20.In case the conductor is a heating appliance, then this energy(w) is converted into heat(H) i.e

Determine the electrical force of attraction between two balloons with separate charges of +3.5 × 10-8 C and -2.9 × 10-8 C when

kenny6666 [7]

Answer:

The electrical force of attraction between the balloons is $2.16\times 10^{-5}\ N$ .

Explanation:

Given that,

Charge 1, $q_1=+3.5\times 10^{-8}\ C$

Charge 2, $q_2=-2.9\times 10^{-8}\ C$

Distance between the charges, d = 0.65 m

We need to find the electrical force of attraction between two balloons. It is given by the formula as :

$F=k\dfrac{q_1q_2}{d^2}\\\\F=9\times 10^9\times \dfrac{3.5\times 10^{-8}\times 2.9\times 10^{-8}}{(0.65)^2}\\\\F=2.16\times 10^{-5}\ N$

So, the electrical force of attraction between the balloons is $2.16\times 10^{-5}\ N$ . Hence, this is the required solution.

3 0

3 years ago

An electron of mass 9.11 x 10^-31 kg has an initial speed of 4.00 x 10^5 m/s. It travels in a straight line, and its speed incre

yarga [219]

Answer:

a. F = 2.32*10^-18 N

b. The force F is 2.59*10^11 times the weight of the electron

Explanation:

a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:

$v^2=v_o^2+2ax$ (1)

v: final speed of the electron = 6.60*10^5 m/s

vo: initial speed of the electron = 4.00*10^5 m/s

a: acceleration of the electron = ?

x: distance traveled by the electron = 5.40cm = 0.054m

you solve the equation (2) for a and replace the values of the parameters:

$a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}$

Next, you use the second Newton law to calculate the force:

$F=ma$

m: mass of the electron = 9.11*10^-31kg

$F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N$

The magnitude of the force exerted on the electron is 2.32*10^-18 N

b. The weight of the electron is given by:

$F_g=mg=(9.11*10^{-31}kg)(9.8m/s^2)=8.92*10^{-30}N$

The quotient between the weight of the electron and the force F is:

$\frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}$

The force F is 2.59*10^11 times the weight of the electron

8 0

3 years ago

Explain why surface tension occurs

Fynjy0 [20]

Hey there! The cohesive forces between liquid molecules are responsible for the phenomenon known as surface tension. The molecules at the surface do not have other like molecules on all sides of them and consequently they cohere more strongly to those directly associated with them on the surface. Hope this helps! :)

5 0

3 years ago

An atom with 9 protons and 8 electrons is what kind of atom?

dlinn [17]

Answer:

nonmetal

Explanation:

6 0

3 years ago

2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo

seropon [69]

Answer:

The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=2\pi f$

Put the value into the formula

$\omega=2\times\pi\times60$

$\omega=376.9\ rad/s$

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

$B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}$

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

$B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}$

$B_{max}=2.901\times10^{-13}\ T$

Hence, The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

7 0

4 years ago