Ask question

Ask question

All categories

3 years ago

6

At position A within a tube containing fluid that is moving with steady laminar flow, the speed of the fluid is 12.0 m/s and the

tube has a diameter 12.00 cm. At position B, the speed of the fluid is 18.0 m/s and the tube has a diameter 6.00 cm. What is the ratio of the density of the fluid at position A to the density of the fluid at position B

1 answer:

Mama L [17]3 years ago

8 0

Answer:

0.375

Explanation:

For incompressible flow, we know that;

ρ1•v1•A1 = ρ2•v2•A2

Where;

ρ1 = density of fluid at position A

v1 = speed of fluid at position A

A1 = area of tube

ρ2 = density of fluid at position B

v2 = speed of fluid at position B

A2 = area of tube

We want to find ratio of the density of the fluid at position A to the density of the fluid at position B.

Thus;

ρ1/ρ2 = (v2•A2)/(v1•A1)

Now, the tube will have the same height.

But we are given;

diameter of A = 12.00 cm = 0.12 m

diameter of B = 6 cm = 0.06 m

Thus;

A1 = π(d²/4)h = πh(0.12²/4)

A2 = πh(0.06²/4)

We are also given;

v1 = 12 m/s

v2 = 18 m/s

Thus;

ρ1/ρ2 = (18 × πh(0.06²/4))/(12 × πh(0.12²/4))

πh/4 will cancel out to give;

ρ1/ρ2 = (18 × 0.06²)/(12 × 0.12²)

ρ1/ρ2 = 0.375

You might be interested in

You’re standing at the highest point on the Moon, 10,786 mm above the level of the Moon’s mean radius. You’ve got a golf club an

Anna71 [15]

Answer:

A) v = 1,675 10³ m / s , B) r₂ = 11,673 10⁶ m

Explanation:

A) This exercise we must use Newton's second law, where the forces of gravity are the Moon

F = m a

acceleration is centripetal

a = v² / r

force is the force of universal attraction

F = G m M / r²

we substitute

G m M / r² = m v² / r

v² = G M / r

distance

r = R_moon + h

r = 1.74 10⁶ +1.0786 10⁴

r = 1,750786 10⁶ m

we calculate

v = √ (6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶)

v = √ (2,8052 10⁶)

v = 1,675 10³ m / s

B) let's use energy conservation

Starting point. In the mountain

Em₀ = K + U = ½ m v² + G m M / r

Final point. Where the speed is zero

$Em_{f}$ = U = G mM / r₂

Em₀ = Em_{f}

½ m v² + G m M / r = G mM / r₂

1 / r₂ = (½ v₂ + G M / r) / GM

let's calculate

1 / r₂ = (½ (1,675 10³)² + 6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶) /(6.67 10⁻¹¹ 7.36 10²²)

1 / r₂ = (1,4028 10⁶ + 2,805 10⁶) / 49.12 10¹¹

1 / r₂ = 8.5664 10⁻⁷

r₂ = 11,673 10⁶ m

6 0

3 years ago

Each roller under a conveyor belt has a radius of 0.5 meters. the rollers turn at a rate of 30 revolutions per minute. what is t

svp [43]

Distance in a minute=<span>0.5 times 30=15 meters
distance in a second</span><span>=15 divided by 60=0.25 meters per second
hope it helps</span>

3 0

3 years ago

Read 2 more answers

The ball is partially white, is that an observation or an inference??

fenix001 [56]

Answer:

this is an observation.

Explanation:

you are observing that the ball is partially white

4 0

3 years ago

A sphere of mass m" = 2 kg travels with a velocity of magnitude υ") = 8 m/s toward a sphere of mass m- = 3 kg initially at rest,

aleksklad [387]

a) 6.4 m/s

b) 2.1 m

c) $61.6^{\circ}$

d) 14.0 N

e) 4.6 m/s

f) 37.9 N

Explanation:

a)

Since the system is isolated (no external forces on it), the total momentum of the system is conserved, so we can write:

$p_i = p_f\\m_1 u_1 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 2 kg$ is the mass of the 1st sphere

$m_2 = 3kg$ is the mass of the 2nd sphere

$u_1 = 8 m/s$ is the initial velocity of the 1st sphere

$v_1$ is the final velocity of the 1st sphere

$v_2$ is the final velocity of the 2nd sphere

Since the collision is elastic, the total kinetic energy is also conserved:

$E_i=E_k\\\frac{1}{2}m_1 u_1^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Combining the two equations together, we can find the final velocity of the 2nd sphere:

$v_2=\frac{2m_1}{m_1+m_2}u_1=\frac{2(2)}{2+3}(8)=6.4 m/s$

b)

Now we analyze the 2nd sphere from the moment it starts its motion till the moment it reaches the maximum height.

