Answer:
Time, t = 8 seconds
Explanation:
An object is thrown upward from the top of a 128-foot building with an initial velocity of 112 feet per second. The height h as a function of time t is given by :
![h=-16t^2+112t+128](https://tex.z-dn.net/?f=h%3D-16t%5E2%2B112t%2B128)
We need to find the time when the object will hit the ground. When it will hit the ground, h = 0
So,
![-16t^2+112t+128=0](https://tex.z-dn.net/?f=-16t%5E2%2B112t%2B128%3D0)
On solving the above quadratic equation, we get the value of t = 8 seconds. So, after 8 seconds the object will hit the ground. Hence, this is the required solution.
Answer:
Between 120 and 180 seconds
1) Focal length
We can find the focal length of the mirror by using the mirror equation:
![\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bf%7D%3D%20%5Cfrac%7B1%7D%7Bd_o%7D%2B%20%5Cfrac%7B1%7D%7Bd_i%7D%20%20%20)
(1)
where
f is the focal length
![d_o](https://tex.z-dn.net/?f=d_o)
is the distance of the object from the mirror
![d_i](https://tex.z-dn.net/?f=d_i)
is the distance of the image from the mirror
In this case,
![d_o = 18 cm](https://tex.z-dn.net/?f=d_o%20%3D%2018%20cm)
, while
![d_i=-4 cm](https://tex.z-dn.net/?f=d_i%3D-4%20cm)
(the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
![\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bf%7D%3D%20%5Cfrac%7B1%7D%7B18%20cm%7D-%20%5Cfrac%7B1%7D%7B4%20cm%7D%3D-%20%5Cfrac%7B7%7D%7B36%20cm%7D%20%20%20%20)
from which we find
![f=- \frac{36}{7} cm=-5.1 cm](https://tex.z-dn.net/?f=f%3D-%20%5Cfrac%7B36%7D%7B7%7D%20cm%3D-5.1%20cm%20)
2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.
3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.
4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
Answer:
The length of the Mercury column of thermometer at ice point is 20mm and 220mm at steam point . when the same Thermometer is placed in contact with another body ,it reads 5°c.what will the length of the Mercury column at the temperature?
Answer:
![x=\dfrac{r}{\sqrt2}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7Br%7D%7B%5Csqrt2%7D)
Explanation:
Given that
Radius =r
Electric filed =E
Q=Charge on the ring
The electric filed at distance x given as
![E=K\dfrac{Q}{(r^2+x^2)^{3/2}}](https://tex.z-dn.net/?f=E%3DK%5Cdfrac%7BQ%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B3%2F2%7D%7D)
For maximum condition
![\dfrac{dE}{dx}=0](https://tex.z-dn.net/?f=%5Cdfrac%7BdE%7D%7Bdx%7D%3D0)
![E=K{Q}{(r^2+x^2)^{-3/2}}](https://tex.z-dn.net/?f=E%3DK%7BQ%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B-3%2F2%7D%7D)
![\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}](https://tex.z-dn.net/?f=%5Cdfrac%7BdE%7D%7Bdx%7D%3DK%7BQ%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B-3%2F2%7D%7D-%5Cdfrac%7B3%7D%7B2%7D%5Ctimes%202%5Ctimes%20x%5Ctimes%20K%7BQ%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B-5%2F2%7D%7D)
For maximum condition
![\dfrac{dE}{dx}=0](https://tex.z-dn.net/?f=%5Cdfrac%7BdE%7D%7Bdx%7D%3D0)
![K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0](https://tex.z-dn.net/?f=K%7BQ%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B-3%2F2%7D%7D-%5Cdfrac%7B3%7D%7B2%7D%5Ctimes%202%5Ctimes%20x%5Ctimes%20K%7BQ%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B-5%2F2%7D%7D%3D0)
![r^2+x^2-3x^2=0](https://tex.z-dn.net/?f=r%5E2%2Bx%5E2-3x%5E2%3D0)
![x=\dfrac{r}{\sqrt2}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7Br%7D%7B%5Csqrt2%7D)
At
the electric field will be maximum.