The circuit component the symbol represents is: C) Battery
Explanation:
Given that,
Terminal voltage = 3.200 V
Internal resistance ![r= 5.00\ \Omega](https://tex.z-dn.net/?f=r%3D%205.00%5C%20%5COmega)
(a). We need to calculate the current
Using rule of loop
![E-IR-Ir=0](https://tex.z-dn.net/?f=E-IR-Ir%3D0)
![I=\dfrac{E}{R+r}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7BE%7D%7BR%2Br%7D)
Where, E = emf
R = resistance
r = internal resistance
Put the value into the formula
![I=\dfrac{3.200}{1.00\times10^{3}+5.00}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B3.200%7D%7B1.00%5Ctimes10%5E%7B3%7D%2B5.00%7D)
![I=3.184\times10^{-3}\ A](https://tex.z-dn.net/?f=I%3D3.184%5Ctimes10%5E%7B-3%7D%5C%20A)
(b). We need to calculate the terminal voltage
Using formula of terminal voltage
![V=E-Ir](https://tex.z-dn.net/?f=V%3DE-Ir)
Where, V = terminal voltage
I = current
r = internal resistance
Put the value into the formula
![V=3.200-3.184\times10^{-3}\times5.00](https://tex.z-dn.net/?f=V%3D3.200-3.184%5Ctimes10%5E%7B-3%7D%5Ctimes5.00)
![V=3.18\ V](https://tex.z-dn.net/?f=V%3D3.18%5C%20V)
(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf
![\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }](https://tex.z-dn.net/?f=%5Cdfrac%7BTerminal%5C%20voltage%7D%7Bemf%7D%3D%5Cdfrac%7B3.18%7D%7B3.200%20%7D)
![\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}](https://tex.z-dn.net/?f=%5Cdfrac%7BTerminal%5C%20voltage%7D%7Bemf%7D%3D%20%5Cdfrac%7B159%7D%7B160%7D)
Hence, This is the required solution.
Answer:
10. 36 g ZnCl2
Explanation:
Zn + 2HCl -> ZnCl2 + H2
0.076 mol Zn
1.37 mol HCl
3 mol H2
Limiting reactant: Zn
1 mol Zn -> 1 mol ZnCl2
0.076 mol Zn ->x x= 0.076 mol ZnCl2=10.36 g
Sum of all forces = mass * acceleration
Ft= tension force
Fw= force of gravity (Fw= mass* acceleration of gravity which is 9.8 this only applies to force of gravity)
Ft- Fw = 0 (there is no acceleration)
Ft = Fw
Ft= m*g
Ft= 0.250kg*9.8m/s
Ft= 2.45N
Explanation:
Given that,
Initial velocity, u = 11.3 m/s
Angle above the horizontal, ![\theta=35^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D35%5E%7B%5Ccirc%7D)
Time of flight :
![t=\dfrac{2u\sin\theta}{g}\\\\t=\dfrac{2\times 11.3\times \sin(35)}{9.8}\\\\t=1.32\ s](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B2u%5Csin%5Ctheta%7D%7Bg%7D%5C%5C%5C%5Ct%3D%5Cdfrac%7B2%5Ctimes%2011.3%5Ctimes%20%5Csin%2835%29%7D%7B9.8%7D%5C%5C%5C%5Ct%3D1.32%5C%20s)
Horizontal distance traveled is given by :
x = ut
x = 11.3 m/s × 1.32 s
x = 14.916 m
Maximum height is given by :
![H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(11.3)^2\times \sin^2(35)}{2\times 9.8}\\\\H=2.14\ m](https://tex.z-dn.net/?f=H%3D%5Cdfrac%7Bu%5E2%5Csin%5E2%5Ctheta%7D%7B2g%7D%5C%5C%5C%5CH%3D%5Cdfrac%7B%2811.3%29%5E2%5Ctimes%20%5Csin%5E2%2835%29%7D%7B2%5Ctimes%209.8%7D%5C%5C%5C%5CH%3D2.14%5C%20m)
Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.