ANSWER FOR BRAINLIEST

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$

And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

Learn more about flow rate:

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7 0

3 years ago

A 46.5-kg ball has a momentum of 57.2 kg m/s. What is the ball's speed?

Serhud [2]

Answer:

1.23 m/s

Explanation:

p=mv

57.2 = 46.5v

v= 57.2/46.5

v= 1.23

If you want to verify your answer, just insert the value of v in the equation.

5 0

3 years ago

A 15.9 kg concrete block is being pulled 0.90 m to the right In 2.5 s

storchak [24]

Answer:

the average is 0.36 m/s

8 0

3 years ago

A planet has a gravitational acceleration on its surface of 2.2 times Earth's gravitational acceleration on its surface. The pla

lesantik [10]

Answer:

The mass of the planet is 55 times the mass of earth.

Explanation:

From the inverse-square gravitation law,

F = (GMm/r²)

If the weight of a body (the force with which the earth attracts a body to its centre) is to be calculated,

F = mg

m = mass of the body,

g = acceleration due to gravity

mg = (GMm/r²)

G = Gravitational constant

M = mass of the earth

m = mass of body

r = distance between the body and the centre of the earth = radius of the earth

The acceleration due to gravity is given by

g = (GM/r²)

Making the mass of the earth, the subject of formula

M = (gr²/G) (eqn 1)

So, the planet described,

Let the acceleration due to gravity on the planet be g₁

Mass of the planet be M₁

Radius of the planet be r₁

g₁ = 2.2g

r₁ = 5r

M₁ = ?

Note that the gravitational constant is the same for both planets.

So, we can write a similar expression for the planet's acceleration due to gravity

g₁ = (GM₁/r₁²)

Substituting all the parameters known in terms of their corresponding earth values

2.2g = [GM₁/(5r)²]

2.2g = [GM₁/25r²]

M₁ = (55gr²/G)

Recall the expression for the mass of the earth

M = (gr²/G)

M₁ = 55 M

The mass of the planet, in terms of Earth masses = 55M

The mass of the planet is 55 times the planet of earth.

Hope this Helps!!!

5 0

4 years ago

Suzie (of mass 42 kg) is roller-blading down the sidewalk going 32 miles per hour. She notices a group of workers down the walkw

Allisa [31]

Answer:

a) Suzie’s average acceleration = -6.46 m/s²

b) Force exerted to stop Suzie = 271.52 N

Explanation:

a) We have equation of motion, v = u + at

Final velocity, v = 0 m/s

Initial velocity, u = 32 mph = 14.22 m/s

Time, t =2.2 s

Substituting

0 = 14.22 + a x 2.2

a = -6.46 m/s²

Suzie’s average acceleration = -6.46 m/s²

b) Mass of Suzie = 42 kg

Force = Mass x Acceleration

F = Ma

F = 42 x -6.46 =-271.52 N

Force exerted to stop Suzie = 271.52 N

5 0

3 years ago