By how much does the pressure of a gas in a rigid vessel decrease when the temperature drops from 0 ∘C to –1∘C?
1 answer:
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Answer:
λ = 482.05 nm
Explanation:
The diffraction phenomenon and the diffraction grating is described by the expression
d sin θ = m λ
where d is the distance between two consecutive slits, λ the wavelength and m an integer representing the order of diffraction
in this case they indicate the distance between slits, the angle and the order of diffraction
λ = d sin θ / m
let's calculate
λ = 1.00 10⁻⁶ sin 74.6 / 2
λ = 4.82048 10⁻⁷ m
Let's reduce to nm
λ = 4.82048 10⁻⁷ m (10⁹ nm / 1 m)
λ = 482.05 nm
Answer:
52.5°C
Explanation:
The final enthalpy is determined from energy balance where initial enthalpy and specific volume are obtained from A-12 for the given pressure and state
mh1 + W = mh2
h2 = h1 + W/m
h1 + Wα1/V1
242.9 kJ/kg + 2.35.0.11049kJ/ 0.35/60kg
=287.4 kJ/kg
From the final enthalpy and pressure the final temperature is obtained A-13 using interpolation
i.e T2 = T1 + T2 -T1/h2 -h1(h2 - h1)
= 50°C + 60 - 50/295.15 - 284.79
(287.4 - 284.79)°C
= 52.5°C
Answer:
t = 1.58 s
Explanation:
given,
Speed of ranger, v = 56 km/h
v = 56 x 0.278 = 15.57 m/s
distance, d = 65 m
deceleration,a = 3 m/s²
reaction time = ?
using stopping distance formula
t is the reaction time
t = 1.58 s
hence, the reaction time of the ranger is equal to 1.58 s.