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3 years ago

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a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column

is 16.0 cm high the oil air interfdace is 4.5 cm above the liquid level in the left arm of the u tube algrebraic expression to deter,ine the density of the unknown fluid

1 answer:

scoray [572]3 years ago

8 0

Answer:

The answer is " $1155\ \frac{kg}{m^3}$ "

Explanation:

Please find the complete question in the attached file.

$p = p_0 + ?gh$

pi = pressure only at two liquids' devices

PA = pressure atmosphere.

1 = oil density

2 = uncertain fluid density

$h_1 = 11 \ cm\\\\h_2= 3 \ cm$

The pressures would be proportional to the quantity $11-3 = 8$ cm from below the surface at the interface between both the oil and the liquid.

$\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\$

$= \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}$

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Answer:

Explanation:

a. The equation of Lorentz transformations is given by:

x = γ(x' + ut')

x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.

x' = 0

t' = 5.00 s

u =0.800 c,

c is the speed of light = 3×10⁸ m/s

Then,

γ = 1 / √ (1 - (u/c)²)

γ = 1 / √ (1 - (0.8c/c)²)

γ = 1 / √ (1 - (0.8)²)

γ = 1 / √ (1 - 0.64)

γ = 1 / √0.36

γ = 1 / 0.6

γ = 1.67

Therefore, x = γ(x' + ut')

x = 1.67(0 + 0.8c×5)

x = 1.67 × (0+4c)

x = 1.67 × 4c

x = 1.67 × 4 × 3×10⁸

x = 2.004 × 10^9 m

x ≈ 2 × 10^9 m

Now, to find t we apply the same analysis:

but as x'=0 we just have:

t = γ(t' + ux'/c²)

t = γ•t'

t = 1.67 × 5

t = 8.35 seconds

b. Mavis reads 5 s on her watch which is the proper time.

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such that

∆t = γ ∆to = (5/3)(5) = 25/3 = 8.3 sec which is the same as part (b)

c. According to Stanley,

dist = u ∆t = 0.8c (8.3) = 2 x 10^9 m

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Answer:

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Explanation:

Given that,

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Using formula of total acceleration

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Put the value into the formula

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Hence, (a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

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