How is force, work, and distance the related
1 answer:
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Answer:
The velocity of the launcher after the projectile is launched is -5.011 m/s
Explanation:
Here we have the mass of the cannon and cart, m₁ = 4.65 kg
Velocity of cannon and cart, v₁ = 2.00 m/s
Mass of projectile, m₂ = 50.0 g = 0.05 kg
Velocity of projectile, v₂ = 647 m/s
Velocity of the launcher, v₃ = Required
Mass of cannon and cart, launcher after launching projectile m₃ = 4.65-0.05
= 4.6 kg
Therefore, from the principle of the conservation of linear momentum, we have
Total initial momentum = Total final momentum
m₁ × v₁ = m₂ × v₂ + m₃ × v₃
Substituting gives
4.65 kg × 2.00 m/s = 0.05 kg × 647 m/s + 4.6 kg × v₃
4.65 kg × 2.00 m/s - 0.05 kg × 647 m/s = 4.6 kg × v₃
-23.05 kg·m/s = 4.6 kg × v₃
v₃ = -5.011 m/s.
Answer: B) Zinc will act as Anode
Explanation:
Standard reduction potential of zinc and nickel are:
Here Zinc undergoes oxidation by loss of electrons, thus act as anode as it has more negative reduction potential.
Nickel undergoes reduction by gain of electrons and thus act as cathode. as it has less negative reduction potential.
Where both are standard reduction potentials.
As the emf is positive, the reaction is spontaneous and reaction will occur.
Thus Zinc will act as anode.
Answer:
The answers are
The average stress = 20000 lb/in²
The maximum tensile stress immediately adjacent to the hole
= 31076.92 lb/in²
Explanation:
To solve the question we have
Weight of tensile load = 30,000 lb
Width of steel bar = 4 in
Thickness of steel bar = 1/2 in
Average Stress = Force/Area
Size of hole drilled = 1.0 in diameter
Available width at cross section where the 1.00 in diameter hole is drilled =
(4 - 1) in = 3 inches
Cross sectional area at the point of reduced cross section due to the drilled hole = Width × Thickness (Since the item is a flat bar)
= 3 in × 1/2 in = 1.5 in²
Therefore Stress = (30000 lb)/(1.5 in²) = 20000 lb/in²
the maximum tensile stress immediately adjacent to the hole.
Bending stress = where
0.5*30000/I = 11076.92 lb/in²
Max stress = 31076.92 lb/in²