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3 years ago

QUICK PLEASE! WILL GIVE BRAINLIEST!

Physics

2 answers:

Crank3 years ago

7 0

Answer:

hi!

Explanation:

i think the answer is x=0m , y=0,55m

Elanso [62]3 years ago

4 0

Answer:

x = 0, y = 1.5 m

Explanation:

The magnitude of the electric field strength is given by the formula

E= $\frac{Kq }{r}$

Where

E is the magnitude of the electric field at the a point in space

k is the universal Coulomb constant

q is the charge of the particle

r is the distance that the point in space, at which we want to know the E, is from the point charge that is causing E

Two positive charges in the question can be referred to as q₁ and q₂.

q₁ =6μC

q₂ =4μC

Each of the positive charges exert electric field of magnitude of E₁ and E₂ respectively. For positive charge, the electric field direction is away from the positive charge, unlike in negative charge where the electric field has a direction towards the negative charge. The resultant magnitude of the electric field exerted by both positive charges is E₀.

E₀=E₁ +E₂

Since the direction of the electric field is outward from each of the positive charge q1 and q2, their electric field strength will be opposite, hence

E₀ = E₁ - E₂

From the question, we were ask to find point where E₀ is zero, so E₁=E₂. Using the origin as point of reference, let the point where the electric field strength is zero x from position of q₁ and 1-x from q₂. So that

E₁ = $\frac{kq_{1} }{x^{2} }$ and E₂ = $\frac{kq_{2} }{(1-x)^{2} }$

Now E₁=E₂

$\frac{kq_{1} }{x^{2} }$ = $\frac{kq_{2} }{(1-x)^{2} }$

Divide both sides of the equation by k

$\frac{q_{1} }{x^{2} }$ = $\frac{q_{2} }{(1-x)^{2} }$

q₁(1-x)²=q₂x²

q₁*(1 - 2x - x²) = q₂x

Remember that q₁ = 6μC and q₂ = 4μC

6(1 - 2x - x²) = 4x²

3(1 - 2x - x²) = 2x²

3 - 6x - 3x² = 2x²

3 - 6x - 5x² = 0

This is a Quadratic Equation of the form ax² + bx + c

x = $\frac{-b +/- \sqrt{b^{2}-4ac } }{2a}$

From our quadratic equation 3 - 6x - 5x² = 0

a = -5 , b = -6 and c = 3

x = $\frac{-(-6) +/- \sqrt{(-6)^{2} -(4*-5*3)} }{2*-5}$

x = $\frac{6 +/- \sqrt{36 - (-60)} }{-10}$

x= $\frac{6 +/- 9.792}{-10}$

x = $\frac{15.797}{-10}$ or $\frac{-3.797}{-10}$

x = -1.5 or 0.379

The negative sign in -1.5 simply indicate the direction of the electric field. Note that since both charges are the same, there would be repulsion which is likely going to increase the distance between the two positive charges. Therefore the correct point where the electric field strength is zero is 1.5 m along the y-axis.

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What energy is directly dependent upon velocity and mass?

erastova [34]

Answer:

Kinetic energy

Explanation:

3 0

3 years ago

An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ outward

3) y-component of the electric field at r =8.7 cm: 0

4) x-component of the electric field at r = 1.15 cm: $1.28\cdot 10^7 N/C$ outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

The linear charge density of the cylindrical insulating shell can be found by using

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

in the outward direction.

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

$E_y=0$

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

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4 0

3 years ago

While playing baseball with your friends your hands begin to sting after you ctach several fast balls.

muminat

Answer:

Explanation:

To stop a ball with high momentum in a small-time imparts a high amount of impact on hands. This is the reason for the stinging of hands.

The momentum of the ball is due to the mass and velocity. To prevent stinging in the hand one needs to lower his hands to increase the time of contact. In this way, the momentum transfer to the hands will be lesser.

7 0

3 years ago

A double-slit experiment uses light of wavelength 650 nm with a slit separation of 0.100 mm and a screen placed 4.0 m away. a) W

dezoksy [38]

Answer:

Explanation:

a ) Slit separation d = .1 x 10⁻³ m

Screen distance D = 4 m

wave length of light λ = 650 x 10⁻⁹ m

Width of central fringe = λ D / d

= $\frac{650\times10^{-9}\times4}{.1\times10^{-3}}$

= 26 mm

b ) Distance between 1 st and 2 nd bright fringe will be equal to width of dark fringe which will also be equal to 26 mm

c ) Angular separation between the central maximum and 1 st order maximum will be equal to angular width of fringe which is equal to

λ / d

= $\frac{650\times10^{-9}}{.1\times10^{-3}}$

= 6.5 x 10⁻³ radian.

8 0

3 years ago

If the volume of a sample of gas is reduced at constant temperature, the average velocity of the molecules _______, the average

VLD [36.1K]

Answer:

Explanation:

If the volume of a sample of gas is reduced at constant temperature, the average velocity of the molecules increases, the average force of an individual collision increases, and the average number of collisions with the wall, per unit area, per second increases.

As volume is reduced, the gas molecules come closer together, which increases the number of collisions between them and their collisions with the container walls. Also, since the distance traveled by each molecule between successive collision decreases, the molecule velocity doesn't decrease much within collisions as a result of which, the average velocity is higher compared to when the gas is stored in a larger volume. Finally, due to constant collisions, the direction of molecule travel changes rapidly owing to which the acceleration of molecules increases.

6 0

3 years ago