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lina2011 [118]
2 years ago
7

A point charge of 9.00 × 10−9 C is located at the origin of a coordinate system. A positive charge of 3.00 × 10−9 C is brought i

n from infinity to a point such that the electrical potential energy associated with the two charges is 8.09 × 10−7 J.
Physics
1 answer:
dlinn [17]2 years ago
8 0

A point charge is located at the origin of a coordinate system. A positive charge is brought in from infinity to a point. The charges are at distance for given electrical potential energy is 3.34 x  10⁷ m.

<h3>What is electric potential energy?</h3>

The electric potential energy is the work done by a test charge to bring it from infinity to a particular location.

The electric potential energy is given by the relation,

V = kQ/r

where k = 9 x 10⁹ J.m/C ,Q = 3 x 10⁻⁹ C, V =8.09 × 10⁻⁷ J.

Substitute the values into the expression to get the distance between the charges.

8.09 × 10⁻⁷ =  9 x 10⁹ x  3 x 10⁻⁹ / r

r =3.34 x  10⁷ m

Thus, the distance between the charges will be 3.34 x  10⁷ m.

Learn more about  electric potential energy.

brainly.com/question/12645463

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A space station, in the form of a wheel 140 m in diameter, rotates to provide an "artificial gravity" of 3.90 m/s2 for persons w
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a_{rad}= \frac{v^2}{r}
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a_{rad}= \frac{r^2 \omega^2}{r}=r\omega^2

Now

70\omega^2=3.90 \frac{m}{s^2}  \\  \\ \omega= \sqrt{ \frac{3.9}{70} }

Using the relation

\omega=2 \pi f

2 \pi f= \sqrt{ \frac{3.9}{70} }\\  \\ f= \frac{1}{2 \pi}\sqrt{ \frac{3.9}{70} }Hz

Putting into rpm

\frac{60}{2 \pi}\sqrt{ \frac{3.9}{70}} =2.254rpm

8 0
3 years ago
An object of mass m is traveling in a circle with centripetal force Fc. If the velocity of the object is v, what is the radius o
borishaifa [10]

Hi there!

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We can rearrange the equation to solve for 'r'.

Multiply both sides by r:
r * F_c = mv^2

Divide both sides by Fc:
\boxed{ r= \frac{mv^2}{F_c}}

7 0
2 years ago
A point charge of magnitude q is at the center of a cube with sides of length L
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Answer is in the attachment below:

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3 years ago
A new planet is discovered beyond Pluto at a mean distance to the sun of 4004 million miles. Using Kepler's third law, determine
AVprozaik [17]

Answer:

103239.89 days

Explanation:

Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

                               a³ / T² = 7.496 × 10⁻⁶  (a.u.³/days²)

where,

a is the distance of the semi-major axis in a.u

T is the orbit time in days

Converting the mean distance of the new planet to astronomical unit (a.u.)

                       1 a.u = 9.296 × 10⁷ miles        

                                      \frac{4004 * 10^{6}}{9.296 * 10^{7}}  =  43.07\ a.u.

Substituting the values into Kepler's third law equation;

                                    \frac{(43.07)^{3}}{T^{2}}  =  7.496 * 10^{-6}  

                                    T^{2} = \frac{(43.07)^{3}}{7.496 * 10^{-6}} (days)²

                                    T^{2} = \sqrt{\frac{(43.07)^{3}}{7.496 * 10^{-6}}}

                                    T = 103239.89 days

An estimate time T for the new planet to travel around the sun in an orbit is 103239.89 days

7 0
3 years ago
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