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IceJOKER [234]
3 years ago
10

The voltage in a circuit is given by the equation V= IR.in this equation v is the voltage Iis correct and R is the resistance wh

ich answer shows this equation solved for current?
Physics
1 answer:
hammer [34]3 years ago
7 0

Answer:I=V/R

Explanation:

V=IR

Divide both sides by R

V/R=IR/R

V/R=I

I=V/R

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4 years ago
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A block that has a mass M = 500 g is hanging at rest on a light string. A projectile with a mass m = 15 g moving horizontally wi
stira [4]

Answer:

2.8m

Explanation:

Convert to metric:

500g = 0.5 kg

15g = 0.015 kg

Assuming elastics collision, we could use the law of momentum conservation to calculate the speed of black and projectile post-collision V:

mv = (M + m)V

V = \frac{mv}{m+M} = \frac{0.015*255}{0.5 + 0.015} = 7.43 m/s

This would make the system moves up, which convert its kinetic energy to potential energy. By the law of energy conservation when the mass reaches its highest point (aka speed = 0)

E_k = E_p

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8 0
4 years ago
Two particles with charges of 2nC and 5nC are separated by a distance of 3m. The charge 2nC is placed on the left. Find the forc
Vanyuwa [196]

Answer:

1\cdot 10^{-8} N to the left

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The magnitude of the electrostatic force between two charges is given by the following equation:

F=k\frac{q_1 q_2}{r^2}

where:

k=9\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the magnitude of the two charges

r is the distance between the two charges

Moreover, the force is:

- Attractive if the charges have opposite sign

- Repulsive if the charges have same sign

In this problem, we have:

q_1=2nC=2\cdot 10^{-9}C is the magnitude of charge 1

q_2=5nC =5\cdot 10^{-9}C is the magnitude of charge 2

r = 3 m is the distance between the two charges

Substituting, we find the force on both charges:

F=\frac{(9\cdot 10^9)(5\cdot 10^{-9})(2\cdot 10^{-9}}{3^2}=1\cdot 10^{-8} N

Here, the two charges are both positive, so the force is repulsive; since the 2 nC charge is on the left, this means that the force on this charge is to the left (away from the 5 nC charge).

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