I don't understand what you mean by "depth" of the steps. The flat part of the step has a front-to-back dimension, and the 'riser' has a height. I don't care about the horizontal dimension of the step because it doesn't add anything to the climber's potential energy. And if the riser of each step is 20cm high, then 3,234 of them only take him (3,234 x 0.2) = 646.8 meters up off the ground. So something is definitely fishy about the steps.
Fortunately, we don't need to worry at all about the steps in order to derive a first approximation to the answer ... one that's certainly good enough for high school Physics.
In order to lift his bulk 828 meters from the street to the top of the Burj, the climber has to provide a force of 800 newtons, and maintain it through a distance of 828 meters. The work [s]he does is (force) x (distance) = <em>662,400 joules. </em>
Answer:
Reproducibility of research
Explanation:
The principle of science that explains why similar experimental investigations conducted in different parts of the world could result in the same outcome is referred to as reproducibility.
<em>A good research or experiment in science must be reproducible, otherwise, the outcome of such an experiment might become inadmissible within the scientific community. It is a core principle of the scientific method that similar results should be obtained when an experiment or observational study conducted in one place is repeated in another place with the same procedure. Hence, an experiment must be reproducible in science in order for the outcome of such an experiment to be part of the general scientific knowledge. </em>
The answer is B(t) = constants x I(t)
Please take precaution on the point that it is an independent field of its radial position, if the point is measured well in the solenoid. (also the radial position is the axis of its symmetry)
Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is 
Explanation:
From the question we are told that
The diameter of the wire is 
The radius of the wire is 
The resistivity of aluminum is 
The electric field change is mathematically defied as

Generally the charge is mathematically represented as

Where A is the area which is mathematically represented as

So

Therefore

substituting values
![Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5Cint%5Climits%5E%7Bt%7D_%7B0%7D%20%7B%20%5B%200.0004t%5E2%20-%200.0001t%20%2B0.0004%5D%20%7D%20%5C%2C%20dt)
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%20t%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
From the question we are told that t = 5 sec
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%205%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004%285%29%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20%285%29%5E2%7D%7B2%7D%20%2B0.0004%285%29%5D%20%7D)
