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Fantom [35]
3 years ago
10

The load across a 120-V power supply consists of three resistances are R1 = 10 Ω, R2 = 30 Ω, and R3 is unknown. If R2 has a diff

erence of potential of 45 V across it, find the other voltage drops across the other resistors, the current through each resistor, and R3.
Physics
1 answer:
patriot [66]3 years ago
3 0
Given R2’s resistance and voltage we can find the current through it of 1.5 amps. Due to Kirchhoff’s junction rule the current going in must match the current going out so the current through them all is 1.5 amps. Using this we can find the voltage through R1 of 15v. Then we subtract V1+V2 from 120 to find that R3 has a voltage of 60v. Next we find that R3 has a resistance of 40 ohms
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Two small plastic spheres are given positive electrical charges. When they are 16.0 cm apart, the repulsive force between them h
katrin2010 [14]

Answer:

a) 7.5425 * 10^-7 C

b) 3.7712 * 10^-7 C ; 1.5085 * 10^-6 C

Explanation:

Part a)

F_{elect} = \frac{k*q_{1}*q_{2} }{R^2} \\q_{1} = q_{2} = q\\q = \sqrt{\frac{F_{elect} * R^2}{k} }\\ = \sqrt{\frac{0.2 * 0.16^2}{9*10^9} }\\\\ q = 7.5425 * 10 ^(-7) C

Part b)

F_{elect} = \frac{k*q_{1}*q_{2} }{R^2} \\q_{1} = 4*q_{2} \\q = \sqrt{\frac{F_{elect} * R^2}{k*4} }\\ = \sqrt{\frac{0.2 * 0.16^2}{4*9*10^9} }\\\\ q_{2}  = 3.7712 * 10 ^(-7) C\\\\q_{1}  = 1.5085 * 10 ^(-6) C

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3 years ago
PLZ HURRY GOTTA HAVE IT DONE BEFORE MY MOM GETS HOME SHES 5 MINS AWAY!!!!!!!
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3 0
3 years ago
On a very muddy football field, a 110kg linebacker tackles an 85kg halfback. Immediately before the collision, the linebacker is
ololo11 [35]

Answer:

A. the magnitude of the velocity at which the two players move together immediately after the collision is 7.9m/s

B. The direction of this velocity is due north as the linebacker since he has obviously has more momentum

Explanation:

This problem bothers on the inelastic collision

Given data

Mass of linebacker m1= 110kg

Mass of halfbacker m2= 85kg

Velocity of linebacker v1= 8.8m/s

Velocity of halfbacker v2= 7.2m/s

Applying the principle of conservation of momentum for inelastic collision we have

m1v1 +m2v2= (m1+m2)v

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Substituting our data into the expression we have

110*8.5+85*7.2= (110+85)v

935+612=195v

1547=195v

v=1547/195

v=7.9m/s

Momentum of linebacker after impact = 110*7.9= 869Ns

Momentum of halfbacker after impact = 85*7.9= 671.5Ns

the direction after impact is due north since the linebacker has greater momentum

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