Don't know if this is a True/False questions but that is true
Work = Force x Distance
Assuming that this work is being done parallel to the displacement that is, but under that assumption:
W = (50)(10)
W = 500 J
Answer: because there is no displacement or movement in the watchman's work. according to science when displacement or movement take place it is said to be work. hope this helps you.
Answer:
The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)
Explanation:
Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.
Step 2: I must calculate the magnitude of the forces acting on the third charge.
F13: Force exerted by charge 1 on charge 3.
F23: Force exerted by charge 2 on charge 3.
K: Constant of Coulomb's law.
d13: distance from charge 1 to charge 3.
d23: distance from charge 2 to charge 3
Fr: Resulting force.
q1=+2.06 x 10-9 C
q2= -3.27 x 10-9 C
q3= +1.05 x 10-9 C
K=9-10^9 N-m^2/C^2
d13= 0,20 m
d23= 0,10 m
F13= K * (q1 * q3)/(d13)^2
F13=9,7335*10^(-8) N
F23=K * (q2 * q3)/(d23)^2
F23= -3,09 * 10^(-7)
Step 3: We calculate the resultant force on charge 3.
Fr=F13+F23= -2,11665 * 10^(-7)
Answer:
The final temperature of the two objects is the same.
Explanation:
The expression for the heat energy in terms of mass, specific heat and the change in the temperature is as follows:
![Q=mc(T_{f} -T_{i})](https://tex.z-dn.net/?f=Q%3Dmc%28T_%7Bf%7D%20-T_%7Bi%7D%29)
Here, Q is the heat energy, m is the mass of the object, c is the specific heat and
are the final temperature and initial temperature.
According to the given question, Two objects of the same mass, but made of different materials, are initially at the same temperature. Equal amounts of heat are added to each object.
............(1)
.............(2)
From (1) and (2),
![mc(T_{f}-T_{i})=mc(T'_{f}-T_{i})](https://tex.z-dn.net/?f=mc%28T_%7Bf%7D-T_%7Bi%7D%29%3Dmc%28T%27_%7Bf%7D-T_%7Bi%7D%29)
![T_{f}-T_{i}=T'_{f}-T_{i}](https://tex.z-dn.net/?f=T_%7Bf%7D-T_%7Bi%7D%3DT%27_%7Bf%7D-T_%7Bi%7D)
![T_{f}=T'_{f}](https://tex.z-dn.net/?f=T_%7Bf%7D%3DT%27_%7Bf%7D)
Therefore, the final temperature of the two objects is the same.