Answer:
the thickness of the glass divided by thickness of water is going to be 1.333 divided by 1.52, which is 0.877. So, the height of this glass, in order to have the same number of wavelengths as in water, the height of the glass will be 0.877 times the height of the water, and so it will be smaller.
Answer:
That is not true all objects fall at the same speed excepts things like feathers or paper.
Answer:
Explanation:
This is a problem based on time dilation , a theory given by Albert Einstein .
The formula of time dilation is as follows .
t₁ = ![\frac{t}{\sqrt{1-\frac{v^2}{c^2} } }](https://tex.z-dn.net/?f=%5Cfrac%7Bt%7D%7B%5Csqrt%7B1-%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%20%7D%20%7D)
t is time measured on the earth and t₁ is time measured by man on ship .
A ) Given t = 20 years , t₁ = ? v = .4c
![\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }](https://tex.z-dn.net/?f=%5Cfrac%7B20%7D%7B%5Csqrt%7B1-%5Cfrac%7B.16c%5E2%7D%7Bc%5E2%7D%20%7D%20%7D)
=1.09 x 20
t₁= 21.82 years
B ) Given t = 5 years , t₁ = ? v = .2c
![\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B%5Csqrt%7B1-%5Cfrac%7B.04c%5E2%7D%7Bc%5E2%7D%20%7D%20%7D)
=1.02 x 5
t₁= 5.1 years
C ) Given t = 10 years , t₁ = ? v = .8c
![\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }](https://tex.z-dn.net/?f=%5Cfrac%7B10%7D%7B%5Csqrt%7B1-%5Cfrac%7B.64c%5E2%7D%7Bc%5E2%7D%20%7D%20%7D)
=1.67 x 10
t₁= 16.7 years
D ) Given t = 10 years , t₁ = ? v = .4c
![\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }](https://tex.z-dn.net/?f=%5Cfrac%7B10%7D%7B%5Csqrt%7B1-%5Cfrac%7B.16c%5E2%7D%7Bc%5E2%7D%20%7D%20%7D)
=1.09 x 10
t₁= 10.9 years
E ) Given t = 20 years , t₁ = ? v = .8c
![\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }](https://tex.z-dn.net/?f=%5Cfrac%7B20%7D%7B%5Csqrt%7B1-%5Cfrac%7B.64c%5E2%7D%7Bc%5E2%7D%20%7D%20%7D)
=1.67 x 20
t₁= 33.4 years
Answer:
442.5 rad
Explanation:
w₀ = initial angular velocity of the disk = 7.0 rad/s
α = Constant angular acceleration = 3.0 rad/s²
t = time period of rotation of the disk = 15 s
θ = angular displacement of the point on the rim
Angular displacement of the point on the rim is given as
θ = w₀ t + (0.5) α t²
inserting the values
θ = (7.0) (15) + (0.5) (3.0) (15)²
θ = 442.5 rad