Answer to Question #210247 in Atomic and Nuclear Physics for qwerty

Question #210247

1.Find I when V = 120V and R = 3O Ω.

2. Suppose a resistor has a value of 110 Ω, and the measured current is 10 mA, what is the

voltage?

3. What is the resistance of a 50-m tungsten (ρ=5.5x10⁻⁸ Ω-m) wire having a diameter of

0.8 mm.

4. An electric light bulb draws 1 A when operating at 120 V circuit. What is the resistance

of the bulb?

5. If the voltage across a 25 000-Ω resistor is 500 V, what is the power (hp) dissipated in

the resistor?

6. Find the shortest time that 120 C can flow through a 20 A circuit breaker without

tripping it?

7. How much chemical energy must a 1.25 V flashlight battery expend in producing a

current flow of 130 mA for 5 minutes?

8. Find the charge in coulombs of (a) 6.28 x 10²¹ electrons and (b) 8.76 x 10²⁰ protons.

9. How much power does an electric clock require if it draws 27.3 mA from a 110-V line?

10. Find the average input power to a radio that consumes 4500 J in 3 min.




1
Expert's answer
2021-06-24T11:59:28-0400

Part (1)

I=VR=12030=4AI=\frac{V}{R}=\frac{120}{30}=4A

Part(2)


V=IR=110×10×103=1.1VV=IR=110\times 10\times10^{-3}=1.1V

Part(3)


r=0.4mm,ρ=5.5×108Ωm,l=50mr=0.4mm,\rho=5.5\times10^{-8}\Omega m,l=50m

R=ρlA=5.5×108×50π×(0.4×103)2R=\frac{\rho l}{A}=\frac{5.5\times10^{-8}\times50}{\pi\times{(0.4\times10^{-3})^2}} =5.47Ω\Omega

Part(4)

R=VI=1201=120ΩR=\frac{V}{I}=\frac{120}{1}=120\Omega

Part(5)


P=V2R=(25000)2500=125×104WP=\frac{V^2}{R}=\frac{(25000)^2}{500}=125\times10^4W

1hp=746W

P=125×104746=1675WP=\frac{125\times10^4}{746}=1675W

Part(6)

q=itq=it

t=qi=12020=6sect=\frac{q}{i}=\frac{120}{20}=6sec

Part(7)

Chemical energy


E=i×V×t=130×103×1.25×5=0.82JE=i\times V \times t=130\times10^{-3}\times1.25\times5=0.82J

Part(8)

(a)

q=ne=6.28×1021×1.6×1019=1004.8Cq=ne=-6.28\times10^{21}\times1.6\times10^{-19}=-1004.8C

(b)

q=ne=8.76×1020×1.6×1019=140.26Cq=ne=8.76\times10^{20}\times1.6\times10^{-19}=140.26C

Part(9)


P=iV=27.3×103×110=3.003WP=iV=27.3\times10^{-3}\times110=3.003W

Part (10)

P=Wt=4500180=25WP=\frac{W}{t}=\frac{4500}{180}=25W



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