Answer:
month = input("Input the month (e.g. January, February etc.): ")
day = int(input("Input the day: "))
if month in ('January', 'February', 'March'):
season = 'winter'
elif month in ('April', 'May', 'June'):
season = 'spring'
elif month in ('July', 'August', 'September'):
season = 'summer'
else:
season = 'autumn'
if (month == 'March') and (day > 19):
season = 'spring'
elif (month == 'June') and (day > 20):
season = 'summer'
elif (month == 'September') and (day > 21):
season = 'autumn'
elif (month == 'December') and (day > 20):
season = 'winter'
print("Season is",season)
Explanation:
Answer:
16-bit wide
Explanation:
In order to find the width of the address bus, we need first to know how many memory cells it is needed to address.
If the size memory is 64 KB, this means that the memory size, in bytes, is equal to the following quantity:
64 KB = 2⁶ * 2¹⁰ bytes = 2¹⁶ bytes.
In order to address this quantity of cell positions, the address bus must be able to address 2¹⁶ bytes, so it must have 16-bit wide.
Answer:
50421.6 m³
Explanation:
The river has an average rate of water flow of 59.6 m³/s.
Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:
Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s
The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken
time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds
The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³
Answer:
false jdbebheuwowjwjsisidhhdd
Answer:
a) V = 0.354
b) G = 25.34 GPA
Explanation:
Solution:
We first determine Modulus of Elasticity and Modulus of rigidity
Elongation of rod ΔL = 1.4 mm
Normal stress, δ = P/A
Where P = Force acting on the cross-section
A = Area of the cross-section
Using Area, A = π/4 · d²
= π/4 · (0.0020)² = 3.14 × 10⁻⁴m²
δ = 50/3.14 × 10⁻⁴ = 159.155 MPA
E(long) = Δl/l = 1.4/600 = 2.33 × 10⁻³mm/mm
Modulus of Elasticity Е = δ/ε
= 159.155 × 10⁶/2.33 × 10⁻³ = 68.306 GPA
Also final diameter d(f) = 19.9837 mm
Initial diameter d(i) = 20 mm
Poisson said that V = Е(elasticity)/Е(long)
= - <u>( 19.9837 - 20 /20)</u>
2.33 × 10⁻³
= 0.354,
∴ v = 0.354
Also G = Е/2. (1+V)
= 68.306 × 10⁹/ 2.(1+ 0.354)
= 25.34 GPA
⇒ G = 25.34 GPA
