I believe that a light bulb releases visible light and a radio antenna releases a radio waves
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
Answer:
λ = 5940 Angstroms
Explanation:
This is an exercise of the relativistic Doppler effect
f’= f √((1- v / c) / (1 + v / c))
Where the speed in between the strr and the observer is positive if they move away
Let's use the relationship
c = λ f
f = c /λ
We replace
c /λ’ = c /λ √ ((1- v / c) / (1 + v / c))
λ = λ’ √ ((1- v / c) / (1 + v / c))
Let's calculate
v = 0.01 c
v = 0.01 3 10⁸
v= 3 10⁶ m / s
λ = 6000 √ [(1- 3 10⁶/3 10⁸) / (1+ 3 10⁶/3 10⁸)]
λ = 6000 √ [0.99 / 1.01]
λ = 5940 Angstroms
13-16 that is where they’re located at
The S.I. unit for the measure of the pressure is the Pascal (Pa). 1 Pascal corresponds to
We can convert the number given by the problem into Pascal:
And since
, we have
