Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.?
Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.
(a) What is the electric potential at point a due to q1 and q2?
(b) What is the electric potential at point b?
(c) A point charge q3 = -6.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?
Answer:
a) the potential is zero at the center .
Explanation:
a) since the two equal-magnitude and oppositely charged particles are equidistant
b)(b) Electric potential at point b, v = Σ kQ/r
r = 5cm = 0.05m
k = 8.99*10^9 N·m²/C²
Q = -2 microcoulomb
v= (8.99*10^9) * (2*10^-6) * (1/√2m - 1) / 0.0500m
v = -105 324 V
c)workdone = charge * potential
work = -6.00µC * -105324V
work = 0.632 J
Answer : B. 0 m/s∧2
Now we know that velocity = distance/ time
And acceleration = velocity/time.
From the graph we see that at the time interval 2 secs,the distance travelled is 10 m. Hence the velocity(v1) =10/2 = 5m/s.
From the graph we see that at the time interval 5 secs,the distance travelled is 25 m. Hence the velocity (v2) =25/5 = 5m/s.
Now acceleceration between the time interval of 2 ad 5 secs is calculated as
(v1-v2)/t2-t1 = (5-5)/(5-2) = 0 m/s∧2
Answer:
-1786.5J
Explanation:
Temperature 1=T1=25°c
Temperature 2=T2=200°c
Pressure P1=1bar
Pressure P2=0.5bars
T=37°c+273=310k
Note number if moles=1
Recall work done =2.3026RTlogp2/P1
2.3026*8.314*310log(0.5/1)
-1786.5J
The sun is a huge ball of gas held together by gravity.
It does not burn the way wood does, due to oxygen, but it gets energy by a process called nuclear fusion, where Hydrogen is converted to Helium.
The sun will cease to "burn" when it runs out of Hydrogen, but that has a long way to go.
