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2 years ago

The argument against your claim (what the other side would say if they disagreed with your claim.) is:

Physics

1 answer:

Vlad1618 [11]2 years ago

3 0

Answer:

a counterclaim

Explanation:

authors purpose is what an author wrote somthing for

opinion is someones thoughts or "side" on a argument

an arguement is a battle of opinions if that makes sense

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A 6.47-mm-high firefly sits on the axis of, and 13.1 cm in front of, the thin lens A, whose focal length is 6.19 cm. Behind lens

bagirrra123 [75]

Answer:

Explanation:

For lens A

object distance u = - 13.1 cm , focal length f = 6.19 cm

From lens formula

1/v - 1/u = 1/f

1 / v + 1/13.1 = 1/6.19

1/v = 1/6.19 - 1/13.1

= .16155 - .07633

= .08522

v = 11.7 3 cm

For lens B

object distance u = - ( 55.7 - 11.73) = - 43.97 cm , focal length f = 27.9 cm

From lens formula

1/v - 1/u = 1/f

1 / v + 1/43.97 = 1/27.9

1/v = 1/27.9 - 1/43.97

= .03584 - .022742

= .013098

v = 76.35 cm

Image will be formed 76.35 cm behind lens B .

magnification of lens system

= m₁ x m₂ , m₁ is magnification by lens A and m₂ is magnification by lens B

= (11.73 / 13.1) x (76.35 / 43.97)

= .8954 x 1.73

= 1.5547

size of image = total magnification x size of object

= 1.5547 x 6.47

= 10 cm approx. The first image will be real and inverted and second image will be erect with respect to object.

6 0

3 years ago

Nuclear fusion has not yet been harnessed as an energy source because

maksim [4K]

Before getting an answer for it first we have to understand nuclear fusion.

Nuclear fusion is a thermo-nuclear reaction in which two light unstable nuclei will form a heavy stable nuclei. In this process there will be some mass defect which will be converted into energy as per Einstein's mass energy equivalence theorem.

The theorem is stated as $E = mc^{2}$ where c is the velocity of light and m is the mass converted into energy.

One take an example of fusion in sun where 4 hydrogen atoms combine to form a helium nucleus which are explained below-

$H_{1}^{1}+H_{1} ^{1} =H_{1} ^{2} +e_{+1} ^{0} +energy$

$H_{1} ^{2} +H_{1} ^{1} =He_{2} ^{3} +energy$

$He_{2} ^{3}+H_{1} ^{1} =He_{2} ^{4} +e_{+1} ^{0} +energy$

-----------------------------------------------------------------------

$4H_{1} ^{1} = He_{2} ^{4} +2e_{+1} ^{0} +energy$

Here $e_{+1} ^{0}$ is the positron.

In this process very high temperature is needed which is approximately equal to the temperature of the sun i.e $10^{6} K$

Such temperature is very difficult to initiate the reaction on the earth surface. It should be carried out in an sustainable way also .Otherwise It will cause nuclear hazards.

6 0

3 years ago

Read 2 more answers

How do we get heat on Earth? Does thermal energy travel directly from the sun?

ExtremeBDS [4]

We get heat on earth from the sun. Energy from the sun is transferred through space and through earths atmosphere to the earths surface. And since this warms the earth surface it creates heat.

7 0

3 years ago

A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ

Elena L [17]

Answer:

The ocean is 6485.6m deep when measured from the vessel

Explanation:

v=1474m/s

t=8.88s

let d represent distance from the vessel to the ocean bottom.

an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

$velocity=\frac{distance}{time}\\ v=\frac{2d}{t} \\vt=2d\\d=\frac{vt}{2}$

$d=\frac{1474*8.8}{2}$

d= 6485.6m

7 0

2 years ago

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0

3 years ago