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gayaneshka [121]
3 years ago
13

When water particles in their gaseous state (X) lose enough energy they

Physics
1 answer:
e-lub [12.9K]3 years ago
3 0
They turn into a liguid. it is the opposite of evaporation. it didn't get hotter, it got cooler
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Bright and dark fringes are seen on a screen when light from a single source reaches two narrow slits a short distance apart. Th
Dima020 [189]

Answer:

Explanation:

Let the thickness of the film is t and the refractive index of the material of film is n.

When light travels through a sheet of thickness t, the optical path traveled is nt.

When the path of one of slit is covered by a sheet of thickness t, the optical path becomes

x = ( n - 1) t

As the one fringe is shift, so the optical path changed by one wavelength.

i.e., x = λ

So, λ = ( n - 1) t

t=\frac{\lambda }{n-1}

7 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
1.33 m^3 of fluid flows out of a pipe in 24.5 s. the fluid leaves the pipe at 3.55 m/s. what is the area of the pipe. (unit=m^2)
Margaret [11]

Answer:

0.015meter^{2}

Explanation:

Total volume of water coming out = 1.33meter^{3}

Also volume = Cross sectional area*Length covered

Length covered = Velocity *time

                           =24.5*3.55

                           =86.97 meter

Let the cross sectional area be A.

1.33 = 86.97*A

A =0.015meter^{2}

3 0
3 years ago
Read 2 more answers
Samples of different materials, A and B, have the same mass, but the sample
madam [21]

Answer:

<u><em>A.)</em></u> The particles that make up material A have more mass than the

particles that make up material B.

HOPE that helps!!! :)

4 0
1 year ago
6. A person lifts a package weighing 75 N. If she lifts it 1.2 m off the floor,
balandron [24]

Answer:

90 Joules

Explanation:

75×1.2=90.

i believe it is this

7 0
3 years ago
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