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3 years ago

Work equals force the times

1 answer:

soldier1979 [14.2K]3 years ago

6 0

<span>Work equals force times distance. When you move an object, you are exerting a force onto it. By exerting a force on the object, you are actually displacing it from its initial position. You cannot apply force to the object without altering its position. Keep in mind that when you exert work, you are exerting energy too. So the work must have a unit in joules in SI units. Force is in newtons or kilogram meter per second squared. And distance is in meters. So you will have newton-meter or joules. </span>

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Determine the Reynolds number for a flow of 0.2 m^3/s through a 203 mm inner diameter circular pipe of a fluid with rho=680 kg/m

IrinaK [193]

To find a solution to this problem it is necessary to apply the concepts related to the Reynolds number and its definitions on the type of fluid.

A Reynolds number less than 2000 considers the laminar fluid, while a Reynolds number greater than 4000 is considered a turbulent fluid. (The intermediate between the two values would be a transient fluid)

The mathematical equation that defines the Reynolds number is given by

$Re = \frac{\rho V D}{\mu}$

Where

$\rho =$ Density

V= Velocity

D= Diameter

$\mu =$ Viscosity

Our values are given as

$Q = 0.2m^3/s$

$D = 203*10^{-3}m$

$\rho = 680kg/m^3$

$\mu = 3.1*10^{-4}Ns/m^2$

$\sigma = 0.022N/m$

The velocity can be find through the Discharge equation,

Q = VA

Where

V = Velocity

A = Area

Replacing,

$0.2 = V* (2\pi*(\frac{203*10^{-3}}{2})^2)$

$V = 3.08m/s$

Replacing at the Reynolds equation,

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

$Re = \frac{680*3.08*203*10^{-3}}{3.1*10^{-4}}$

$Re = 1.37*10^6$

Since Reynolds' number is greater than 4000, then we consider this a turbulent fluid.

3 0

3 years ago

A steel cable of diameter 3.0 cm supports a load of 2.0kN. What is the fractional length increase of the cable compared with the

zhannawk [14.2K]

Answer:

1.429*10^-5 m

Explanation:

From the question, we are given that

Diameter of the cable, d = 3 cm = 0.03 m

Force on the cable, F = 2 kN

Young Modulus, Y = 2*10^11 Pa

Area of the cable = πd²/4 = (3.142 * 0.03²) / 4 = 0.0028 / 4 = 0.0007 m²

The fractional length = Δl/l

Δl/l = F/AY

Δl/l = 2000 / 0.0007 * 2*10^11

Δl/l = 2000 / 1.4*10^8

Δl/l = 1.429*10^-5 m

Therefore, the fractional length is 1.429*10^-5 m long

6 0

2 years ago

A tank is 10 m long, 6 m wide, 3 m high, and contains kerosene with density 820 kg/m3 to a depth of 2.5 m. (Use 9.8 m/s2 for the

Valentin [98]

Answer:

Explanation:

a )

Pressure on the bottom

= hdg

= 2.5 x 820 x 9.8

= 20090 N / m²

b )

force = area x pressure

= 20090 x 10 x 6 N

= 1205400 N

c )

Force on 10m x 2.5 m face

center of gravity of the tank will lie at height 2 .5 / 2 = 1.25 m

Pressure = hdg

= 1.25 x 820 x 9.8

= 10045 N/m²

force = pressure x area

= 10045 x 10 x 2.5

= 251125 N .

7 0

3 years ago

Read 2 more answers

2. A liquid that is twice as dense as water is used in a barometer. With this barometer, atmospheric pressure would push the liq

Scrat [10]

Answer:

The reason why the height for the liquid is 5.18 m is because, for constant pressure, height is inversely proportional to density.

Explanation:

We know that pressure, P = ρgh where ρ = density of material, g = 9.8 m/s and h = height.

Since density of liquid = 2 × density of water = 2 × 1000 kg/m = 2000 kg/m³

Since atmospheric pressure = 101500 N/m²,

The height of liquid for this pressure is h = P/ρg = 101500/(2000 × 9.8) = 5.18 m

For the same pressure, the height for water is h = P/ρg = 101500/(1000 × 9.8) = 10.36 m

Since pressure = constant,

P = ρgh ⇒ P/ρg = h since P and g are constant, h ∝ 1/ρ

ρ₁h₁ = ρ₂h₂

So, the reason why the height for the liquid is 5.18 m is because, for constant pressure, height is inversely proportional to density.

8 0

2 years ago

Don’t need to answer a question just say something and you’ll get Marked brainliest (Free points WONT GET BRAINLIEST RIGHT AWAY)

stiks02 [169]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers