<span>b. less climatic variation between the summer and winter seasons in the middle and high latitudes
As the tilt becomes higher (approaches 24 degrees) there is greater variation between the summer and winter months, due to the fact that the tilt toward the sun in the summer and away from the sun in the winter are more pronounced. </span>
Solution:
We have,
Power [P] = 25000 Watt
Mass [m] = 6000 kg
Height [h] = 20 metres
Time [t] = ?
Now,
P = W/t = F x d/t = mxgx h/t
Or, 25000 = 6000 x 10 x 20/25000 [.......g = 10
m/s^2]
Or, t = 6000 x 10 x 20/25000
Or, t = 1200/25
Therefore, t = 48 second
Hence, the required time for the crane to lift the load is 48 seconds.
The answer should be B. Fault.
Answer:
d. changing refractive index of gases in the atmosphere
Explanation:
light from a star passes through the atmosphere to reach our eyes. however, the layers of gases in the atmosphere are in constant motion, so the light gets bent and appears to be flickering.
i hope this helps! :D
Answer:
Broglie wavelength: electron 1.22 10⁻¹⁰ m
, proton 2.87 10⁻¹² m
, hydrogen atom 7.74 10⁻¹² m
Explanation:
The equation given by Broglie relates the momentum of a particle with its wavelength.
p = h /λ
In addition, kinetic energy is related to the amount of movement
E = ½ m v²
p = mv
E = ½ p² / m
p = √2mE
If we clear the first equation and replace we have left
λ = h / p =
λ = h / √2mE
Let's reduce the values that give us SI units
1 ev = 1,602 10⁻¹⁹ J
E1 = 100 eV (1.6 10⁻¹⁹ J / 1eV) = 1.6 10⁻¹⁷ J
We look in tables for the mass of the particle and the Planck constant
h = 6,626 10-34 Js
me = 9.1 10-31 Kg
mp = 1.67 10-27 Kg
Now let's replace and calculate the wavelengths
a) Electron
λ1 = 6.6 10⁻³⁴ / √(2 9.1 10⁻³¹ 1.6 10⁻¹⁷) = 6.6 10⁻³⁴ / 5.39 10⁻²⁴
λ1 = 1.22 10⁻¹⁰ m
b) Proton
λ2 = 6.6 10-34 / √(2 1.67 10⁻²⁷ 1.6 10⁻¹⁷) = 6.6 10⁻³⁴ / 2.3 10⁻²²
λ2 = 2.87 10⁻¹² m
c) Bohr's first orbit
En = 13.606 / n2 [eV]
n = 1
E1 = 13.606 eV
E1 = 13,606 ev (1.6 10⁻¹⁹ / 1eV) = 21.77 10⁻¹⁹ J
λ3 = 6.6 10⁻³⁴ /√(2 1.67 10⁻²⁷ 21.77 10⁻¹⁹) = 6.6 10⁻³⁴ / 8.52 10⁻²³
λ3 = 0.774 10⁻¹¹ m = 7.74 10⁻¹² m
