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3 years ago

A cylinder of radius R and height H is floating upright in

Physics

1 answer:

emmainna [20.7K]3 years ago

6 0

Answer:

Pressure difference between Top and Bottom of the cylinder is given as

$\Delta P = \frac{gH}{2}(\rho_A + \rho_B)$

Explanation:

As we know that the force due to pressure is balanced by the weight of the cylinder

So we will have

$F = mg$

so we have

$\Delta P \pi R^2 = mg$

so we have

$\Delta P \pi R^2 = \pi R^2(\rho_A(\frac{H}{2}) + \rho_B(\frac{H}{2}))g$

so we have

$\Delta P = \frac{gH}{2}(\rho_A + \rho_B)$

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What does the diagram represent?

solong [7]

Answer:

Atoms in a element on the periodic table I dont know which one tho so try google that sorry I am not much of a help

Explanation:

4 0

2 years ago

A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its

m_a_m_a [10]

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

x² = 0.0002142

x = √0.0002142

x = 0.0146m

5 0

3 years ago

This is a machine consisting of two wheels of different sizes that rotate together. It is a modified lever.

tiny-mole [99]

Your answer is a Wheel and Axel.

3 0

2 years ago

A race car makes one lap around a track of radius 50 m in 9.0 s. What is the average velocity? *

Oksi-84 [34.3K]

Given that,

Radius of track, r = 50 m

time , t = 9 s

velocity, v = ?

Distance covered by car in one lap around a track is equal to the circumference of the track.

C = 2 π r = 2 * 3.14 * 50

C = 314.159 m

Distance covered by car, s = 314.159 m

Velocity = distance/ time

V = 314.159 / 9

V = 34.9 m/s

The average velocity of car is 34.9 m/s.

7 0

2 years ago

P7.16 A thin flat plate 55 by 110 cm is immersed in a 6-m/s stream of SAE 10 oil at 20C. Compute the total friction drag if the

timofeeve [1]

Answer:

The total friction drag for the long side of the plate is 107 N

The total friction drag for the long side of the plate is 151.4 N

Explanation:

The first question is to obtain the friction drag when the fluid i parallel to the long side of the plate

The block representation of the this problem is shown on the first uploaded image

Where the U is the initial velocity = 6 m/s

So the equation we will be working with is

$F = \frac{1}{2} \rho C_fAU^2$

Where $\rho$ is the density of SAE 10W = $870\ kg/m^3$ This is obtained from the table of density at 20° C

$C_f$ is the friction drag coefficient

This coefficient is dependent on the Reynolds number if the Reynolds number is less than $5*10^5$ then the flow is of laminar type and

$C_f$ = $\frac{1.328}{\sqrt{Re} }$

But if the Reynolds number is greater than $5*10^5$ the flow would be of Turbulent type and

$C_f = \frac{0.074}{Re_E^{0.2}}$

Where Re is the Reynolds number

To obtain the Reynolds number

$Re = \frac{\rho UL}{\mu}$

where L is the length of the long side = 110 cm = 1.1 m

and $\mu$ is the Dynamic viscosity of SAE 10W oil $= 1.04*10^{-1} kg /m.s$

This is gotten from the table of Dynamic viscosity of oil

$Re = \frac{870 *6*1.1}{1.04*10^{-1}}$

$= 55211.54$

Since 55211.54 < $5.0*10^5$

Hence

$C_f = \frac{1.328}{\sqrt{55211.54} }$

$= 0.00565$

$A$ is the area of the plate = $\frac{ (110cm)(55cm)}{10000}$ = $0.55m^2$

Since the area is immersed totally it should be multiplied by 2 i.e the bottom face and the top face are both immersed in the fluid

$F = \frac{1}{2} \rho C_f(2A)U^2$

$F =\frac{1}{2} *870 *0.00565*(2*0.55)*6^2$

$F = 107N$

Considering the short side

To obtain the Reynolds number

$Re = \frac{\rho U b}{\mu}$

Here b is the short side

$Re =\frac{870*6*0,55}{1.04*10^{-1}}$

$=27606$

Since the value obtained is not greater than $5*10^5$ then the flow is laminar

And

$C_f = \frac{1.328}{\sqrt{Re} }$

$= \frac{1.328}{\sqrt{27606} }$

$= 0.00799$

The next thing to do is to obtain the total friction drag

$F = \frac{1}{2} \rho C_f(2A)U^2$

Substituting values

$F = \frac{1}{2} * 870 * 0.00799 * 2( 0.55) * 6^2$

$= 151.4 N$

4 0

2 years ago