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3 years ago

Write the poem titled is Handcuffs and write a dairy entry over two days as the speaker who is in prison.

Physics

1 answer:

Reika [66]3 years ago

7 0

Explanation:

The slum children in an elementary school look pathetic. Their hairs are uncombed. They look pale and shabby. They are undernourished and diseased

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Problem 1. Cylinders and a pendulum A uniform solid cylinder of radius r and mass m can roll inside a hollow cylinder of radius

Nataliya [291]

Answer:

The two answers are in the explanation

Explanation:

Please find the attached files for the solution

7 0

4 years ago

What are cirrus.cirrostratus. And cirrocumulus clouds made up of

Irina-Kira [14]

1. All clouds are made up of tiny ice crystals or small water droplets that form together with a dust on the atmosphere.
Cirrus, cirrostratus and cirrocumulus are made up of ice crystals. It forms when the water vapor goes condensation at high altitudes in where the pressure on the atmosphere reaches the point from 600 mbar at 4000 meter above the sea level up to 200mbar at 12 000m above sea level. These clouds form thunderstorms, cyclones and contrails (a manmade cloud).

3 0

3 years ago

In a setup like that in Figure 27.7, a wavelength of 625 nm is used in a Young's double-slit experiment. The separation between

kap26 [50]

The separation between the slits is d = 8.96

What is fringe width?

Fringe width is the distance between two consecutive bright spots (maximas, where constructive interference take place)
Or two consecutive dark spots (minimas, where destructive interference take place).

Fringe width is given by β = λL/d

In the first case fringe width is β1 = λLA /d = 625 x 10-9 x 0.36 / ( 1.4 x 10-5 ) = 0.016071428 m

The total width of the screen is 0.2 m . So, on one side of the central maximum, the width is 0.1 m

No. of fringes in this 0.1m = 0.1 / 0.016071428 = 6.222

So, since there is a bright fringe after every fringe width, the number of bright fringes on one side of central maximum is 6.

In the second case fringe width is β1 = λLAB /d = 625 x 10-9 x 0.25 / ( 1.4 x 10-5 ) = 0.011160714 m

The total width of the screen is 0.2 m . So, on one side of the central maximum, the width is 0.1 m

No. of fringes in this 0.1m = 0.1 / 0.011160714 = 8.96

So, since there is a bright fringe after every fringe width, the number of bright fringes on one side of central maximum is 8. The ninth one will not be seen since the screen is less a little less in width.

Learn more about fringe width

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The complete question is -

In a setup like that in Figure 27.7, a wavelength of 625 nm is used in a Young's double-slit experiment. The separation between the slits is d = 1.4 × 10-3 m. The total width of the screen is 0.20 m. In one version of the setup, the separation between the double slit and the screen is LA = 0.36 m, whereas in another version it is LB = 0.25 m. On one side of the central bright fringe, how many bright fringes lie on the screen in the two versions of the setup? Do not include the central bright fringe in your counting. --Tm = 3 (Bright fringe) ++m = 0 (Bright fringe) -m = 3 (Bright fringe) Figure 27.7

5 0

1 year ago

A single conservative force acts on a 5.30-kg particle within a system due to its interaction with the rest of the system. The e

cupoosta [38]

Answer:

Given that

m = 5.3 kg

Fx = 2x + 4

We know that work done by force F given as

w= ∫ F. dx

Given that x=1.08 m to x=6.5 m

Fx = 2x + 4

w= ∫ F. dx

$w=\int_{1.08}^{6.5}(2x+4) .dx$

$w=\left [x^2+4x \right ]_{1.08}^{6.5}$

$w=(6.5^2-1.08^2)+4(6.5-1.08)\ J$

w=62.7 J

We know that potential energy given as

$F=-\dfrac{dU}{dx}$

∫ dU = -∫F.dx ( w= ∫ F. dx)

ΔU= -62.7 J

We know that form work power energy theorem

Net work = Change in kinetic energy

W= KE₂ - KE₁

62.7 =KE₂ - (1/2)x 5.3 x 3²

KE₂ = 86.55 J

This is the kinetic energy at 6.5m

8 0

3 years ago

PLEASE HURRY

Sophie [7]

Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)

• The net force in the parallel direction is

∑ F (para) = -mg sin(21°) - f = ma

• The net force in the perpendicular direction is

∑ F (perp) = n - mg cos(21°) = 0

Solving the second equation for n gives

n = mg cos(21°)

n = (0.200 kg) (9.80 m/s²) cos(21°)

n ≈ 1.83 N

Then the magnitude of friction is

f = µn

f = 0.25 (1.83 N)

f ≈ 0.457 N

Solve for the acceleration a :

-mg sin(21°) - f = ma

a = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)

a ≈ -5.80 m/s²

so the block is decelerating with magnitude

a = 5.80 m/s²

down the ramp.

5 0

3 years ago