Answer:
c) The wavelength decreases but the frequency remains the same.
Explanation:
Light travels at different speed in different mediums.
Refractive index is equal to velocity of the light 'c' in empty space divided by the velocity 'v' in the substance.
Or ,
n = c/v.
<u>The frequency of the light does not change but the wavelength of the light changes with change in the speed.</u>
c = frequency × Wavelength
Frequency is constant,
The formula can be written as:
n = λ / λn.
Where,
λn is the wavelength in the medium
λ is the wavelength in vacuum
<u>When the light travels to glass, it speed slows down and also the wavelength decreases as both are directly proportional. There will be no effect on frequency.</u>
Answer:
1) El diámetro es de aproximadamente 913,987 cm.
2) La fuerza del cilindro es 5576850 kgf
Explanation:
1) Los parámetros dados son;
El volumen del aire = 13,122 litros = 13122000 cm³
La presión de trabajo = 8.5 kgf / cm²
La longitud del cilindro = 20 cm.
Por lo tanto, tenemos;
El área de la base del cilindro = π · r² = 13122000 cm³ / (20 cm) = 656100 cm²
r = √ (656100 / π) ≈ 456,994 cm
El diámetro = 2 × r ≈ 2 × 456.994 ≈ 913.987 cm
El diámetro ≈ 913,987 cm
2) La fuerza del cilindro = El área de la base del cilindro × La presión de trabajo
∴ La fuerza del cilindro = 656100 cm² × 8.5 kgf / cm² = 5576850 kgf
La fuerza del cilindro = 5576850 kgf
Answer:
i)-6.25m/s
ii)18 metres
iii)26.5 m/s or 95.4 km/hr
Explanation:
Firstly convert 90km/hr to m/s
90 × 1000/3600 = 25m/s
(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)
0 = (25)^2 + 2A(50)
0 = 625 + 100A....then moved the other value to one
-625 = 100A
Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)
(ii) Firstly convert 54km/hr to m/s
In which this is 54 × 1000/3600 = 15m/s
then apply the same formula as that in (i)
0 = (15)^2 + 2(-6.25)s
-225 = -12.5s
Hence the stopping distance = 18metres
(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question
0 = u^2 + 2(-6.25)(56)
u^2 = 700
Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s
In km/hr....26.5 × 3600/1000 = 95.4 km/hr
