Answer:
f'=5.58kHz
Explanation:
This is an example of the Doppler effect, the formula is:

Where f is the actual frequency,
is the observed frequency,
is the velocity of the sound waves,
the velocity of the observer (which is negative if the observer is moving away from the source) and
the velocity of the source (which is negative if is moving towards the observer). For this problem:


Answer:
103.57 Km/h
Explanation:
From the question given above, the following data were obtained:
Distance = 725 Km
Time = 7 hours
Speed =?
Speed can be defined as the distance travelled per unit time. Mathematically, it is expressed as:
Speed = Distance /time
With the above formula, we can calculate how fast he will drive (i.e the speed) in order to get there on time. This is illustrated below:
Distance = 725 Km
Time = 7 hours
Speed =?
Speed = Distance /time
Speed = 725 / 7
Speed = 103.57 Km/h
Thus, to get there on time, he will drive with a speed of 103.57 Km/h
Answer:
1 cm⁻¹ =1.44K 1 ev = 1.16 10⁴ K
Explanation:
The relationship between temperature and thermal energy is
E = K T
The relationship of the speed of light
c =λ f = f / ν 1/λ= ν
The Planck equation is
E = h f
Let's start the transformations
c = f λ = f / ν
f = c ν
E = h f
E = h c ν
E = KT
h c ν = K T
T = h c ν / K =( h c / K) ν
Let's replace the constants
h = 6.63 10⁻³⁴ J s
c = 3 10⁸ m / s
K = 1.38 10⁻²³ J / K
v = 1 cm-1 (100 cm / 1 m) = 10² m-1
T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 1 10²
A = h c / K = 1,441 10⁻²
T = 1.44K
ν = 103 cm⁻¹ = 103 10² m
T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 103 10²
T = 148K
1 Rydberg = 1.097 10 7 m
As we saw at the beginning the λ=1 / v
T = (h c / K) 1 /λ
T = 1,441 10⁻² 1 / 1,097 10⁷
T = 1.3 10⁻⁹ K
E = 1Ev (1.6 10⁻¹⁹ J /1 eV) = 1.6 10⁻¹⁹ J
E = KT
T = E/K
T = 1.6 10⁻¹⁹ /1.38 10⁻²³
T = 1.16 10⁴ K
Answer:
The final speed of the crate is 12.07 m/s.
Explanation:
For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:


Now, we can calculate the final speed of the crate at the end of 10.0 m:
For the next 10.5 meters we have frictional force:


So, the acceleration is:
The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:
Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.
I hope it helps you!