To solve this problem, it is necessary to apply the concepts related to Work according to the Force and distance, as well as the concepts related to energy lost-or gained-by heat. Mathematically the energy corresponding to heat is given as:
![Q = mC_p(\Delta T)](https://tex.z-dn.net/?f=Q%20%3D%20mC_p%28%5CDelta%20T%29)
Where,
m = mass
= Specific heat
= Change in Temperature
At the same time the Work made by the Force and the distance is given as:
![W = F*d \rightarrow W=mg*d](https://tex.z-dn.net/?f=W%20%3D%20F%2Ad%20%5Crightarrow%20W%3Dmg%2Ad)
As the force is applied at an angle of 30 degrees, the efficient component would be given by the vertical then the work / energy would be determined as:
![W = mg*dsin(30)](https://tex.z-dn.net/?f=W%20%3D%20mg%2Adsin%2830%29)
![W = (15m)(0.2kg)(9.81)(sin30)](https://tex.z-dn.net/?f=W%20%3D%20%2815m%29%280.2kg%29%289.81%29%28sin30%29)
![W = 13.24J](https://tex.z-dn.net/?f=W%20%3D%2013.24J)
Now this energy is used to heat the aluminum. We can find the change at the temperature as follow:
![Q = mC_p(\Delta T)](https://tex.z-dn.net/?f=Q%20%3D%20mC_p%28%5CDelta%20T%29)
![13.24 = (0.2)(900)(\Delta T)](https://tex.z-dn.net/?f=13.24%20%3D%20%280.2%29%28900%29%28%5CDelta%20T%29)
![\Delta T = 0.0736 \°C](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%200.0736%20%5C%C2%B0C)
Therefore the correct answer is B.
|V| = 10.33 units and the direction θ = -47.35° or 312.65°.
Given the x and y components of a vector, we can calculate the magnitude and direction from these components.
Applying the Pythagorean theorem we have that the magnitude of the vector is:
|V| = ![\sqrt{Vx^{2}+Vy^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7BVx%5E%7B2%7D%2BVy%5E%7B2%7D%20%20%7D)
|V| = ![\sqrt{(7.00units)^{2}+(-7.60units)^{2}} = \sqrt{106units^{2}} = 10.33units](https://tex.z-dn.net/?f=%5Csqrt%7B%287.00units%29%5E%7B2%7D%2B%28-7.60units%29%5E%7B2%7D%7D%20%3D%20%5Csqrt%7B106units%5E%7B2%7D%7D%20%3D%2010.33units)
The expression for the direction of a vector comes from the definition of the tangent of an angle:
tan θ =
------> θ = arc tan ![\frac{Vy}{Vx}](https://tex.z-dn.net/?f=%5Cfrac%7BVy%7D%7BVx%7D)
θ = arc tan ![\frac{-7.60units}{7.00units}](https://tex.z-dn.net/?f=%5Cfrac%7B-7.60units%7D%7B7.00units%7D)
θ = -47.35° or 312.65°
<span>an error having a nonzero mean, so that its effect is not reduced when observations are averaged.</span>
The area of the Earth that is most similar to the Sun's convection zone would be the mantle. The convection zone of the sun is its outermost layer where heat transfer by convection happens which is similar to the Earth's mantle. It would be the crust because it is the outer most layer.