Answer:
Explanation:
This image is the example of Newton's first law of motion because:
As per Newton's first law, football will remain in the state of rest until a player applies an external force by kicking the ball.
And the ball will keep on moving until unless the net of a goal post exerts the external force to stop the ball.
Acceleration is the rate of change in an object's velocity
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:
250,000
Explanation:
<h2> </h2>
<h2>formula = ( F=ma </h2>
- F=1500N
- a=6m/s^2
- F= ma
- m=?
- 1500/6 = m
- m=250 kg
- 1kg =1000gm so 250kg =250,000gm
- m =250×10^3 gm