Answer:
The correct one is that the force on B is half of the force on A
Explanation:
Because radius for the inside of the curve is half the radius for the outside and Car A travels on the inside while car B, travels at equal speed on the outside of the curve. Thus force on B will be half on A
Question: The force between a pair of 0.005 C is 750 N. What is the distance between them?
Answer:
17.32 m
Explanation:
From coulomb's Law,
F = kqq'/r²........................... Equation 1
Where F = Force between the force, q' and q = both charges respectively, k = coulomb's constant, r = distance between both charges.
make r the subject of the equation above
r = √(kqq'/F)..................... Equation 2
From the question,
Given: q = q' = 0.005 C, F = 750 N
Constant: k = 9.0×10⁹ Nm²/C².
Substitute these values into equation 2
r = √(9.0×10⁹×0.005×0.005/750)
r = √(300)
r = 17.32 m.
Hence the distance between the pair of charges = 17.32 m
Answer:
Pretty sure its -4.8%
Explanation:
Others asked this question and got this as an answer.
Answer:633.8 KJ
Explanation:
Given
mass of water![\left ( m\right )=250gm](https://tex.z-dn.net/?f=%5Cleft%20%28%20m%5Cright%20%29%3D250gm)
Initial temperature![\left ( T_i\right )=20^{\circ}C](https://tex.z-dn.net/?f=%5Cleft%20%28%20T_i%5Cright%20%29%3D20%5E%7B%5Ccirc%7DC)
Final temperature ![\left ( T_f\right )=100^{\circ}C](https://tex.z-dn.net/?f=%5Cleft%20%28%20T_f%5Cright%20%29%3D100%5E%7B%5Ccirc%7DC)
Specific heat of water
=4190 J/kg-k
heat of vaporization![\left ( L\right )=22.6\times 10^5 J/kg](https://tex.z-dn.net/?f=%5Cleft%20%28%20L%5Cright%20%29%3D22.6%5Ctimes%2010%5E5%20J%2Fkg)
Heat required for process
=heat to raise water temperature from 20 to 100 +Heat to vapourize water completely
Q=mc![\left ( T_f-T_i\right )+mL](https://tex.z-dn.net/?f=%5Cleft%20%28%20T_f-T_i%5Cright%20%29%2BmL)
Q=![0.25\times 4190\times \left ( 100-20\right )+0.25\times 22\times 10^5](https://tex.z-dn.net/?f=0.25%5Ctimes%204190%5Ctimes%20%5Cleft%20%28%20100-20%5Cright%20%29%2B0.25%5Ctimes%2022%5Ctimes%2010%5E5)
Q=![\left ( 0.838+5.5\right )\times 10^5](https://tex.z-dn.net/?f=%5Cleft%20%28%200.838%2B5.5%5Cright%20%29%5Ctimes%2010%5E5)
Q=![6.338\times 10^5J=633.8 KJ](https://tex.z-dn.net/?f=6.338%5Ctimes%2010%5E5J%3D633.8%20KJ)