Question 15 When moving oxidation numbers down as subscripts in an ionic bond, do you include the charge (+/-)?
1 answer:
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If we neglect frictional force, the total mechanical energy of the ball is conserved.
The total mechanical energy of the ball is the sum of its kinetic energy K and its potential energy U:
where the kinetic energy depends on the speed v of the ball:
while the potential energy depends on the height h at which the ball is:
As the ball travels along the roller coaster, there is a continuous conversion between kinetic and potential energy, because the total mechanical energy E has always the same value. Therefore, when the ball goes on top of a hill, its height h increases and its potential energy U increases as well, while the speed v decreases and K decreases. Vice-versa, when the ball reaches the bottom of a hill, its height h decreases and therefore the potential energy U decreases, while the speed v increases and therefore the kinetic energy K of the ball increases as well.
The total mechanical energy of the ball is the sum of its kinetic energy K and its potential energy U:
where the kinetic energy depends on the speed v of the ball:
while the potential energy depends on the height h at which the ball is:
As the ball travels along the roller coaster, there is a continuous conversion between kinetic and potential energy, because the total mechanical energy E has always the same value. Therefore, when the ball goes on top of a hill, its height h increases and its potential energy U increases as well, while the speed v decreases and K decreases. Vice-versa, when the ball reaches the bottom of a hill, its height h decreases and therefore the potential energy U decreases, while the speed v increases and therefore the kinetic energy K of the ball increases as well.
Answer:
Distance: -30.0 cm; image is virtual, upright, enlarged
Explanation:
We can find the distance of the image using the lens equation:
where:
f = 15.0 cm is the focal length of the lens (positive for a converging lens)
p = 10.0 cm is the distance of the object from the lens
q is the distance of the image from the lens
Solving for q,
The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).
The magnification can be found as
The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.
Also, the fact that the magnification is positive tells us that the image is upright.