Answer:
Explanation:
The most important thing to remember about parabolic motion in physics is that when an object reaches its max height, the velocity right there at the highest point is 0. Use this one-dimensional motion equation to solve this problem:
v = v₀ + at and filling in:
0 = v₀ + (-9.8)(4.0) **I put in 4.0 for time so we have more than just 1 sig fig here**
0 = v₀ - 39 and
-v₀ = -39 so
v₀ = 39 m/s
' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.
That's all the physics we need to know to answer this question.
The rest is just arithmetic.
(60 joules/sec) · (30 days) · (8 hours/day) · (3600 sec/hour)
= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)
= 51,840,000 joules
__________________________________
Wait a minute ! Hold up ! Hee haw ! Whoa !
Excuse me. That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's
(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)
= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)
= 14.4 kW·hour
Rounded to the nearest whole number:
14 kWh
1)
HCl: hydrogen, chloride
3CO2: carbon, oxygen
2Na2SO4:sodium, sulphur, oxygen.
2)
-HCl: 1 hydrogen atom, 1 chlorine atom
-CO2: 1 carbon atom, 2 oxygen atoms
-Na2SO4: 2 sodium atoms, 1 sulphur atom, 4 oxygen atoms.
3)
-HCl: 2 atoms
-3CO2: 9 atoms
-2Na2SO4: 14 atoms.
Answer:
The angular momentum of a cylinder, when it is rotating with constant angular velocity is Lini =Iωi
. When two cylinders are added to the rotating cylinder, which are identical in their dimensions, the moment of inertia of the entire system increases (since mass increases). The final moment of inertia will be 3I
Since friction exist, all the cylinders start rotating with same angular velocity, the new angular velocity can be calculated using conservation of angular momentum
Thus, Iωi =3Iωf ⟹ωf =ωi/3 = 0.33ωi
Answer:
A. the left half becomes neutral while the right half remains negatively charged
Explanation:
This is because wherever light strikes the photoconductor, it transforms from an insulator into a conductor. The charge will then migrate through it and leaves its surface. By exposing the left half of the photoconductor to light, you allow its local charge to leave and it becomes neutral.
