Answer:
30 cm
Explanation:
To solve this problem, we use the lens equation:
where
f is the focal length of the lens
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem, we know that
q = -30 cm is the distance of the image from the lens; it is negative because the image is formed in front of the lens, so it is a virtual image
We also know that the size of the image is twice that of the object, so the magnification is 2:
where M is the magnification of the lens. Solving this equation for p,
So, the distance of the object from the lens is 15 cm.
Now we can finally solve the lens equation to find f, the focal length:
Answer:
SATA tiene las siguientes características importantes: Voltaje reducido. Los estándares ATA actuales utilizan 5,0 V o 3,3 V (ATA-100/133). Estos voltajes relativamente altos junto con altas densidades de pines hacen que 100 MB / s sea la velocidad de datos más alta que se puede alcanzar de manera realista.
Answer:
t = 2s
Explanation:
When you're looking for instantaneous portions of a graph, of any sort really, it means you're observing a rate at a single point in time [or possibly some other variable]. It's sorta like a snapshot of a rate as opposed to an average rate over an interval. After choosing this rate we'll typically draw a straight, tangent line through it to indicate it's slope. (Tangent lines are just lines that only touch a single point on a graph or shape.)
Another thing to take note of are the values of the graph's major axes. The "y-axis" corresponds to velocity in meters per second, while the "x-axis" corresponds to time in seconds. Normally when relating the two we put "y" over the "x" and say that at any point there are "y[units]" per "x[units]". Though with instantaneous rates, we say the value of "x" is "1"; for reasons I can try to further explain later if you'd like.
With the above information in mind we can turn our attention to your graph. You're told to find the point on this graph where the instantaneous rate of acceleration is -2 m/s². The only place where the graph reflects an instantaneous rate of -2m/s² is at t = 2s. At t = 2, the rate comes out to (2[m/s]/1s), which simplifies to 2m/s². If you then draw the tangent line through the point, you'll find that the line is decreasing (going down from left to right) which means that the instantaneous rate is negative.
So at t = 2s, we have an instantaneous acceleration of -2m/s².
The correct answer is "wavelength".
In fact, this is exactly the definition of wavelength: in a wave, the wavelength is defined as the distance between two succesive crests or two successive troughs, therefore the correct answer is
D) wavelength.
