Answer:
A key element is powering economies with clean energy, replacing polluting coal - and gas and oil-fired power stations - with renewable energy sources, such as wind or solar farms. This would dramatically reduce carbon emissions. Plus, renewable energy is now not only cleaner, but often cheaper than fossil fuels
Explanation:
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This question is incomplete, the complete question is;
A 1,040 N force is recorded on a hemispherical vane as it redirects a 2.5 cm- blade diameter water jet through a 180 angle.
Determine the velocity of the flowing water jet if the blade is assumed to be frictionless.
Answer: the velocity of the flowing water jet is 32.55 m/s assuming the blade is frictionless
Explanation:
Given that;
Force Ft = 1040 N
diameter d = 2.5 cm = 0.025 m
we know that; force acting on Hemispherical plate is;
Ft = 2δav²
where
a is area = π/4(0.025)²
δ is density of water = 1000 kg/m³
v is velocity = ?
now we substitute
1040 = 2 × 1000 × (π/4(0.025)²) × v²
1040 = 0.9817v²
v² = 1040 / 0.9817
v² = 1059.3867
v = √1059.3867
v = 32.5482 ≈ 32.55 m/s
Therefore the velocity of the flowing water jet is 32.55 m/s assuming the blade is frictionless
Solution :
The nuclear reaction for boron is given as :
And the reaction for Cadmium is :
![$^{113}\textrm{Cd}_48 + ^{1}\textrm{n}_0 \rightarrow ^{114}\textrm{Cd}_48 + \gamma [5 \ \textrm{MeV}]$](https://tex.z-dn.net/?f=%24%5E%7B113%7D%5Ctextrm%7BCd%7D_48%20%2B%20%5E%7B1%7D%5Ctextrm%7Bn%7D_0%20%5Crightarrow%20%5E%7B114%7D%5Ctextrm%7BCd%7D_48%20%2B%20%5Cgamma%20%5B5%20%5C%20%5Ctextrm%7BMeV%7D%5D%24)
We know that it is easier that to shield or stop an alpha particle (i.e. He nucli) as they can be stopped or obstructed by only a few centimetres of the material. However, the gamma rays ( γ ) can penetrate through the material to a greater distance. Therefore, we can choose the first one.
Explanation & answer:
Assuming a smooth transition so that there is no abrupt change in slopes to avoid frictional loss nor toppling, we can use energy considerations.
Initially, the cube has a kinetic energy of
KE = mv^2/2 = 10 lbm * 20^2 ft^2/s^2 / 2 = 2000 lbm-ft^2 / s^2
At the highest point when the block stops, the gain in potential energy is
PE = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2
By assumption, there was no loss in energies, we equate PE = KE
322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2
=>
h = 2000 /322 = 6.211 (ft)
distance up incline = h / sin(30) = 12.4 ft
