At the 1 s mark, which objects have equal momentum?

Physics

1 answer:

Rudik [331]3 years ago

3 0

C. The bowling ball and the bicycle

p = mv
P of bike = 12x5 = 60
P of rock = 2x20 = 40
P of ball = 5 x 10 = 60

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kumpel [21]

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3 0

3 years ago

*example included* Two uncharged spheres are separated by 3.50 m. If 1.30 ✕ 10¹² electrons are removed from one sphere and place

likoan [24]

Considering the Coulomb's Law, the magnitude of the Coulomb force is 3.1865 N.

<h3>Coulomb's Law</h3>

Charged bodies experience a force of attraction or repulsion on approach.

From Coulomb's Law it is possible to predict what the electrostatic force of attraction or repulsion between two particles will be according to their electric charge and the distance between them.

From Coulomb's Law, the electric force with which two point charges at rest attract or repel each other is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

$F=k\frac{Qq}{d^{2} }$

where:

F is the electrical force of attraction or repulsion. It is measured in Newtons (N).
Q and q are the values of the two point charges. They are measured in Coulombs (C).
d is the value of the distance that separates them. It is measured in meters (m).
K is a constant of proportionality called the Coulomb's law constant. It depends on the medium in which the charges are located. Specifically for vacuum k is approximately 9×10⁹ $\frac{Nm^{2} }{C^{2} }$ .

The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.

In this case, you know that:

The two uncharged sphere are separated by the distance of d= 3.50 m
The number of electrons are 1.30×10¹².
Electrons is elementary charge and charges on both the sphere is same. The value of electron is 1.602×10⁻¹⁹ C. This is, Q=q=1.30×10¹²×1.602×10⁻¹⁹ C= 2.0826×10⁻⁷ C

Replacing in Coulomb's Law:

$F=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{(2.0826x10^{-7} C)x(2.0826x10^{-7} C)}{(3.50 m)^{2} }$