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Marizza181 [45]
3 years ago
14

HELP PLEASE!

Physics
1 answer:
seraphim [82]3 years ago
6 0

Answer:

ᵒᵒᵒᵒᵏᵏ

ᵏᵏᵏ

Explanation:

ᵒᵒᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏ

ᵏᵒᵒᵒᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵐᵐ

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When you connect an unknown resistor across the terminals of a 1.50 V D-cell battery having negligible internal resistance, you
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Explanation:

Using ohm's law

a) V = IR where V is voltage in Volt, I is current in Ampere and R is resistance in ohms

R = V / I = 1.50 V/ ( 2.05 /1000) A = 731.71 ohms

b) Power = IV = \frac{V}{R} × v = \frac{V^{2} }{R} = \frac{9^{2} }{731.71} = 0.1107 W

c) E = IR + Ir = ( 731.71 × 0.0036) + ( 35 × 0.0036) = 2.76 V

d) Power use by the resistor = I²R = 0.0036² × 731.71 = 0.00948 W = 0.00948 W = 0.000009483 kw × ( 18 / 60 ) H = 2.84 × 10⁻⁶ KW-H

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​

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