All categories

2 years ago

A piston having a diameter of 5.48 inches and a length of 9.50 in slides downward with a

Engineering

1 answer:

inna [77]2 years ago

4 0

Answer:

$V = 4.585 \times 10^{-3} ft/s$

Explanation:

surface Area is given as $= \pi dL$

here , d is diameter of piston

L is distance travelled by the piston

$a = \pi 5.48\times 9.5$

velocity of piston

$W = \mu A \frac{V}{t}$

where,

w is weight of piston, $\mu$ dynamic viscosity of fluid, t is thickness of fluid

$0.5 = 0.016 \times \frac{163.55}{12^2} \frac{V}{\frac{0.002}{12}}$

$0.5 = 109.033 V$

SOLVING FOR V

$V = 4.585 \times 10^{-3} ft/s$

You might be interested in

When trying to solve a frame problem it will typically be necessary to draw many free body diagrams. a)-True b)-False

klasskru [66]

Answer:

True

Explanation:

When trying to solve a frame problem in Engineering or Physics, it will typically be necessary to draw more than one body diagram.

When we have several parts of the frame or a set of frames, we have the anchor point, as well as the intersections of frames. Besides that, usually, there is a particle or rigid body together with the frame system. In this sense, usually, it is required to analyze a body diagram for the particle or rigid body suspended, as well as the intersections of the frames. So, usually, it will be required a minimum of two body diagrams.

If the system is more complex, or there are many intersections points, it will be required more than two body diagrams.

Finally, indeed, it will typically be necessary to draw many-body diagrams.

6 0

3 years ago

Suppose that a wireless link layer using a CSMA-like protocol backs off 1ms on average. A packet’s link and physical layer heade

Liula [17]

Tell me why i got this question got it right and now won’t remember but i’ll get back at you when i remember

5 0

3 years ago

3–102 One of the common procedures in fitness programs is to determine the fat-to-muscle ratio of the body. This is based on the

gayaneshka [121]

Answer:

x_fat = [ 0.5*(Wsa + Wsw) - p_muscle*V ] / V*( p_fat - p_muscle )

Explanation:

Given:

- The total volume of body = V

- The average density of the body = p_avg

- The density of muscle = p_muscle

- The density of fat = p_fat

Find:

Obtain a relation for the volume fraction of body fat x_fat

Solution:

- The volume of the fat is given by:

V_fat = x_fat*V

- The volume of the muscle is given by:

V_muscle = V - V_fat

= V - x_fat*V

=V*( 1 - x_fat )

- We will use the conservation of mass for the body related as:

mass_fat + mass_muscle = Total average mass

p_fat*V_fat + p_muscle*V_muscle = p_avg*V

p_fat*x_fat*V + p_muscle*V*( 1 - x_fat ) = p_avg*V

p_fat*x_fat + p_muscle*( 1 - x_fat ) = p_avg

- To determine p_1 we weigh the body in air:

Weight reading (Wsa) = m = p_1*V

p_1 = Wsa / V*g

- To determine p_2 we weigh the body in water:

Weight reading (Wsw) = m - p_w*V= p_1*V - p_w*V

Weight reading (Wsw) = V*(p_1 - p_w) = V*(p_2)

Where, p_2 = p_1 - p_water

p_2 = Wsw / V

- The average density p_avg:

p_avg = 0.5*(p_1 + p_2)

p_avg = 0.5*(Wsa / V + Wsw / V)

p_avg = 0.5*(Wsa + Wsw) / V

- Plug in the mass equation:

p_fat*x_fat + p_muscle*( 1 - x_fat ) = 0.5*(Wsa + Wsw) / V

x_fat*( p_fat - p_muscle ) = 0.5*(Wsa + Wsw) / V - p_muscle

x_fat = [ 0.5*(Wsa + Wsw) - p_muscle*V ] / V*( p_fat - p_muscle )

6 0

3 years ago

How many sets of equations (V and M equations) would you need to describe shear and moment as functions of x for this beam? In o

den301095 [7]

Shear and moment as functions of x is described below .

Explanation:

1. Beam is the slender bar that carries transverse

loading; that is, the applied force are perpendicular to the bar.

2. In a beam, the internal force system consist of a shear force and

a bending moment acting on the cross section of the bar.

3. The shear force and the bending moment usually vary continuously

along the length of the beam.

4. The internal forces give rise to two kinds of stresses on a

transverse section of a beam:

(1) normal stress that is caused by

bending moment and

(2) shear stress due to the shear force.

Knowing the distribution of the shear force and the bending

moment in a beam is essential for the computation of stresses

and deformations.

Shear- Moment Equations

The determination of the internal force system acting at a given

section of a beam : draw a free-body diagram that expose these

forces and then compute the forces using equilibrium equations.

The goal of the beam analysis －determine the shear force V and the bending moment M at every cross section of the beam.

To derive the expressions for V and M in terms of the distance x

measured along the beam. By plotting these expressions to scale,

obtain the shear force and bending moment diagrams for the

beam.

The shear force and bending moment diagrams are convenient

visual references to the internal forces in a beam; in particular, they identify the maximum values of V and M

5 0

3 years ago

The fan blades suddenly experience an angular acceleration of 2 rad/s2. If the blades are rotating with an initial angular veloc

madreJ [45]

Answer:

Option B

$116 ft/s^{2}$

Explanation:

$\theta=2 rev=2(2\pi)=4\pi$

$\alpha \theta=0.5(\omega_f^{2}-\omega_i^{2})$

$\alpha (4\pi)= 0.5(\omega_f^{2}-\omega_i^{2})$

$\alpha (8\pi)= (\omega_f^{2}-\omega_i^{2})$

$(2) (8\pi)= (\omega_f^{2}-\omega_i^{2})$

$(2) (8\pi)= (\omega_f^{2}-4^{2})$

$\omega_f=8.14 rads/s$

$v=r\omega=1.75*8.14=14.245 ft/s$

Centripetal acceleration $=\omega_f^{2} r=8.14^{2}*1.75=115.95 ft/s^{2}$

Tangential component=dr=2*1.75=3.5

$Resultant=\sqrt{3.5^{2}+115.95^{2}}\approx 116 ft/s^{2}$

5 0

3 years ago