A vector from the starting point to the end point.
Answer:
I = 0.25 [amp]
Explanation:
To solve this problem we must use ohm's law which tells us that the voltage is equal to the product of the current by the resistance.
V = I*R
where:
V = voltage [Volt]
I = amperage or current [amp]
R = resistance [ohm]
Since all resistors are connected in series, the total resistance will be equal to the arithmetic sum of all resistors.
Rt = 2 + 8 + 14
Rt = 24 [ohm]
Now clearing I for amperage
I = V/Rt
I = 6/24
I = 0.25 [amp]
Answer:
Net electric field, ![E_{net}=91406.24\ N/C](https://tex.z-dn.net/?f=E_%7Bnet%7D%3D91406.24%5C%20N%2FC)
Explanation:
Given that,
Charge 1, ![q_1=7.5\ nC=7.5\times 10^{-9}\ C](https://tex.z-dn.net/?f=q_1%3D7.5%5C%20nC%3D7.5%5Ctimes%2010%5E%7B-9%7D%5C%20C)
Charge 2, ![q_2=-2.9\ nC=-2.9\times 10^{-9}\ C](https://tex.z-dn.net/?f=q_2%3D-2.9%5C%20nC%3D-2.9%5Ctimes%2010%5E%7B-9%7D%5C%20C)
distance, d = 3.2 cm = 0.032 m
Electric field due to charge 1 is given by :
![E_1=\dfrac{kq_1}{r^2}](https://tex.z-dn.net/?f=E_1%3D%5Cdfrac%7Bkq_1%7D%7Br%5E2%7D)
![E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}](https://tex.z-dn.net/?f=E_1%3D%5Cdfrac%7B9%5Ctimes%2010%5E9%5Ctimes%207.5%5Ctimes%2010%5E%7B-9%7D%7D%7B%280.032%29%5E2%7D)
![E_1=65917.96\ N/C](https://tex.z-dn.net/?f=E_1%3D65917.96%5C%20N%2FC)
Electric field due to charge 2 is given by :
![E_2=\dfrac{kq_2}{r^2}](https://tex.z-dn.net/?f=E_2%3D%5Cdfrac%7Bkq_2%7D%7Br%5E2%7D)
![E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}](https://tex.z-dn.net/?f=E_2%3D%5Cdfrac%7B9%5Ctimes%2010%5E9%5Ctimes%202.9%5Ctimes%2010%5E%7B-9%7D%7D%7B%280.032%29%5E2%7D)
![E_2=25488.28\ N/C](https://tex.z-dn.net/?f=E_2%3D25488.28%5C%20N%2FC)
The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :
![E_{net}=E_1+E_2](https://tex.z-dn.net/?f=E_%7Bnet%7D%3DE_1%2BE_2)
![E_{net}=65917.96+25488.28](https://tex.z-dn.net/?f=E_%7Bnet%7D%3D65917.96%2B25488.28)
![E_{net}=91406.24\ N/C](https://tex.z-dn.net/?f=E_%7Bnet%7D%3D91406.24%5C%20N%2FC)
So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.
Answer:
the center of the flower or the ovary
Answer:
Explanation:
Given
length of window ![h=2.9\ m](https://tex.z-dn.net/?f=h%3D2.9%5C%20m)
time Frame for which rock can be seen is ![\Delta t=0.134\ s](https://tex.z-dn.net/?f=%5CDelta%20t%3D0.134%5C%20s)
Suppose h is height above which rock is dropped
Time taken to cover ![h+2.9 is t_1](https://tex.z-dn.net/?f=h%2B2.9%20is%20t_1)
so using equation of motion
![y=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=y%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where y=displacement
u=initial velocity
a=acceleration
t=time
time taken to travel h is
![h=0+0.5\times g\times (t_2)^2---2](https://tex.z-dn.net/?f=h%3D0%2B0.5%5Ctimes%20g%5Ctimes%20%28t_2%29%5E2---2)
Subtract 1 and 2 we get
![2.9=0.5g(t_1^2-t_2^2)](https://tex.z-dn.net/?f=2.9%3D0.5g%28t_1%5E2-t_2%5E2%29)
![5.8=g(t_1+t_2)(t_1-t_2))](https://tex.z-dn.net/?f=5.8%3Dg%28t_1%2Bt_2%29%28t_1-t_2%29%29)
and from equation ![t_1-t_2=0.134\ s](https://tex.z-dn.net/?f=t_1-t_2%3D0.134%5C%20s)
so ![t_1+t_2=\frac{5.8}{9.8\times 0.134}](https://tex.z-dn.net/?f=t_1%2Bt_2%3D%5Cfrac%7B5.8%7D%7B9.8%5Ctimes%200.134%7D)
![t_1+t_2=4.416\ s](https://tex.z-dn.net/?f=t_1%2Bt_2%3D4.416%5C%20s)
and ![t_1=t_2+\Delta t](https://tex.z-dn.net/?f=t_1%3Dt_2%2B%5CDelta%20t)
so ![t_2+\Delta t+t_2=4.416](https://tex.z-dn.net/?f=t_2%2B%5CDelta%20t%2Bt_2%3D4.416)
![2t_2+0.134=4.416](https://tex.z-dn.net/?f=2t_2%2B0.134%3D4.416)
![t_2=0.5\times 4.282](https://tex.z-dn.net/?f=t_2%3D0.5%5Ctimes%204.282)
![t_2=2.141\ s](https://tex.z-dn.net/?f=t_2%3D2.141%5C%20s)
substitute the value of
in equation 2
![h=0.5\times 9.8\times (2.141)^2](https://tex.z-dn.net/?f=h%3D0.5%5Ctimes%209.8%5Ctimes%20%282.141%29%5E2)
![h=22.46\ m](https://tex.z-dn.net/?f=h%3D22.46%5C%20m)