Since its total mechanical energy is conserved, its initial kinetic energy is entirely converted into gravitational potential energy at the highest point.

So we can write:

$KE_i = PE_f$

$\frac{1}{2}mv^2 = mgh$

where

m = 3 kg is the mass of the sphere

v = 6.4 m/s is the initial speed of the sphere

$g=9.8 m/s^2$ is the acceleration due to gravity

h is the maximum height reached

Solving for h, we find

$h=\frac{v^2}{2g}=\frac{(6.4)^2}{2(9.8)}=2.1 m$

c)

Here the 2nd sphere is tied to a rope of length

L = 4 m

We know that the maximum height reached by the sphere in its motion is

h = 2.1 m

Calling $\theta$ the angle that the rope makes with the vertical, we can write

$h = L-Lcos \theta$

Which can be rewritten as

$h=L(1-cos \theta)$

Solving for $\theta$ , we can find the angle between the rope and the vertical:

$cos \theta = 1-\frac{h}{L}=1-\frac{2.1}{4}=0.475\\\theta=cos^{-1}(0.475)=61.6^{\circ}$

d)

The motion of the sphere is part of a circular motion. The forces acting along the centripetal direction are:

- The tension in the rope, T, inward

- The component of the weight along the radial direction, $mg cos \theta$ , outward

Their resultant must be equal to the centripetal force, so we can write:

$T-mg cos \theta = m\frac{v^2}{r}$

where r = L (the radius of the circle is the length of the rope).

However, when the sphere is at the highest point, it is at rest, so

v = 0

Therefore we have

$T-mg cos \theta=0$

So we can find the tension:

$T=mg cos \theta=(3)(9.8)(cos 61.6^{\circ})=14.0 N$

e)

We can solve this part by applying again the law of conservation of energy.

In fact, when the sphere is at a height of h = 1 m, it has both kinetic and potential energy. So we can write:

$KE_i = KE_f + PE_f\\\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + mgh'$

where:

$KE_i$ is the initial kinetic energy

$KE_f$ is the kinetic energy at 1 m

$PE_f$ is the final potential energy

v = 6.4 m/s is the speed at the bottom

v' is the speed at a height of 1 m

h' = 1 m is the height

m = 3 kg is the mass of the sphere

And solving for v', we find:

$v'=\sqrt{v^2-2gh'}=\sqrt{6.4^2-2(9.8)(1)}=4.6 m/s$

f)

Again, since the sphere is in circular motion, the equation of the forces along the radial direction is

$T-mg cos \theta = m\frac{v^2}{r}$

where

T is the tension in the string

$mg cos \theta$ is the component of the weight in the radial direction

$m\frac{v^2}{r}$ is the centripetal force

In this situation we have

v = 4.6 m/s is the speed of the sphere

$cos \theta$ can be rewritten as (see part c)

$cos \theta = 1-\frac{h'}{L}$

where in this case,

h' = 1 m

L = 4 m

And $r=L=4 m$ is the radius of the circle

Substituting and solving for T, we find:

$T=mg cos \theta + m\frac{v^2}{r}=mg(1-\frac{h'}{L})+m\frac{v^2}{L}=\\=(3)(9.8)(1-\frac{1}{4})+(3)\frac{4.6^2}{4}=37.9 N$

4 0

3 years ago

A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary

Archy [21]

Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

m = 10 kg
g = 9.8 m/s²
T = ?

Use formula:

$\boxed{\bold{T=m*g}}$

Replace and solve:

$\boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}$

$\boxed{\boxed{\bold{T=98\ N}}}$

The tension in the rope is <u>98 Newtons.</u>

Greetings.

5 0

3 years ago