The answer to your question is Meiosis.
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-vf
Answer:
48
Explanation:
you basically divide 1200 into 25
Incomplete Question.The Complete question is
The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass of the Earth: 5.97 × 10^24 kg (assume a uniform mass distribution) Radius of the Earth: 6371 km Distance of Earth from Sun: 149,600,000 km
(i)Calculate the rotational kinetic energy of the Earth due to rotation about its axis, in joules.
(ii)What is the rotational kinetic energy of the Earth due to its orbit around the Sun, in joules?
Answer:
(i) KE= 2.56e29 J
(ii) KE= 2.65e33 J
Explanation:
i) Treating the Earth as a solid sphere, its moment of inertia about its axis is
I = (2/5)mr² = (2/5) * 5.97e24kg * (6.371e6m)²
I = 9.69e37 kg·m²
About its axis,
ω = 2π rads/day * 1day/24h * 1h/3600s
ω= 7.27e-5 rad/s,
so its rotational kinetic energy
KE = ½Iω² = ½ * 9.69e37kg·m² * (7.27e-5rad/s)²
KE= 2.56e29 J
(ii) About the sun,
I = mR²
I= 5.97e24kg * (1.496e11m)²
I= 1.336e47 kg·m²
and the angular velocity
ω = 2π rad/yr * 1yr/365.25day * 1day/24h * 1h/3600s
ω= 1.99e-7 rad/s
so
KE = ½ * 1.336e47kg·m² * (1.99e-7rad/s)²
KE= 2.65e33 J
Answer:
Cp = 4756 [J/kg*°C]
Explanation:
In order to calculate the specific heat of water, we must use the equation of energy for heat or heat transfer equation.
Q = m*Cp*(T_f - T_i)/t
where:
Q = heat transfer = 2.6 [kW] = 2600[W]
m = mass of the water = 0.8 [kg]
Cp = specific heat of water [J/kg*°C]
T_f = final temperature of the water = 100 [°C]
T_i = initial temperature of the water = 18 [°C]
t = time = 120 [s]
Now clearing the Cp, we have:
Cp = Q*t/(m*(T_f - T_i))
Now replacing
Cp = (2600*120)/(0.8*(100-18))
Cp = 4756 [J/kg*°C]