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ANEK [815]
3 years ago
11

I NEED THIS ANSWER ASAP PLEASE!!! What happens when a light ray passes through the focal point, and then reflects off of a conve

x mirror? A.The reflected ray will run parallel to the center of curvature. B. The reflected ray will run parallel to the principal axls. C. The reflected ray will pass through the center of curvature. D. The reflected ray will pass through a point between the center of curvature and the focal point. E. The reflected ray will pass through a point between the focal point and the vertex
Physics
1 answer:
Alexandra [31]3 years ago
8 0

What happens when a light ray passes through the focal point, and then reflects off of a convex mirror?

E. The reflected ray will pass through a point between the focal point and the vertex

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du
Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time t is

\theta=\dfrac\alpha2t^2

It rotates 44.5 rad in this time, so we have

44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}

b. Since acceleration is constant, the average angular velocity is

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2

where \omega_f is the angular velocity achieved after 6.00 s. The velocity of the disk at time t is

\omega=\alpha t

so we have

\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}

making the average velocity

\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}

Another way to find the average velocity is to compute it directly via

\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}

c. We already found this using the first method in part (b),

\omega=14.8\dfrac{\rm rad}{\rm s}

d. We already know

\theta=\dfrac\alpha2t^2

so this is just a matter of plugging in t=12.0\,\mathrm s. We get

\theta=179\,\mathrm{rad}

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2

Then for t=6.00\,\rm s we would get the same \theta=179\,\rm rad.

7 0
3 years ago
Which description of energy changes in nuclear reactions is correct?
Kryger [21]
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4 0
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dimaraw [331]

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Explanation:

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Note no fossils contain Uranium 238. Uranium 238 is only found in igneous or volcanic rocks. So no fossils can be dated directly using U 238.

Because of the huge differences in the half lives of Carbon 14 and Uranium238 they cannot be used together. Carbon 14 can only be used to date fossils of a very recent age. Uranium 238 can only be used to date volcanic rocks of a very old age.

3 0
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