To solve this problem we will use the equilibrium conditions in the Electrostatic Forces. In turn, we will use the concept formulated from Coulomb's laws to determine the intensity of the Forces and make the respective considerations.
Our values for the two charges are:
As a general consideration we will start by determining that they are at a unit distance (1) separated from each other. And considering that both are negative charges, they will be subjected to repulsive force. Said equilibrium compensation will be achieved only by placing a third force between the two.
Let the third charge be 
is placed at a distance x from 
The force on 
due to 
is
The condition of equilibrium is
from
To find the magnitude of we use 
The magnitude of the third charge must be 0.43 the first charge 
Answer:
No temperature change occurs from heat transfer if ice melts and becomes liquid water (i.e., during a ... to change 1 kg of liquid water at the normal boiling point (100ºC at atmospheric pressure) to steam (water vapor).
The melting point of lead is 327.3o C. Assume the final temperature of the system is T. Then the amount of energy released by the lead as it solidifies is. ΔQ = mleadLlead = 0.09 kg*(2.45*104 J/kg) = 2205 J
Answer:
(a) The work done is 0.05 J
(b) The force will stretch the spring by 3.8 cm
Explanation:
Given;
work done in stretching the spring from 30 cm to 45 cm, W = 3 J
extension of the spring, x = 45 cm - 30 cm = 15 cm = 0.15 m
The work done is given by;
W = ¹/₂kx²
where;
k is the force constant of the spring
k = 2W / x²
k = (2 x 3) / (0.15)²
k = 266.67 N/m
(a) the extension of the spring, x = 37 cm - 35 cm = 2 cm = 0.02 m
work done is given by;
W = ¹/₂kx²
W = ¹/₂ (266.67)(0.02)²
W = 0.05 J
(b) force = 10 N
natural length L = 30 cm
F = kx
x = F / k
x = 10 / 266.67
x = 0.0375 m
x = 3.75 cm = 3.8 cm
Thus a force of 10 N will stretch the spring by 3.8 cm
Answer: 
Explanation:-
As we know that,
![m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
where,
= mass of iron horseshoe = 0.35 kg = 350 g (1kg=1000g[/tex]
= mass of water = 21.9 kg = 21900 g
= final temperature = ?
= temperature of iron horseshoe = 
= temperature of water = 
= specific heat of iron horseshoe = 
= specific heat of water = 
Now put all the given values in equation (1), we get
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
![350\times 0.450\times (T_{final}-600)^0C=-[21900g\times 4.184\times (T_{final}-21.8)]](https://tex.z-dn.net/?f=350%5Ctimes%200.450%5Ctimes%20%28T_%7Bfinal%7D-600%29%5E0C%3D-%5B21900g%5Ctimes%204.184%5Ctimes%20%28T_%7Bfinal%7D-21.8%29%5D)
Therefore, the final equilibrium temperature is .
Absolute, Atmospheric, Differential, and Gauge Pressure
