A potter's wheel rotates with an angular frequency of 1.54 rad/s. What is its period?

How many electrons must be removed from each of two 4.85-kg copper spheres to make the electric force of repulsion between them

VladimirAG [237]

Answer:

$n=2.611*10^{9}electrons$

Explanation:

Given data

mass of copper m=4.85 kg

charge of electron qe= -1.6×10⁻¹⁹C

To find

Number of electron n must be removed

Solution

Equate the magnitude of electric force Fe of repulsion between two two spheres to the magnitude of gravitational force of attraction between them

So

$F_{e}=F_{g}\\ k\frac{q_{e}q_{e}}{r^{2} }=G\frac{m^{2} }{r^{2} }\\ k(q_{e})^{2}=Gm^{2}\\$

where q is charge of each sphere which is equal to number n of removed electrons multiplied by each charge qe

So

$k(nq_{e})^{2}=Gm^{2}\\n=\sqrt{\frac{Gm^{2}}{k(q_{e})^{2}} }\\ n=\sqrt{\frac{(6.67*10^{-11}N.m^{2}/kg^{2})(4.85kg)^{2}}{(8.99*10^{9}N.m^{2}/C^{2} )(1.6*10^{-16}C)^{2}} }\\n=2.611*10^{9}electrons$

3 0

3 years ago

Which of the following is not a velocity?*

ira [324]

Answer:

50 mph

Explanation:

doesn’t have direction

3 0

3 years ago

Read 2 more answers

You've been called in to investigate a construction accidentin

KengaRu [80]

Answer:

57300 N

Explanation:

The container has a mass of 5300 kg, the weight of the container is:

f = m * a

w = m * g

w = 5300 * 9.81 = 52000 N

However this container was moving with more acceleration, so dynamic loads appear.

w' = m * (g + a)

w' = 5300 * (9.81 + 1) = 57300 N

The rating for the cable was 50000 N

The maximum load was exceeded by:

57300 / 50000 - 1 = 14.6%

5 0

3 years ago

Physics help please..............

adoni [48]

Answer:

3.68 m/s

Explanation:

Full answer in the attached picture

7 0

3 years ago

In the most common isotope of Hydrogen the nucleus is made out of a single proton. When this Hydrogen atom is neutral, a single

FinnZ [79.3K]

Answer:

The ratio of electric force to the gravitational force is $2.27\times 10^{39}$

Explanation:

It is given that,

Distance between electron and proton, $r=4.53\ A=4.53\times 10^{-10}\ m$

Electric force is given by :

$F_e=k\dfrac{q_1q_2}{r^2}$

Gravitational force is given by :

$F_g=G\dfrac{m_1m_2}{r^2}$

Where

$m_1$ is mass of electron, $m_1=9.1\times 10^{-31}\ kg$

$m_2$ is mass of proton, $m_2=1.67\times 10^{-27}\ kg$

$q_1$ is charge on electron, $q_1=-1.6\times 10^{-19}\ kg$

$q_2$ is charge on proton, $q_2=1.6\times 10^{-19}\ kg$

$\dfrac{F_e}{F_g}=\dfrac{kq_1q_2}{Gm_1m_2}$

$\dfrac{F_e}{F_g}=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}$

$\dfrac{F_e}{F_g}=2.27\times 10^{39}$

So, the ratio of electric force to the gravitational force is $2.27\times 10^{39}$ . Hence, this is the required solution.

3 0

3 years ago