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Molodets [167]
2 years ago
6

Hey can anyone please help me with this it’s due in few hours and I’m stuck with ittt

Physics
1 answer:
Ray Of Light [21]2 years ago
4 0

Answer:

Check body of the explanation

Explanation:

Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about 10 ms^-^2, the density of whatever you're sinking in, and the depth at which you are. In formula, p(h) = p_0 + \rho g h, and the pressure is the same for every point of the tank at the same depth.

At this point, we can start answering!

1a. The pressure at A is - not counting atmosferic pressure - 1000 * 10 * 1 = 10^4 Pa, while in B is 1000*10*2 = 2*10^4 Pa, so it's half of it.

1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is 800*10*2 = 1,6*10^4 Pa: Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

2a. The volume is simply the product, so 4m*3m*2m = 24m^3

2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: 800* 24 = 1,92 *10^4kg

2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's 1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N

2d. \rho gh again, what a surprise! 800 {kg \over m^3} * 10 {N \over kg}} * 2 m = 1,6* 10^4 {N\over m^2} =1.6*10^4 Pa

3. Yet again, \rho gh. 1000 {kg \over m^3} * 10 {N \over kg}} * 2 m = 2* 10^4 {N\over m^2} =2*10^4 Pa

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A) 4.7 cm

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sin \theta=\frac{n \lambda}{d}

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a is the distance between the slits

In this problem,

n = 5

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a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

D=75.0 cm = 0.75 m

The distance of the 5th  bright fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

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The measured redshift is z =2

Explanation:

Since the object is traveling near light speed, since v/c = 0.8, then we have to use a redshift formula for relativistic speeds.

z= \sqrt{\cfrac{c+v}{c-v}}-1

Finding the redshift.

We can prepare the formula by dividing by lightspeed inside the square root to both numerator and denominator to get

z= \sqrt{\cfrac{1+\cfrac vc}{1-\cfrac vc}}-1

Replacing the given information

z= \sqrt{\cfrac{1+0.8}{1-0.8}}-1

z= \sqrt{\cfrac{1.8}{0.2}}-1\\z= \sqrt{9}}-1\\z=3-1\\z=2

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A cook preparing a meal for a group of people is an example of ______
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Answer:

Option B is correct.

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