3.1 Punktmasse

3.1.1 Kinematik

3.1.1.A (Master)

3.1.1.G

3.1.1.D

3.1.1.E

3.1.1.C

3.1.1.K

3.1.1.I

3.1.1.J

3.1.1.L

3.1.2 Energie & Arbeit

3.1.2.A

3.1.2.B

3.1.2.C

3.1.2.D (Master)

Etwas Interessantes zum Thema Energie und Arbeit

Wieviel Energie ist das?

Erwärmen von kaltem Wasser für eine ausgiebige heiße Dusche.
Mit dem Partner auf dem Rücken von Meereshöhe auf den Mount Everest steigen.
Energie eines Balkonkraftwerks mit 850 W Peak Power an einem schönen Sommertag.

SymPy

Source Code eines Programms dazu:

from sympy import pprint
from sympy.physics.units import kg, K, J, m, s


print("\n1 J = 1 Ws = 1 Nm")
print("1 kJ = 1 kWs = 1 kNm")
print("3600 kJ = 1 kWh = 3600 kNm")

k = 1000
N = kg*m/s**2
Nm = N*m
kJ = k*J        # = kWs
kWh = 3600 * kJ
kNm =  k*Nm

# Rounding:
def iso_round(x, pv):
    try:
        x = float(x)
        tmp = round(x, pv)
    except TypeError:
        from functools import partial
        func = partial(round, ndigits = pv)
        tmp = x.applyfunc(func)
    return tmp

print("\n1. Enjoying a hot shower (80 l):")
print("Energy in kWh to heat 80 kg of water by 40 K (from 10°C to 50°C):")
mass = 80 *kg
c = 4.18 * kJ / kg / K
dT = 40 *K
tmp = c * mass * dT
tmp /= kWh
tmp = iso_round(tmp, 2)
pprint(tmp)

print("\n2. Climbing Mount Everest with your partner on your back:")
print("Energy in kWh to lift 2 x 80 kg by 8849 m:")
mass = 2 * 80 *kg
g = 9.81 * m/s**2
h = 8849 * m
tmp = mass*g*h
tmp /= 3600 * kNm
tmp = iso_round(tmp, 2)
pprint(tmp)

print("\n3. Using a small PV system (850 W Peak Power) in the summer:")
print("Energy in kWh produced per day by PV system:")
tmp = 6 *kWh
tmp /= kWh
tmp = iso_round(tmp, 2)
pprint(tmp)