Answer:
4m
Explanation:
Using the kinematic equation;
For the first stage when the car is initially moving at 0.5m/s
definitely u = 0.5 m/s and v = 0
The cart comes to rest after traveling I m, ∴ (s) = 1m
the acceleration of the car can be expressed by applying the kinematic equation:
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
making "a" the subject of the formula; we have:
![a = \frac{v^2-u^2}{2s}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bv%5E2-u%5E2%7D%7B2s%7D)
![a= \frac{0-(0.5 m/s)^2}{2(1m)}](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%7B0-%280.5%20m%2Fs%29%5E2%7D%7B2%281m%29%7D)
a = 0.125 m/s²
The experiment is repeated on the same track, but now the cart is initially moving at I m/s
i.e v = 1 m/s and u=0
![S=\frac{v^2-u^2}{2a}](https://tex.z-dn.net/?f=S%3D%5Cfrac%7Bv%5E2-u%5E2%7D%7B2a%7D)
![S= \frac{(1m/s)^2-0}{2(0.125m/s^2)}](https://tex.z-dn.net/?f=S%3D%20%5Cfrac%7B%281m%2Fs%29%5E2-0%7D%7B2%280.125m%2Fs%5E2%29%7D)
S = 4 m
∴ the cart traveled 4m before coming to rest.
It is so because the steel is more colder than our body.
Answer:
The distance is 1.69 m.
Explanation:
Given that,
First charge ![q_{1}= 3.8\times10^{-6}\ C](https://tex.z-dn.net/?f=q_%7B1%7D%3D%203.8%5Ctimes10%5E%7B-6%7D%5C%20C)
Second charge ![q_{2}=3.2\times10^{-6}\ C](https://tex.z-dn.net/?f=q_%7B2%7D%3D3.2%5Ctimes10%5E%7B-6%7D%5C%20C)
Distance = 3.25 m
We need to calculate the distance
Using formula of electric field
![E_{1}=E_{2}](https://tex.z-dn.net/?f=E_%7B1%7D%3DE_%7B2%7D)
![\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bkq_%7B1%7D%7D%7Bx%5E2%7D%3D%5Cdfrac%7Bkq_%7B2%7D%7D%7B%28d-x%29%5E2%7D)
![\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bq_%7B1%7D%7D%7Bq_%7B2%7D%7D%3D%5Cdfrac%7B%28x%29%5E2%7D%7B%28d-x%29%5E2%7D)
![\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cdfrac%7Bq_%7B1%7D%7D%7Bq_%7B2%7D%7D%7D%3D%5Cdfrac%7Bx%7D%7Bd-x%7D)
![x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}](https://tex.z-dn.net/?f=x%3D%28d-x%29%5Ctimes%5Csqrt%7B%5Cdfrac%7Bq_%7B1%7D%7D%7Bq_%7B2%7D%7D%7D)
Put the value into the formula
![x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}](https://tex.z-dn.net/?f=x%3D%283.25-x%29%5Ctimes%5Csqrt%7B%5Cdfrac%7B3.8%5Ctimes10%5E%7B-6%7D%7D%7B3.2%5Ctimes10%5E%7B-6%7D%7D%7D)
![x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}](https://tex.z-dn.net/?f=x%2Bx%5Ctimes%5Csqrt%7B%5Cdfrac%7B3.8%5Ctimes10%5E%7B-6%7D%7D%7B3.2%5Ctimes10%5E%7B-6%7D%7D%7D%3D3.25%5Ctimes%5Csqrt%7B%5Cdfrac%7B3.8%5Ctimes10%5E%7B-6%7D%7D%7B3.2%5Ctimes10%5E%7B-6%7D%7D%7D)
![x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}](https://tex.z-dn.net/?f=x%281%2B%5Csqrt%7B%5Cdfrac%7B3.8%5Ctimes10%5E%7B-6%7D%7D%7B3.2%5Ctimes10%5E%7B-6%7D%7D%7D%29%3D3.25%5Ctimes%5Csqrt%7B%5Cdfrac%7B3.8%5Ctimes10%5E%7B-6%7D%7D%7B3.2%5Ctimes10%5E%7B-6%7D%7D%7D)
![x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B3.25%5Ctimes%5Csqrt%7B%5Cdfrac%7B3.8%5Ctimes10%5E%7B-6%7D%7D%7B3.2%5Ctimes10%5E%7B-6%7D%7D%7D%7D%7B%281%2B%5Csqrt%7B%5Cdfrac%7B3.8%5Ctimes10%5E%7B-6%7D%7D%7B3.2%5Ctimes10%5E%7B-6%7D%7D%7D%29%7D)
![x=1.69\ m](https://tex.z-dn.net/?f=x%3D1.69%5C%20m)
Hence, The distance is 1.69 m.
Answer:
C. Photosphere
Explanation:
The lights shown in the figure comes from the outermost layer of the Sun. This layer is called photosphere.
This is the layer from where the light of the Sun is radiated, before travelling through space and reaching us.
The photosphere is the coldest layer of the Sun: its surface temperature is between 4500 and 6000 K. Its width is approximately 100 km.
A characteristic of the photosphere is the presence of the sunspots, which appear as darker spots, and are regions of lower temperature caused by a concentration of magnetic flux.