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1 year ago

PLSS HELP AND I WILL GIVE BRAINLY TO THE ONE WHO ANSWERS CORRECTLY

1 answer:

7 0

The potential difference,electric current ,resistance and new electric current will be 12 V,4 A,3 Ω,2 A.

<h3>What is resistance?</h3>

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.

The energy in terms of the charge and potential difference is;

E= qV

60=5 C × V

V= 12 V

The electric current is found as;

$\rm I= \frac{q}{t} \\\\ I= \frac{5}{1.25}\\\\ I=4 \ A$

From the ohm's law;

V=IR

12=4 ×R

R=3Ω

If the voltage is constant and the resistance is doubled, then the new electric current is half of the previous condition;

$\rm I'=\frac{I}{2} \\\\ I'=\frac{4}{2} \\\\\ I'=2 \ A$

Hence, the potential difference,electric current ,resistance and new electric current will be 12 V,4 A,3 Ω,2 A.

To learn more about the resistance, refer to the link;

brainly.com/question/20708652

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2 years ago

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

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Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

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also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

$2(V_1+9.0)^2=(2V_1+9.0)^2$

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

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$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

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Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

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