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3 years ago

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The spring has a stiffness k = 200 N>m and an unstretched length of 0.5 m. If it is attached to the 3-kg smooth collar and th

e collar is released from rest at A, determine the speed of the collar when it reaches B. Neglect the size of the collar.

1 answer:

ruslelena [56]3 years ago

4 0

Answer:

15.467 m/s

Explanation:

See attached picture.

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Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh

Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

$R_a=e^{-\left ( \frac{T-r}{n}\right )B}$

$R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}$

$R_a=e^{-4.4}=0.01227$

For B

$R_b=e^{-\left ( \frac{300-336}{100}\right )3}$

$R_b=e^{1.08}=2.944$

B is better for 100 hours

(b)For 750 hours

$R_a=e^{-0.5866}=0.55621$

$R_b=e^{0.144}=1.154$

So here B is more Reliable.

3 0

3 years ago

olya-2409 [2.1K]

The thickness is thick

5 0

3 years ago

determine the position d of the 6- kn load so that the average normal stress in each rod is the same.

Zinaida [17]

The load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2}$ area.

<h3>What is meant by torque?</h3>

The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.

Let the beam is of length L

Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero

$N_1 * x=N_2 *(L-x)$

also, we know that stress at both ends are same

$\frac{N_1}{12}=\frac{N_2}{8}$

$2 * N_1=3 * N_2$

Now from two equations we have

$$\frac{3}{2} N_2 * x=N_2 *(L-x)$

solving the above equation we have

$$x=\frac{2}{5} L$

so the load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2}$ area.

The complete question is:

47. the beam is supported by two rods ab and cd that have cross-sectional areas of $$$12mm^2$ and $$$8mm^2$ , respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.

To learn more about torque refer to:

brainly.com/question/20691242

#SPJ4

7 0

2 years ago

2. (Problem 4.60 on main book, diameters different) Water flows steadily through a fire hose and nozzle. The hose is 35 mm diame

Viefleur [7K]

Answer:

coupling is in tension

Force = -244.81 N

Explanation:

Diameter of Hose ( D1 ) = 35 mm

Diameter of nozzle ( D2 ) = 25 mm

water gage pressure in hose = 510 kPa

stream leaving the nozzle is uniform

exit speed and pressure = 32 m/s and atmospheric

<u>Determine the force transmitted by the coupling between the nozzle and hose </u>

attached below is the remaining part of the detailed solution

Inlet velocity ( V1 ) = V2 ( D2/D1 )^2

= 32 ( 25 / 35 )^2

= 16.33 m/s

4 0

3 years ago

The lift on a spinning circular cylinder in a freestream with a velocity of 30 m/s and at standard sea level conditions is 6 N/m

Evgesh-ka [11]

Answer:

The circulation around the cylinder is 0.163 $\frac{m^{2} }{s}$

Explanation:

Given :

Velocity of spinning cylinder $v = 30$ $\frac{m}{s}$

Sea level density $\rho = 1.23$ $\frac{kg}{m^{3} }$

Sea level span $L = 6$ $\frac{N}{m}$

Lift per unit circulation is given by,

$L = \rho v c$

Where $c =$ circulation around cylinder

$c = \frac{L}{\rho v}$

$c = \frac{6}{1.23 \times 30}$

$c = 0.163$ $\frac{m^{2} }{s}$

Therefore, the circulation around the cylinder is 0.163 $\frac{m^{2} }{s}$

5 0

3 years ago