Answer: Leandra puts on her mittens because if you do not you will burn your self, due to extremely high temperatures.
Explanation:
D = 1/f, where D is the power in diopters and f is the focal length in meters.
D=1/20
<u>D=0.05</u>
Answer:
This is how I figured it out:
- 215.5 rounded to one significant figure is 200
- 101.02555 rounded to one significant figure is 100.
- 200 + 100 = 300.
Hope this helps!
Explanation:
It stands for Unified Computing System
I hope I helped :3
Let
be the average acceleration over the first 2.46 seconds, and
the average acceleration over the next 6.79 seconds.
At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let
be the velocity of the car after the first 2.46 seconds.
By definition of average acceleration, we have
![a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}](https://tex.z-dn.net/?f=a_1%3D%5Cdfrac%7Bv-30.0%5C%2C%5Cfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%7D%7B2.46%5C%2C%5Cmathrm%20s%7D)
![a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}](https://tex.z-dn.net/?f=a_2%3D%5Cdfrac%7B15.2%5C%2C%5Cfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D-v%7D%7B6.79%5C%2C%5Cmathrm%20s%7D)
and we're also told that
![\dfrac{a_1}{a_2}=1.66](https://tex.z-dn.net/?f=%5Cdfrac%7Ba_1%7D%7Ba_2%7D%3D1.66)
(or possibly the other way around; I'll consider that case later). We can solve for
in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:
![1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}](https://tex.z-dn.net/?f=1.66a_2%3D%5Cdfrac%7Bv-30.0%5C%2C%5Cfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%7D%7B2.46%5C%2C%5Cmathrm%20s%7D)
![\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}](https://tex.z-dn.net/?f=%5Cimplies1.66%5Cleft%28%5Cdfrac%7B15.2%5C%2C%5Cfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D-v%7D%7B6.79%5C%2C%5Cmathrm%20s%7D%5Cright%29%3D%5Cdfrac%7Bv-30.0%5C%2C%5Cfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%7D%7B2.46%5C%2C%5Cmathrm%20s%7D)
Now we can solve for
. We find that
![v=20.8\,\dfrac{\mathrm m}{\mathrm s}](https://tex.z-dn.net/?f=v%3D20.8%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D)
In the case that the ratio of accelerations is actually
![\dfrac{a_2}{a_1}=1.66](https://tex.z-dn.net/?f=%5Cdfrac%7Ba_2%7D%7Ba_1%7D%3D1.66)
we would instead have
![\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)](https://tex.z-dn.net/?f=%5Cdfrac%7B15.2%5C%2C%5Cfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D-v%7D%7B6.79%5C%2C%5Cmathrm%20s%7D%3D1.66%5Cleft%28%5Cdfrac%7Bv-30.0%5C%2C%5Cfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%7D%7B2.46%5C%2C%5Cmathrm%20s%7D%5Cright%29)
in which case we would get a velocity of
![v=24.4\,\dfrac{\mathrm m}{\mathrm s}](https://tex.z-dn.net/?f=v%3D24.4%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D)