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3 years ago

When ____________________ equals the force of gravity on a falling object, the object reaches terminal velocity.

Physics

1 answer:

Lilit [14]3 years ago

5 0

The correct answer to the question is : Air resistance.

EXPLANATION:

Before coming into any conclusion, first we have to understand the terminal velocity.

An object is said to be moving with terminal velocity if the net force acting on the object is zero i.e the body is non accelerating.

As per the question, the object is falling under gravity. The force of gravity acts in vertical downward direction.

Except gravity, the object is also under air resistance. It is the force which opposes the motion of the object, and acts in vertical upward direction.

If the air resistance becomes equal to force of gravity, then the net force acting on the object is zero . It is so because two forces act in opposite direction.

In case of liquids, the resistance force is called as viscosity.

Hence, the correct answer to the question is air resistance.

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A positive point charge (q = +5.45 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 1.49 m. A p

ruslelena [56]

Answer:

$r_{B} = 3.34\ m$

Solution:

As per the question:

Point charge, q = $5.45\times 10^{- 8}\ C$

Test charge, $q_{o} = 4.96\times 10^{- 11}\ C$

Work done by the electric force, $W_{AB} = - 9.05\times 10^{- 9}\ J$

Now,

We know that the electric potential at a point is given by:

$V = \frac{kqq'}{r}$

where

r = separation distance between the charges.

Also,

The work done by the electric force i moving a test charge from point A to B in an electric field:

$W_{AB} = kqq_{o}(\frac{1}{r_{B} - \frac{1}{r_{A}}}$

$- 9.05\times 10^{- 9} = 9\times 10^{9}\times 5.45\times 10^{- 8}\times 4.96\times 10^{- 11}(\frac{1}{r_{B}} - \frac{1}{1.49}}$

$- 0.3719 = (\frac{1}{r_{B}} - 0.67}$

$r_{B} = \frac{1}{0.299} = 3.34\ m$

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