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Mnenie [13.5K]
3 years ago
11

Water flows through a water hose at a rate of Q1 = 860 cm3/s, the diameter of the hose is d1 = 1.85 cm. A nozzle is attached to

the water hose. The water leaves the nozzle at a velocity of v2 = 10.8 m/s.
a. Enter an expression for the cross-sectional area of the hose, A1, in terms of its diameter, d1.
b. Calculate the numerical value of A1, in square centimeters.
c. Enter an expression for the speed of the water in the hose, v1, in terms of the volume flow rate Q1 and cross-sectional area A1.
d. Calculate the speed of the water in the hose, v1 in meters per second
e. Enter an expression for the cross-sectional area of the nozzle, A2, in terms of v1, v2 and A1.
f. Calculate the cross-sectional area of the nozzle, A2 in square centimeters.
Physics
1 answer:
Snowcat [4.5K]3 years ago
3 0

Answer:

a) A1 =  \frac{\pi (d1)^{2} }{4}

b) A1 = 2.688 cm^{2}

c) Q1 = A1 x v1

d) v1 = 3.1994 m/s

e) A2 = \frac{A1 X v1}{v2}

f)  A2 = 0.7963cm^{2}

Explanation:

a) Area = \pi r^{2}

r = \frac{d}{2}

thus,

area = \pi (\frac{d}{2})^{2}

A1 =  \frac{\pi (d1)^{2} }{4}[/tex]b) d1 = 1.85 cmsubstituting in the above equation,A1 =  [tex]\frac{\pi (d1)^{2} }{4}

A1 =  \frac{\pi (1.85)^{2} }{4}

A1 = 2.688 cm^{2}

c) Flow rate = Area x velocity ( refer brainly.com/question/13997998)

Q1 = A1 x v1

d) From the above equation,

v1 = \frac{Q1}{A1} = \frac{860}{2.688} = 319.94 cm/s = 3.1994 m/s

e) Since the flow rate Q1 is constant throughout the hose, Av is a constant.

i.e. A1 x v1 = A2 x v2

thus,

A2 = \frac{A1 X v1}{v2}

f) v2 = 10.8 m/s.

substituting the values in the above equation,

A2 = \frac{2.688 X 3.1994}{10.8}  = 0.7963cm^{2}

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A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym
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Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

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- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

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- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

                                     L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

                                    L_(p,i) = m*V_pi*R

                                    L_(p,i) = 3.6*3.3*0.44

                                    L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

                                    L_(c,i) = I*w_i

                                    L_(c,i) = 0.5*M*R^2*0

                                    L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

                                    L_i = L_(p,i) + L_(c,i)

                                    L_i = 5.2272 + 0

                                    L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

                                   L_(p,f) = I_p*w_f

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                                   L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

                                  L_f = L_(p,f) + L_(c,f)

                                  L_f = I_p*w_f + I_c*w_f

                                  L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

                                  L_i = L_f

                                  5.2272 = w_f*(I_p + I_c)

                                  w_f =  5.2272/ R^2*(m + 0.5M)

Plug in values:

                                  w_f =  5.2272/ 0.44^2*(3.6 + 0.5*45)

                                  w_f =  5.2272/ 5.05296

                                  w_f = 1.0345 rad/s

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