Answer:
Explanation:
For critical angle C the relation is
SinC = 1 / μ where μ is refractive index of denser medium with respect to rarer medium .
In case of A , refractive index of denser medium with respect to rarer medium .is equal to 1.33 / 1 = 1.33
In case of B , refractive index of denser medium with respect to rarer medium .is equal to 2.42 / 1 = 2.42
In case of C , refractive index of denser medium with respect to rarer medium .is equal to 2.42 / 1.33 = 1.82
In case of D , refractive index of denser medium with respect to rarer medium .is equal to 1.50 / 1.33 = 1.127
In case of D , refractive index is lowest , critical angle will be highest .
In case of B , refractive index of denser medium with respect to rarer medium .is highest so critical angle will be lowest.
Ranking from largest to smallest
D > A >C >B .
Power = Work done / time
Work done = Force * Distance
= 300 N * 1.5 m = 450 J
Power = 450 / 0.75 = 600 Watts.
Answer:
<h3>The answer is 0.47 kg</h3>
Explanation:
The mass of the object given it's momentum and velocity can be found by using the formula
where
p is the momentum
v is the velocity
We have
We have the final answer as
<h3>0.47 kg</h3>
Hope this helps you
It is an example of reactivity
iron usually rusts due to oxygen in the air
Answer:
5.23 C
Explanation:
The current in the wire is given by I = ε/R where ε = induced emf in the wire and R = resistance of wire.
Now, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = AΔB and A = area of loop and ΔB = change in magnetic field intensity = B₂ - B₁
B₁ = 0.670 T and B₂ = 0 T
ΔB = B₂ - B₁ = 0 - 0.670 T = - 0.670 T
A = πD²/4 where D = diameter of circular loop = 13.2 cm = 0.132 m
A = π(0.132 m)²/4 = 0.01368 m² = `1.368 × 10⁻² m²
ε = -ΔΦ/Δt = -AΔB/Δt = -1.368 × 10⁻² m² × (-0.670 T)/Δt= 0.9166 × 10⁻² Tm²/Δt
Now, the resistance R of the circular wire R = ρl/A' where ρ = resistivity of copper wire = 1.68 x 10⁻⁸ Ω.m, l = length of wire = πD and A' = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m
R = ρl/A' = 1.68 x 10⁻⁸ Ω.m × π × 0.132 m÷π(2.25 × 10⁻³ m)²/4 = 0.88704/5.0625 = 0.1752 × 10⁻² Ω = 1.752 × 10⁻³ Ω
So, I = ε/R = 0.9166 × 10⁻² Tm²/Δt1.752 × 10⁻³ Ω
IΔt = 0.9166 × 10⁻² Tm²/1.752 × 10⁻³ Ω = 0.5232 × 10 C
Since ΔQ = It = 5.232 C ≅ 5.23 C
So the charge is 5.23 C