Sustainable buildings are generally less expensive over the lifetime of the building, due to lower energy costs in production of the materials and lower energy bills. Option A is correct.
<h3>What is sustainable buildings? </h3>
Sustainable buildings refer to both a structure and the implementation of environmentally responsible and resource-efficient methods.
Sustainable buildings are generally less expensive over the lifetime of the building, due to lower energy costs in production of the materials and lower energy bills,
Hence,option A is correct.
To learn more about the sustainable buildings, refer;
brainly.com/question/8434592
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Answer:
option D is correct
Explanation:
It is important to note that equipotential lines are always perpendicular to electric field lines. No work is required to move a charge along an equipotential, since ΔV = 0. Thus the work is :
W = −ΔPE = −qΔV = 0.
Work is zero if force is perpendicular to motion. Force is in the same direction as E, so that motion along an equipotential must be perpendicular to E. More precisely, work is related to the electric field by:
W = Fd cos θ = qEd cos θ = 0.
- The change in kinetic energy Δ K.E by conservation should be:
Δ K.E = W
Since, W = 0:
Δ K.E = 0
- If change in kinetic energy is zero it means that charge moves at a constant speed. Hence, option D is correct.
<h2>
Answer:</h2>
<em><u>48.189°</u></em>
<h2>
Explanation:</h2>
Let us say,
Radius of curvature of the clown's bald head is = R
Angle where the grape leaves the contact with the head is (with vertical) = θ
Height from the top of the head at which the contact is lost = y
Mass of the grape = m
Velocity of the grape at the point where it loses contact = v
So,
Using the <u>Conservation of Work and Energy,</u> we can say that there is 0 Work done on the system,
W = ΔK + ΔU
So,
![0=(\frac{1}{2}mv^{2}-0)+(mgy-0)\\v^{2}=2gy\\Now,\\y=R-Rcos\theta\\y=R(1-cos\theta)\\So,\\v^{2}=2[R(1-cos\theta)]g](https://tex.z-dn.net/?f=0%3D%28%5Cfrac%7B1%7D%7B2%7Dmv%5E%7B2%7D-0%29%2B%28mgy-0%29%5C%5Cv%5E%7B2%7D%3D2gy%5C%5CNow%2C%5C%5Cy%3DR-Rcos%5Ctheta%5C%5Cy%3DR%281-cos%5Ctheta%29%5C%5CSo%2C%5C%5Cv%5E%7B2%7D%3D2%5BR%281-cos%5Ctheta%29%5Dg)
Now, using this at the point where contact is lost,
![N=-m(\frac{v^{2}}{R})+mg.cos\theta\\N=-m[2g(1-cos\theta)]+mg.cos\theta\\](https://tex.z-dn.net/?f=N%3D-m%28%5Cfrac%7Bv%5E%7B2%7D%7D%7BR%7D%29%2Bmg.cos%5Ctheta%5C%5CN%3D-m%5B2g%281-cos%5Ctheta%29%5D%2Bmg.cos%5Ctheta%5C%5C)
At that point the <u>Normal force will be zero, because the contact is lost.</u>
So,
On putting, N = 0 we get,
![N=-m[2g(1-cos\theta)]+mg.cos\theta\\0=-m[2g(1-cos\theta)]+mg.cos\theta\\2g-2g.cos\theta=g.cos\theta\\3g.cos\theta=2g\\cos\theta=\frac{2}{3}\\\theta=48.189\,degrees](https://tex.z-dn.net/?f=N%3D-m%5B2g%281-cos%5Ctheta%29%5D%2Bmg.cos%5Ctheta%5C%5C0%3D-m%5B2g%281-cos%5Ctheta%29%5D%2Bmg.cos%5Ctheta%5C%5C2g-2g.cos%5Ctheta%3Dg.cos%5Ctheta%5C%5C3g.cos%5Ctheta%3D2g%5C%5Ccos%5Ctheta%3D%5Cfrac%7B2%7D%7B3%7D%5C%5C%5Ctheta%3D48.189%5C%2Cdegrees)
<em><u>Therefore, the angle at which the grape lose contact with the bald head is at 48.189° from vertical.</u></em>
The relation between electric field intensity, electric force and charge is given as follow,

Where,
F = Force
q = Charge
The direction of force on a charge placed in an electric field is same as the direction of electric field if the charge has positive polarity and opposite if charge has negative polarity.
Replacing we have that


As direction of force is upward and polarity of charge is negative, hence the direction of field will be opposite to that of force. Therefore direction of field is downward.