What determines the ideal mechanical advantage of a pulley?

How far does a 1.2g bullet with kinetic energy of 1.2 j go in 2 seconds ?

almond37 [142]

Answer:

To calculate the energy in joules, simply enter the mass of ammunition (in grams) that you use, and the fps that you've read from your Chrono unit.

5 0

1 year ago

Is the refraction different entering medium that has a higher index of refraction compared to entering amedium that has a lower

oee [108]

Answer:

Yes

Explanation:

The speed of light when it travels through glass, diamond, etc, the light travels at different speed from the speed of light. Speed of the light in material is related to the index of refraction.

The change in speed which occurs when the light passes from one medium to the another is responsible for bending of the light which is called as refraction.

When the light goes into a medium with the higher index of the refraction, light bends towards normal. Conversely, if the light traveling goes from higher refractive index to lower refractive index, it will bend away from the normal.

Hence, the refraction is different in both the scenario.

6 0

3 years ago

A point charge is placed at each corner of square with side leanth a. The charges all have same magnitude q. My question, What i

nexus9112 [7]

Answer:

E = k q / a² (1.3535) (- i ^ + j ^)

E = k q / a² 1.914 , θ’= 135

Explanation:

For this exercise we will use Newton's second law where we must add as vectors

E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃

Let's look for the value of each term

On the x axis

E₁₂ = k q / a²

On the y axis

E₁₄ = k q / a²

For the charge in the opposite corner we look for the distance

d = √ (a² + a²) = a √2

let's look for the field

E₁₃ = k q / d²

E₁₃ = k q / 2a²

let's use trigonometry to find the two components of this field

cos 45 = E₁₃ₓ / E₁₃

E₁₃ₓ = E₁₃ cos 45

sin 45 = E_{13y} / E₁₃

E_{13y} = E₁₃ sin 45

E₁₃ₓ = k q / 2a² cos 45

E_{13y} = k q / 2a² sin 45

let's find each component of the electric field

X axis

Eₓ = -E₁₂ - E₁₃ₓ

Eₓ = - k q / a² - k q / 2a² cos 45

Eₓ = - k q / a² (1 + cos 45/2)

cos 45 = sin 45 = 0.707

Eₓ = - k q / a² (1 + 0.707 / 2)

Eₓ = - k q / a² (1.3535)

Y axis

E_y = E₁₄ + E_{13y}

E_y = k q / a² + k q / 2a² sin 45

E_y = k q / a² (1 + sin 45/2)

E_y = k q / a² (1.3535)

we can give the results in two ways

E = k q / a² (1.3535) (- i ^ + j ^)

In modulus and angle form, let's use Pythagoras' theorem for the angle

E = √ (Eₓ² + E_y²)

E = k q / a² 1.3535 √2

E = k q / a² 1.914

we use trigonometry for the angle

tan θ = E_y / Eₓ

θ = tan⁻¹ (E_y / Eₓ)

θ = tan⁻¹ (1 / -1)

θ = 45

in the third quadrant, if we measure the angle of the positive side of the x-axis

θ‘= 90 + 45

θ’= 135

4 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:

1)

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 :

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

2)

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.

3)

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,

$E_y=0$

4)

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,

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

5)

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

brainly.com/question/4273177

#LearnwithBrainly

4 0

3 years ago

A converging lens can produce both real and virtual images depending on the position of the object. Explain when converging lens

inessss [21]

Answer: C

Explanation: