Lithium is more active than aluminium A.True B.false
2 answers:
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Answer:
a. The volume V₁ and V₂
b. The case that involves the greatest momentum change = Case B
c. The case that involves the greatest impulse = Case B
d. b. The case that involves the greatest force = Case B
Explanation:
Here we have
Case A: V₁ = 150-mL, v₁ = 8 m/s
Case B: V₂ = 600-mL, v₁ = 8 m/s
a. The variable that is different for the two cases is the volume V₁ and V₂
b. The momentum change is by the following relation;
ΔM₁ = Mass, m × Δv₁
The mass of the balloon are;
Δv₁ = Change in velocity = Final velocity - Initial velocity
Mass = Density × Volume
Density of water = 0.997 g/mL
Case A, mass = 150 × 0.997 = 149.55 g
Case B, mass = 600 × 0.997 = 598.2 g
The momentum change is;
Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s
Case B: Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s
Therefore Case B has the greatest momentum change
The case that has the gretest momentum change = Case B
c. The momentum change = impulse therefore Case B involves the greatest impulse
d. Here we have;
Impulse = Momentum change = × Δt = mΔV
∴ = m·ΔV/Δt
∴ For Case A = 149.55×8/Δt = 1196.4/Δt N
For Case B = 598.2×8/Δt = 4785.6/Δt
Where Δt is the same for Case A and Case B, for Case B >>
for Case B
Therefore, Case B involves the greatest force.
Complete Question
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to 1/5.
Answer:
The angle is
Explanation:
From the question we are told that
The light emerging from second Polaroid is 1/5 the unpolarized
Generally the intensity of light emerging from the first Polaroid is mathematically represented as
Generally from the Malus law the intensity of light emerging from the second Polaroid is mathematically represented
=>
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From the question
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Answer:
D) The worker will accelerate at 2.17 m/s² and the box will accelerate at 1.08 m/s² , but in opposite directions.
Explanation:
Newton's third law
Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.
F₁₂= -F₂₁
F₁₂: Force of the box on the worker
F₂₁: Force of the worker on the box
Newton's second law
∑F = m*a
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Formula to calculate the mass (m)
m = W/g
Where:
W : Weight (N)
g : acceleration due to gravity (m/s²)
Data
W₁ =1.8 kN : box weight
W₂ = 0.900 kN : worker weight
F₂₁ = 0.200 kN
F₁₂ = - 0.200 kN
g = 9.8 m/s²
Newton's second law for the box
∑F = m*a
F₂₁ = m₁*a₁ m₁=W₁/g
0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁
a₁= 1.08 m/s² : acceleration of the box
Newton's second law for the worker
∑F = m*a
F₁₂ = m₂*a₂ , m₂=W₂/g
- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂
a₂= -2.17 m/s² : acceleration of the worker