, it should not be welded. It would be better to perform the structure with a 6061 aluminum, which has similar characteristics and is not so affected by welding.
Answer:
the average force 11226 N
Explanation:
Let's analyze the problem we are asked for the average force, during the crash, we can find this from the impulse-momentum equation, but this equation needs the speeds and times of the crash that we could look for by kinematics.
Let's start looking for the stack speeds, it has a free fall, from rest (Vo=0)
Vf² = Vo² - 2gY
Vf² = 0 - 2 9.8 7.69 = 150.7
Vf = 12.3 m / s
This is the speed that the battery likes when it touches the beam. They also give us the distance it travels before stopping, let's calculate the time
Vf = Vo - g t
0 = Vo - g t
t = Vo / g
t = 12.3 / 9.8
t = 1.26 s
This is the time to stop
Now let's use the equation that relates the impulse to the amount of movement
I = Δp
F t = pf-po
The amount of final movement is zero because the system stops
F = - po / t
F = - mv / t
F = - 1150 12.3 / 1.26
F = -11226 N
This is the average force exerted by the stack on the vean
A and C Im pretty sure :)
Answer:
The correct answer is Dean has a period greater than San
Explanation:
Kepler's third law is an application of Newton's second law where the force is the universal force of attraction for circular orbits, where it is obtained.
T² = (4π² / G M) r³
When applying this equation to our case, the planet with a greater orbit must have a greater period.
Consequently Dean must have a period greater than San which has the smallest orbit
The correct answer is Dean has a period greater than San
