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nexus9112 [7]
2 years ago
5

If the magnification produced by a lens has a negative value, the image will be

Physics
2 answers:
drek231 [11]2 years ago
8 0

Answer:

I think it's B, not quite sure tho.

damaskus [11]2 years ago
5 0
I believe the answer is A. Virtual and inverted.
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A Michelson interferometer uses light from a sodium lamp. Sodium atoms emit light having wavelengths 589.0 nm and 589.6 nm. The
DIA [1.3K]

The distance mirror M2 must be moved so that one wavelength has produced one more new maxima than the other wavelength is;

<u><em>L = 57.88 mm</em></u>

<u><em /></u>

We are given;

Wavelength 1; λ₁ = 589 nm = 589 × 10⁻⁹ m

Wavelength 2; λ₂ = 589.6 nm = 589.6 × 10⁻⁹ m

We are told that L₁ = L₂. Thus, we will adopt L.

Formula for the number of bright fringe shift is;

m = 2L/λ

Thus;

For Wavelength 1;

m₁ = 2L/(589 × 10⁻⁹)

For wavelength 2;

m₂ = 2L/(589.6)

Now, we are told that one wavelength must have produced one more new maxima than the other wavelength. Thus;

m₁ - m₂ = 2

Plugging in the values of m₁ and m₂ gives;

(2L/589) - (2L/589.6) = 2

divide through by 2 to get;

L[(1/589) - (1/589.6)] = 1

L(1.728 × 10⁻⁶) = 1

L = 1/(1.728 × 10⁻⁶)

L = 578790.67 nm

L = 57.88 mm

Read more at; brainly.com/question/17161594

8 0
2 years ago
What is the effect on the force of gravity between two objects if the mass of one object remains unchanged while the distance to
Vadim26 [7]

Answer:

The correct answer to the question is

B. It always decreases

Explanation:

To solve the question, we note that the foce of gravity is given by

F_G=\frac{Gm_1m_2}{r^2} where

G= Gravitational constant

m₁ = mass of first object

m₂ = mass of second object

r = the distance between both objects

If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have

F_{G2} =\frac{Gm_1(2m_2)}{(2r)^2} = \frac{2}{4} \frac{Gm_1m_2}{r^2}

Therefore the gravitational force is halved. That is it will always decrease

4 0
2 years ago
A 0.73-m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the bar
Tasya [4]

Answer:

A)  B = 5.4 10⁻⁵ T, B) the positive side of the bar is to the West

Explanation:

A) For this exercise we must use the expression of Faraday's law for a moving body

            fem = -  \frac{d \phi }{dt}

            fem = - \frac{d (B l y}{dt}= - B l v- d (B l y) / dt = - B lv

            B = - \frac{fem}{l \ v}

we calculate

             B = - 7.9 10⁻⁴ /(0.73 20)

             B = 5.4 10⁻⁵ T

B) to determine which side of the bar is positive, we must use the right hand rule

the thumb points in the direction of the rod movement to the south, the magnetic field points in the horizontal direction and the rod is in the east-west direction.

Therefore the force points in the direction perpendicular to the velocity and the magnetic field is in the east direction; therefore the positive side of the bar is to the West

4 0
3 years ago
What will occur when the waves peak at the same place at the same time?
dedylja [7]
Hello!

This is a matter of superposition.
When the waves peak at the same time and place, they produce constructive interference, meaning the waves interact together in a positive way, to make a wave with Amplitude of both waves added together. When the peaks differ however, at the same time and place, then it is destructive interference and the waves essentially cancel each other out.

Hope this helps. Any questions please just ask. Thank you kindly.
5 0
2 years ago
A swimming pool has the shape of a right circular cylinder with radius 21 feet and height 10 feet. Suppose that the pool is full
AysviL [449]

Answer:

The water required to pump all the water to a platform 2 feet above the top of the pool is  is 6061310.32 foot-pound.

Explanation:

Given that,

Radius = 21 feet

Height = 10 feet

Weighing = 62.5 pounds/cubic

Work = 4329507.37572

Height = 2 feet

Let's look at a horizontal slice of water at a height of h from bottom of pool

We need to calculate the area of slice

Using formula of area

A=\pi r^2

Put the value into the formula

A=\pi\times21^2

A=441\pi\ feet^2

Thickness of slice t=\Delta h\ ft

The volume is,

V=(441\pi\times\Delta h)\ ft^3

We need to calculate the force

Using formula of force

F=W\times V

Where, W = water weight

V = volume

Put the value into the formula

F=62.5\times(441\pi\times\Delta h)

F=27562.5\pi\times\Delta h\ lbs

We need to calculate the work done

Using formula of work done

W=F\times d

Put the value into the formula

W=27562.5\pi\times\Delta h\times(10-h)\ ft\ lbs

We do this by integrating from h = 0 to h = 10

We need to find the total work,

Using formula of work done

W=\int_{0}^{h}{W}

Put the value into the formula

W=\int_{0}^{10}{27562.5\pi\\times(10-h)}dh

W=27562.5\pi(10h-\dfrac{h^2}{2})_{0}^{10}

W=27562.5\pi(10\times10-\dfrac{100}{2}-0)

W=4329507.37572

To pump 2 feet above platform, then each slice has to be lifted an extra 2 feet,

So, the total distance to lift slice is (12-h) instead of of 10-h

We need to calculate the water required to pump all the water to a platform 2 feet above the top of the pool

Using formula of work done

W=\int_{0}^{h}{W}

Put the value into the formula

W=\int_{0}^{10}{27562.5\pi\\times(12-h)}dh

W=27562.5\pi(12h-\dfrac{h^2}{2})_{0}^{10}

W=27562.5\pi(12\times10-\dfrac{100}{2}-0)

W=1929375\pi

W=6061310.32\ foot- pound

Hence, The water required to pump all the water to a platform 2 feet above the top of the pool is  is 6061310.32 foot-pound.

8 0
3 years ago
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