Answer:
2633.7 s
Explanation:
From the question,
Heat lost by the water heater = Heat gained by the water
Applying,
P = cm(t₂-t₁)/t.................. Equation 1
Where P = power of the heat, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature, t = time
make t the subject of the equation
t = cm(t₂-t₁)/P.............. Equation 2
From the question,
Given: c = 4190 J/kgK, P = 3.5 kW = 3500 W, m = 40 kg, t₁ = 20°C, t₂ = 75°C
Substitute these values into equation 2
t = 4190×40(75-20)/3500
t = 9218000/3500
t = 2633.7 s
If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:
Fg=Fcp
m*g=m*(v²/R),
where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite.
Masses cancel out and we have:
G*(M/r²)=v²/R, R=r so:
G*(M/r)=v²
r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,
r=G*(M/ω²r²),
r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:
r³=G*(M/(4π²/T²)), and finally we take the third root to get r:
r=∛{(G*M*T²)/(4π²)}=4.226*10^7 m= 42 260 km which is the height of a geostationary satellite.
Since each student emits 100 W, so 170 students will emit:
total heat = 100 W * 170 = 17,000 W
Convert minutes to seconds:
time = 50 min * (60 s / min) = 3000 s
The energy is therefore:
E = 17,000 W * 3000 s
<span>E = 51 x 10^6 J = 51 MJ</span>
Answer:
The highest vertical position is where your maximum potential energy lies. At the highest altitude point of course ! This is when the kinetic energy is only due to horizontal motion (since the vertical component reaches zero).
Explanation:
i looked it up ok
