A wind turbine captures the kinetic energy of moving air (known as "wind")
to convert into electrical energy.
Answer:
The correct option is;
b. How high the barbell is being lifted
Explanation:
The work done in lifting the 160 N barbell is equal to the potential energy gained by the barbell in the final elevated position of the barbell
Therefore, in order to find the work done to lift the barbell, we can calculate the potential energy gained by the barbell when it is placed at height
The work done to lift the barbell = The potential energy, P.E., gained by the barbell
The potential energy of an object, P. E. = The mass of the object, m × The acceleration due to gravity, g × The height to which the barbell is lifted, h
∴ Mathematically, P.E. = m × g × h
Weight, W = Mass, m × The acceleration due to gravity, g
The weight, W of the barbell is given as W = 160 N
∴ W = m × g = 60 N
From which we have;
P.E. = m × g × h = W × h
∴ The work done to lift the barbell = P.E. = 60 N × h = 60 N × The height to which the barbell is lifted.
Therefore the information which need be known is the height to which the barbell.
Answer:
The speed of the light ray is halved
Explanation:
The index of refraction of a medium is the ratio between the speed of light in a vacuum (c) and the speed of light in the medium (v):
![n=\frac{c}{v}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7Bc%7D%7Bv%7D)
For medium X, we have
(1)
For medium Y, we have
(2)
Dividing (1) by (2), we find
(3)
In this problem, the index of refraction of medium Y is twice as great as the index of refraction of medium X:
![n_y = 2 n_x](https://tex.z-dn.net/?f=n_y%20%3D%202%20n_x)
Substituting this into eq.(3), we get
![\frac{n_x}{2n_x}=\frac{v_y}{v_x}\\v_y = \frac{v_x}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bn_x%7D%7B2n_x%7D%3D%5Cfrac%7Bv_y%7D%7Bv_x%7D%5C%5Cv_y%20%3D%20%5Cfrac%7Bv_x%7D%7B2%7D)
So, as a light ray travels from medium X into medium Y, the speed of the light ray is halved.
Answer:
Φ = Q/ε0 = q/ε0
Explanation:
The electric flux through any Gaussian surface is equal to the net charge inside the surface over the dielectric permitivity of vacuum. This is the Gauss theorem for electric flux trough a surface. In this case is not important that the suface is a surface of a cube.
I hope this is usefull for you!
High melting and boiling point
ionic bonds are very strong
a lot of energy is needed to break them
conductive when liquid
ionic com. can only conduct electricity if their ions are free to move