A student conducted an experiment to measure the acceleration due to gravity 'g' of a simple pendulum.

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

3 years ago

Which is greater the attraction of the earth for 1 kg of aluminum or aluminum or attraction of 1kg of aluminum for the earth?

OleMash [197]

Those forces are exactly equal.

Gravity always works as a pair of <em>EQUAL</em> forces ... one in each direction
between two masses. Your weight on the Earth is exactly the same as
the Earth's weight on you.

8 0

3 years ago

So, why can a properly executed karate kick break a concrete block without fracturing bones [16]? first, bone is a very strong m

Sav [38]

PM me for full answer, please. If it's not too late.

3 0

3 years ago

How much work does gravity do on the compressor during this displacement?

gayaneshka [121]

Explanation and Examples

let the mass of the compressor be

mass (m):

height in x axis is (h1)

height in y axis be (h2):

Height difference: h2-h1

displacement x force:

mass x gravity x height

(m)*9.8*(height difference) = ___ J

Since gravity is forcing down, it would be negative!

Put the values that you require and get the answer.

6 0

3 years ago

Read 2 more answers

4. Una cuerda de acero de piano mide 1.60 m de longitud y 0.20 cm de diámetro. ¿Cuál es la tensión en la cuerda si se estira 0.2

bazaltina [42]

Answer:

$1030.83\ \text{N}$

Explanation:

$\Delta L$ = Cambio en la longitud de la cuerda = 0.25 cm

T = tensión en cuerda

A = Área de la cadena = $\dfrac{\pi}{4}d^2$

d = Diámetro de la cuerda = 0.2 cm

L = Longitud original de la cuerda = 1.6 m

El cambio de longitud de una cuerda viene dado por

$\Delta L=\dfrac{TL}{AE}\\\Rightarrow T=\dfrac{\Delta LAE}{L}\\\Rightarrow T=\dfrac{0.25\times 10^{-2}\times \dfrac{\pi}{4}(0.2\times 10^{-2})^2\times 210\times 10^9}{1.6}\\\Rightarrow T=1030.84\ \text{N}$

La tensión en la cuerda es $1030.84\ \text{N}$ .

8 0

3 years ago