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3 years ago

A free negative charge released in an electric field will

Physics

1 answer:

SashulF [63]3 years ago

5 0

Answer:

Will experience a force due to electric field.

Explanation:

When a free negative charge is released in an electric field it experiences a force due to the electric field in a direction opposite to the direction of the magnetic field.

According to Coulomb's law this force is mathematically given as:

$F=E.q$

and, electric field due to a charge is given as:

$E=\frac{1}{4\pi.\epsilon_0}.\frac{q}{r^2}$

where:

permittivity of free space $\epsilon_0=8.85\times 10^{-12}\ m^{-3}.kg^{-1}.s^4.A^2$

q = magnitude of charge

r = radial distance from the charge

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The charges of two particles are as follows: Q1=2 x 10 -8 C and Q2 = 3 x 10 -7 C. Find the magnitude of the force between these

Fantom [35]

Answer:

F = 3.86 x 10⁻⁶ N

Explanation:

First, we will find the distance between the two particles:

$r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}\\$

where,

r = distance between the particles = ?

(x₁, y₁, z₁) = (2, 5, 1)

(x₂, y₂, z₂) = (3, 2, 3)

Therefore,

$r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\$

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

$F = \frac{kq_{1}q_{2}}{r^2}\\$

where,

F = magnitude of force = ?

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 2 x 10⁻⁸ C

q₂ = magnitude of second charge = 3 x 10⁻⁷ C

r = distance between the charges = 3.741 m

Therefore,

$F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\$

5 0

3 years ago

In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo

Wittaler [7]

Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

Time = 7.00 s

Suppose we find the magnitude of the acceleration of the 602.0-g block

We need to calculate the acceleration

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Where, s = distance

t = time

a = acceleration

Put the value into the formula

$60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2$

$a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}$

$a=0.0244\ m/s^2$

Hence, The acceleration of the both masses is 0.0244 m/s².

5 0

3 years ago

a) Calculate the magnitude of displacement of the car in 40 seconds. b) During which part of the journey was the car acceleratin

Zielflug [23.3K]

Answer:

a) 600 meters

b) between 0 and 10 seconds, and between 30 and 40 seconds.

c) the average of the magnitude of the velocity function is 15 m/s

Explanation:

a) In order to find the magnitude of the car's displacement in 40 seconds,we need to find the area under the curve (integral of the depicted velocity function) between 0 and 40 seconds. Since the area is that of a trapezoid, we can calculate it directly from geometry:

$Area \,\,Trapezoid=(\left[B+b]\,(H/2)\\displacement= \left[(40-0)+(30-10)\right] \,(20/2)=600\,\,m$

b) The car is accelerating when the velocity is changing, so we see that the velocity is changing (increasing) between 0 and 10 seconds, and we also see the velocity decreasing between 30 and 40 seconds.

Notice that between 10 and 30 seconds the velocity is constant (doesn't change) of magnitude 20 m/s, so in this section of the trip there is NO acceleration.

c) To calculate the average of a function that is changing over time, we do it through calculus, using the formula for average of a function:

$Average\,of\,f(x)=\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx$

Notice that the limits of integration for our case are 0 and 40 seconds, and that we have already calculated the area under the velocity function (the integral) in step a), so the average velocity becomes:

$Avearage=\frac{600\,\,m}{40\,\,s}= 15\,\,\frac{m}s}$