Answer:
How to find the maximum height of a projectile?
if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest. ...
if α = 45°, then the equation may be written as: ...
if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion.
Complete Question
A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is 
Compute U if the bulb remains on for 5h
Answer:
The value is 
Explanation:
From the question we are told that
The power rating of the bulb is 
The resistance is 
The voltage is ![V = V_o sin [2 \pi ft]](https://tex.z-dn.net/?f=V%20%20%3D%20%20V_o%20%20sin%20%5B2%20%5Cpi%20ft%5D)
The energy expanded is
The voltage 
The frequency is 
The time considered is 
Generally power is mathematically represented as
=> ![P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}](https://tex.z-dn.net/?f=P%20%3D%20%20%5Cfrac%7B%28%20110%20%20sin%20%5B2%20%5Cpi%20%2A%2060t%5D%29%5E2%7D%7B%20144%7D)
=> ![P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}](https://tex.z-dn.net/?f=P%20%3D%20%20%5Cfrac%7B%20110%5E2%20%5B%20sin%20%5B120%20%5Cpi%20t%5D%29%5E2%7D%7B%20144%7D)
So
![U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt](https://tex.z-dn.net/?f=U%20%20%3D%20%20%5Cint%5Climits%5ET_0%20%7B%20%5Cfrac%7B%20110%5E2%2A%20%20%5Bsin%20%5B120%20%5Cpi%20t%5D%29%5E2%7D%7B%20144%7D%7D%20%5C%2C%20dt)
=> ![U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt](https://tex.z-dn.net/?f=U%20%20%3D%20%20%5Cfrac%7B110%5E2%7D%7B144%7D%20%5Cint%5Climits%5ET_0%20%7B%20%28%20%20%20sin%5E2%20%5B120%20%5Cpi%20t%5D%7D%20%5C%2C%20dt)
=> 
=> 
=> ![U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.](https://tex.z-dn.net/?f=U%20%3D%20%20%5Cfrac%7B110%5E2%7D%7B144%7D%20%5B%5Cfrac%7Bt%7D%7B2%7D%20%20-%20%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%20%5Cfrac%7Bsin%28240%20%5Cpi%20t%29%7D%7B240%20%5Cpi%7D%20%5D%20%5D%5Cleft%20%20%7C%20T%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
=> ![U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.](https://tex.z-dn.net/?f=U%20%3D%20%20%5Cfrac%7B110%5E2%7D%7B144%7D%20%5B%5Cfrac%7Bt%7D%7B2%7D%20%20-%20%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%20%5Cfrac%7Bsin%28240%20%5Cpi%20t%29%7D%7B240%20%5Cpi%7D%20%5D%20%5D%5Cleft%20%20%7C%2018000%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]](https://tex.z-dn.net/?f=U%20%3D%20%20%5Cfrac%7B110%5E2%7D%7B144%7D%20%5B%5Cfrac%7B18000%7D%7B2%7D%20%20-%20%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%20%5Cfrac%7Bsin%28240%20%5Cpi%20%2818000%29%29%7D%7B240%20%5Cpi%7D%20%5D%20%5D)
=>
<span>R = rate of flow = 0.370 L/s
H = height = 2.9 m
T= time = 3.9 s
V = velocity of water when it hits the bucket = sqrt(2gh) = sqrt(2 x 9.8 x 2.9) =7.539 m/s2
G value = 9.8 m/s2
Wb = weight of bucket = 0.690 kg x 9.8 m/s2 = 6.762 N
Wa = weight of accumulated water after 3.9 s
Fi = force of impact of water on the bucket
S = reading on the scale = Wa + Wb + Fi
mass of water accumulated after 3.9 s = R x T = 0.370 x 3.9 = 1.443 L = 1.443 kg
Therefore, Wa = 1.443 x 9.8 = 14.1414 N
Fi = rate of change of momentum at the impact point = R x V (because R = dm/dt)
= 0.37 x 7.539 = 2.78943 N
S = 14.1414 + 6.762 + 2.78943 = 23.692 N</span>
Answer:
This property could be used to create technologically-advanced tools or machines that could easily locate the mineral deposits.
Explanation:
Mineral deposits are hard to find, unless you have the skill or the proper tools in locating them. This is the reason why many people are mining in order to explore the different areas where they could find these deposits.
If one would consider the property of minerals, such as being good conductors of heat and electricity,<u> then they could create a tool or machine that would aid in their exploration.</u> Inventors could probably come up with a sensitive detector which signals when it reaches an area of high heat and electric conductivity. Since most minerals such as <em>gold, silver, copper, galena, bornite </em>and the like have this property, then miners will have a lesser amount of time looking for them.
If this technology will be implemented, though, regulation policy must be strictly implemented because it might lead to<em> over-mining</em> thus leading to the depletion of mineral deposits.
