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Ludmilka [50]
3 years ago
5

A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.727 g, q = 2.73 µC is locat

ed on the x axis at x = 19.3 cm, moving with a speed of 49.2 m/s in the positive y direction. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)
Physics
1 answer:
Montano1993 [528]3 years ago
5 0

Answer:

Q = 12.466μC

Explanation:

For the particle to execute a circular motion, the electrostatic force must be equal to the centripetal force:

Fe = \frac{K*q*Q}{r^{2}} = \frac{m*V^{2}}{r}

Solving for Q:

Q = \frac{m*V^{2}*r}{K*q}

Taking special care of all units, we can calculate the value of the charge:

Q = 12.466μC

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Answer:

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3 0
3 years ago
A devout halloweener not only dressed as an astronaut, but travelled to the moon for the full experience. The astronaut jumps on
nikdorinn [45]

Answer:

2.78 m

Explanation:

At the peak, the velocity is 0.

Given:

a = -1.6 m/s²

v₀ = 2.98 m/s

v = 0 m/s

x₀ = 0 m

Find:

x

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (2.98 m/s)² + 2(-1.6 m/s²) (x - 0 m)

x = 2.775 m

Rounded to 3 sig-figs, the astronaut halloweener reaches a maximum height of 2.78 meters.

5 0
3 years ago
Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

6 0
3 years ago
Read 2 more answers
Crying is a .........of distress to babies.
RSB [31]
Hey there

Crying is a signal <span>of distress to babies.</span>
5 0
3 years ago
Where is the natural light display called aurora borealis located?
Xelga [282]

The natural light display called aurora borealis is located in the northern

hemisphere.

There are two types of aurora which are called aurora borealis and aurora

australis. The aurora borealis is located in the Northern hemisphere while

the aurora australis is located in the Southern hemisphere.

They receive their energy through the interaction of charged particles

on the Sun and Earth to produce the light display. An example

of the interaction involves solar wind with atoms of the upper atmosphere.

Read more on brainly.com/question/20191244

5 0
2 years ago
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