F=mv^2/R
----> V^2=FR/m=(350x0.9)/2.5=126
----- V=11.22 m/s
Answer with Explanation:
We are given that
Angle of incidence,
Angle of refraction,
a.Refractive index of air,
We know that


b.Wavelength of red light in vacuum,

Wavelength in the solution,

c.Frequency does not change .It remains same in vacuum and solution.
Frequency,
Where 
Frequency,
d.Speed in the solution,

Answer:
T = 712.9 N
Explanation:
First, we will find the speed of the wave:
v = fλ
where,
v = speed of the wave = ?
f = frequency = 890 Hz
λ = wavelength = 0.1 m
Therefore,
v = (890 Hz)(0.1 m)
v = 89 m/s
Now, we will find the linear mass density of the wire:

where,
μ = linear mass density of wie = ?
m = mass of wire = 90 g = 0.09 kg
L = length of wire = 1 m
Therefore,

μ = 0.09 kg/m
Now, the tension in wire (T) will be:
T = μv² = (0.09 kg/m)(89 m/s)²
<u>T = 712.9 N</u>
Answer:
KE₂ = 6000 J
Explanation:
Given that
Potential energy at top U₁= 7000 J
Potential energy at bottom U₂= 1000 J
The kinetic energy at top ,KE₁= 0 J
Lets take kinetic energy at bottom level = KE₂
Now from energy conservation
U₁+ KE₁= U₂+ KE₂
Now by putting the values
U₁+ KE₁= U₂+ KE₂
7000+ 0 = 1000+ KE₂
KE₂ = 7000 - 1000 J
KE₂ = 6000 J
Therefore the kinetic energy at bottom is 6000 J.