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marshall27 [118]

3 years ago

8

The critical angle for a beam of light passing from water into air is 48.8°. this means that all light rays with an angle of inc

idence in the water that is greater than 48.8° will be

2 answers:

Len [333]3 years ago

8 0

totally internally reflected

liraira [26]3 years ago

3 0

totally internally reflected

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An object of height 2.2 cm is placed 5.1 cm in front of a diverging lens of focal length 19 cm and is observed through the lens

monitta

Answer:

a) i = -4.02 cm , b) h’= 1,576 cm

Explanation:

a) The constructor equation is

1 / f = 1 / i + 1 / o

Where f is the focal length, i and o are the distance to the image and the object

Let's clear the distance to the image

1 / i = 1 / f - 1 / o

1 / i = 1 / -19 - 1 / 5.1

1 / i = 0.2487

i = -4.02 cm

b) let's use the expression of magnification

m = h’/ h = - i / o

h’= - h i / o

h’= 2.2 4.02 /5.1

h’= 1,576 cm

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Examine the nuclear reaction: Superscript 1 subscript 1 H plus superscript 1 subscript 0 n right arrow superscript 2 subscript 1

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Answer:

C. A change has occurred in the nucleus.

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Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.

Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge $q_{1}= 3.8\times10^{-6}\ C$

Second charge $q_{2}=3.2\times10^{-6}\ C$

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

$E_{1}=E_{2}$

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

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

$\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}$

$x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}$

Put the value into the formula

$x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}$

$x=1.69\ m$

Hence, The distance is 1.69 m.

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