Answer:
None of the wavelength is in the visible range
Explanation:
Constructive interference of the reflected waves for different wavelengths can be estimated using:
λ = 2nd/m
where m is 1,2,3, ...
Therefore:
m=1, λ = 750 nm
m=2, λ = 750/2 = 375 nm
The limits of eye's sensitivity is between 430 nm and 690 nm. Beyond this range, the eye's sensitivity drops to approximately 1% of its maximum value.
Fc=mv^2/r so we get
2000kg*(25m/s)^2/(80m)= 15625N of force
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Answer:
a)ω₂=127.64 rad/s ( min)
ω₁= 333.32 rad/s ( max)
b) At d₂= 11.75 cm ,ω₂=127.64 rad/s
d₁=4.5 cm,ω₁= 333.32 rad/s
c)α = 7.1 x 10⁻³ rad/s²
Explanation:
Given that
r₂= 11.75 cm
r₁=4.5 cm
v= 7.5 m/s
a)
We know that
v=ω r
ω =Angular speed
r= radius
v= velocity
When d₂= 11.75 cm :
v=ω r
7.5 x 2 =ω₂ x 0.1175
ω₂=127.64 rad/s ( min)
N₂=1219.4 rpm
When d₁=4.5 cm :
v=ω r
7.5 x 2=ω₁ x 0.045
ω₁= 333.32 rad/s ( max)
N₁=3184.58 rpm
The average angular acceleration α given as
ω₁ = ω₂ + α t
333.32 = 127.64 + α x 8 x 60 x 60
α = 7.1 x 10⁻³ rad/s²
Answer:
The force exerted by the child is 38.25 Newton
Explanation:
We use the formula F=mxa (m=mass and a= aceleration):
F= 45kg x 0,85 m/s2=38, 25 kgxm/s2= <em>38, 25 N</em>
Answer:
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