m = mass of the cat being lifted = 5 kg
h = height to which the cat is raised into air = 2 m
g = acceleration due to gravity of earth = 9.8 m/s²
U = gravitational potential energy gained by the cat
Gravitational potential energy gained by the cat is given as
U = m g h
inserting the values in the above equation
U = (5 kg) (9.8 m/s²) (2 m)
U = (5 x 9.8 x 2) (kg m²/s²)
U = 98 J
Answer:
λ ≈ 462 nm
Explanation:
Given:-
- The slit separation, a = 10^-5 m
- The first bright fringe occurs at, y = 0.03 m from central order
- The distance between the screen and slit, L = 0.65 m
Find:-
what is the wavelength of this light?
Solution:-
- The Young's double slit experiment gives us the relation for the interaction of a light with wavelength ( λ ) that pass through 2-slits with separation ( a ) and interfere constructively or destructively at the the screen Hence, forming an interference pattern of bright and dark fringes.
- The relation for separation ( y ) between bright fringes is related to the above mentioned parameters as given below:
y = L*n*λ / a
Where,
n: The order of fringe from central
Re-arrange for wavelength ( λ ).
- The order of first bright fringe is n = 1:
λ = y*a / L
λ = 0.03*0.00001 / 0.65
λ = 4.61538 * 10^-7 m
λ ≈ 462 nm
Answer:
the index of refraction for the transparent material is 1.48
Explanation:
let n1 = 1.0 be the refractive index of air and n2 be the refractive index of the trasnparent material.
then:
Snell's Law state that:
n1×sin(∅1) = n2×sin(∅2)
where ∅1 = 56.0° is the angle of incidence and ∅2 = 90 - 56.0 = 34° be the angle refraction.
n1×sin(∅1) = n2×sin(∅2)
n2 = n1×sin(∅1)/sin(∅2)
= (1.0)×sin(56.0°)/sin(34.0°)
= 1.48
Therefore, the index of refraction for the transparent material is 1.48
Answer:
• As heat is applied to one end of this material, atoms in the hotter region gain vibràtory energy at a maximum amplitude. They transfer it on to their neighboring atoms and heat is transfered along this material in form of vibràtory energy.
Explanation:
