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3 years ago

Compared to a virtual image, a real image

Physics

1 answer:

Rom4ik [11]3 years ago

5 0

Compared to a virtual image, a real image can be observed on a screen.

Explanation

When a beam of light meets at a specific point after the reflection from the mirror, the image formed at the point is said to be a real image.

Real images are generally inverted.

When a beam of light seems to appear to meet at a certain point after the reflection from the mirror, the image formed at the point is said to be a virtual image.

Virtual images are generally erect.

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In the figure shown, the coefficient of kinetic friction between the block and the incline is 0.2. What is the

Maksim231197 [3]

qwert yuiop asdfg hjkl zxcvb nm,./

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2 years ago

Show that any three linear operators A, B, and Ĉ satisfy the following (Ja- cobi) identity (10 pt) [[A, B] Ĉ] + [[B,C), A] + [[C

DerKrebs [107]

Answer:

Three linear operators A,B, and C will satisfy the condition $[[A, B],C] + [[B,C), A] + [[C, A], B] = 0$ .

Explanation:

According to the question we have to prove.

$[[A, B],C] + [[B,C), A] + [[C, A], B] = 0$

Now taking Left hand side of the equation and solve.

$[[A, B],C] + [[B,C), A] + [[C, A], B]$

Now use commutator property on it as,

$=[A,B] C-C[A,B]+[B,C]A-A[B,C]+[C,A]B-B[C,A]\\=(AB-BA)C-C(AB-BA)+(BC-CA)A-A(BC-CB)+(CA-AC)B-B(CA-AC)\\=ABC-BAC-CAB+CBA+BCA-CAB-ABC+ACB+CAB-ACB-BCA+BAC\\=0$

Therefore, it is proved that $[[A, B],C] + [[B,C), A] + [[C, A], B] = 0$ .

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3 years ago

Calculate the weight of the wooden cube.<br>thanks i will rate you 5 star after answer is given...

Lelechka [254]

Answer:

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2 years ago

If a 5000-kg is moving at a speed of 43 m/s, what is its momentum?

Eduardwww [97]

Answer:

215000kgm/s

Explanation:

Given parameters:

Mass of the moving body = 5000kg

Velocity = 43m/s

Unknown:

Momentum = ?

Solution:

The momentum of a body is the amount of motion a body possess.

It is mathematically expressed as:

Momentum = mass x velocity

Now:

Momentum = 5000 x 43 = 215000kgm/s

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3 years ago

Monochromatic light falls on two very narrow slits 0.048 mm apart. successive fringes on a screen 5.00 m away are 6.5 cm apart n

atroni [7]

In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is
$y= \frac{m \lambda D}{d}$
where D=5.00 m is the distance of the screen from the slits, and
$d=0.048 mm=0.048 \cdot 10^{-3}m$ is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
$\lambda = \frac{yd}{mD}= \frac{(0.065 m)(0.048 \cdot 10^{-3}m)}{(1)(5.00 m)}= 6.24 \cdot 10^{-7}m$

And from the relationship between frequency and wavelength, $c=\lambda f$ , we can find the frequency of the light:
$f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{6.24 \cdot 10^{-7}m}=4.81 \cdot 10^{14}Hz$

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3 years ago