a battery of 5v and internal resistance of 5 ohms is connected to a resistor of 20 ohms. calculate the value of terminal voltage
1 answer:
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Answer:
80⁰
Explanation:
triangle AOB = 180
So, a = 50
triangle CAB = a+a+x=180
CAB = 50+50+X=180
CAB = X = 180-100
X=80
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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)
The collision of car 2 is more violent (because more impulse is exerted)
Explanation:
The collision which is more violent is the one in which more impulse is exchanged.
The impulse exerted by each car on the wall is equal to the change in momentum of the car:
where
m is the mass of the car
is the change in velocity
For the car 1,
m = 1000 kg
(the sign is negative because the velocity of the car has changed from 100 km/h to 0 km/h)
So the magnitude of the impulse of car 1 is
For the car 2,
m = 1200 kg
So the magnitude of the impulse of car 2 is
So, car 2 exerts a larger impulse, therefore its impact is more violent.
Learn more about impulse and change in momentum:
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