Answer:
d. 50 C
Explanation:
In this problem, we have to add 800 ml of water at 20 Celsius to 800 ml of water at 80 Celsius.
According to the 2nd law of thermodynamics, heat transfers from hot to cold temperature.
The quantity of both the different waters is equal so this makes it very easy. All we have to do is find the mean of both the temperatures:
Final temperature = (20 C + 80 C)/2
= 50 Celsius
The resistance between A and B is 10 ohms.
Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N

Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
Answer:
285 seconds
Explanation:
Jenny speed is 3.8 m/s
Alyssa speed in 4.0 m/s
Alyssa starts after 15 seconds
Find the distance covered by Jenny, when Alyssa starts
Distance=Speed*time
Distance covered by Jenny in 15 seconds= 3.8×15=57m
Relative speed of the two members heading same direction will be;
4.0m/s-3.8m/s=0.2m/s
To find the time Alyssa catch up with Jenny you divide the distance to be covered by Alyssa by the relative speed of the two
Distance=57m, relative speed=0.2m/s t=57/0.2 =285 seconds
=4.75 minutes