Two particles each have a mass of 6.7
x10-5kg.
one has a charge of +5.0 x 10-6c,
and the other has a charge of -5.0 x 10-6c.
they are initially held at rest at a distance of 0.68 mapart. both
are then released and accelerate toward each other. howfast is each
particle moving when the separation between them isone-half its
initial value?

70.28 m/s²

Explanation:

Step 1: identify the given parameters

mass of each particle (m) = 6.7X10^-5 kg

Charge of the particles = +5X10^-6 C and -5X10^-6 C

Distance between the particles = 0.68 m

Step 2: calculate the force of attraction between the particles

Applying coulomb's law

where k is a constant = 9X10^9 Nm²/c²

F = 0.4866N

Step 3: calculate the acceleration of each particle

F = m*a

a = F/m

a = 0.4866/(6.7X10^-5)

a = 7,262.69 m/s²

Step 4: calculate the velocity of each particle when the separation is one-half the original value

V² = U² + 2as

where U is initial velocity = 0

V² = 0 + 2 X 7,262.69 X 0.34

V² = 4,938.629

V = √4,938.629

V = 70.28 m/s²

the rate of each particle when the separation between them is one-half the original value = 70.28 m/s²

continuous conducting path for current, power supply

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2