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3 years ago

Which image is observe by seeing a distant thing using magnifying glass?

Physics

2 answers:

Snezhnost [94]3 years ago

8 0

Answer: D) Big - straight

<h2>Brainliest pls if correct!</h2>

lesya692 [45]3 years ago

7 0

Answer:

1/f = 1/o + 1 /i thin lens equation

1 / i = 1/f - 1/o = (o - f) / (o f) rearranging

or i = o f / (o - f) for the image distance

This image will be positive (real) with values given

m = -i / o magnification

m = - f / (o - f) shows m is negative and the image inverted

Also, since o here is large m will be small and the image is small

I think they mean A as the right answer

Different texts use different terms but here i is image distance and o the object distance

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A solid block of metal is at its melting temperature. it takes 24j to melt 6kg of a metal. what is the latent heat of fusion of

Marina86 [1]

The Latent heat of fusion is also known as the Enthalpy of Fusion. It is a kind pf change in enthalpy.

<h3>Brief explanation of Latent Heat of Fusion? </h3>

The latent heat of fusion, is the change in enthalpy that takes place when energy, typically heat, is provided to a specific quantity of a substance to causes it to change from a solid state to a liquid state at a constant pressure.

For instance, 1 kilogram of ice melts at 0 °C at a variety of pressures, absorbing 333.55 kJ of energy without causing a change in temperature.

The heat of solidification, which occurs when a substance changes from a liquid to a solid, is equal and opposing.

This energy includes the contribution necessary to displace its surroundings against ambient pressure in order to accommodate any corresponding change in volume.

The melting point or the freezing point, depending on the situation, is the temperature at which the phase transition takes place. Conventionally, unless otherwise stated, the pressure is taken to be 1 atm (101.325 kPa).

<h3>Calculation</h3>

The formula of latent heat of fusion is L= q/m

Here q is the heat provided and m is the mass of the object.

So, L = 24j/6kg

L= 4j/kg

So, the latent heat of fusion is 4 j/kg

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2 years ago

You have a friend who reports that he falls asleep easily around 11 PM but then awakens for about an hour most nights around two

Solnce55 [7]

Answer:

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Explanation:

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3 years ago

a dog pulls on a pillow with a force of 8.4 N at an angle of 31 degrees above the horizontal. what is the x component of this fo

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The x component of the force is 8.4N * cos(31°) = 7.2N (2 s.f.)

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4 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.

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3 years ago

Coulomb’s Law explains which of the following

Travka [436]

The force of attraction or repulsion between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them and acts along the line joining the two charges . F=k Q1Q2/r^2

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