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2 years ago

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A center-point bending test was performed on a 2 in. x d in. wood lumber according to ASTM D198 procedure with a span of 4 ft an

d the 4 in. side is positioned vertically. If the maximum load was 240 kips and the modulus of rupture was 940.3 ksi, what is the value of d

1 answer:

Zigmanuir [339]2 years ago

8 0

Answer:

3.03 INCHES

Explanation:

According to ASTM D198 ;

Modulus of rupture = ( M / I ) * y ----- ( 1 )

M ( bending moment ) = R * length of span / 2

= (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in

I ( moment of inertia ) = bd^3 / 12

= ( 2 )*( d )^3 / 12 = 2d^3 / 12

b = 2 in , d = ?

length of span = 4 * 12 = 48 inches

R = P / 2 = 240 * 10^3 / 2 = 120 * 10^3 Ib

y ( centroid distance ) = d / 2 inches

back to equation ( 1 )

( M / I ) * y

940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2

= ( 288 * 10^4 * 12 ) / 2d^3 ) * d / 2

940300 = 34560000* d / 4d^3

4d^3 ( 940300 ) = 34560000 d ( divide both sides with d )

4d^2 = 34560000 / 940300

d^2 = 9.188 ∴ Value of d ≈ 3.03 in

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A spring-loaded toy gun is used to shoot a ball of mass m = 1.50 kg straight up in the air. The spring has spring constant k = 6

adell [148]

Answer:

1) a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

2) The muzzle velocity of the ball is approximately 5.272 meters per second.

3) The maximum height of the ball is 1.417 meters.

Explanation:

1) Which of the following statements are true?

a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

True, statement indicates that there is no air resistence and no friction between ball and the inside of the gun because the first never touches the latter one.

b) The forces of gravity and the spring have potential energies associated with them.

False, force of gravity do work on the ball and spring receives a potential energy at being deformated by the ball.

c) No conservative forces act in this problem after the ball is released from the spring gun.

False, the absence of no conservative forces is guaranteed for the entire system according to the statement of the problem.

2) According to the statement, we understand that spring is deformed and once released and just after reaching its equilibrium position, the muzzle velocity is reached. As spring deformation is too small in comparison with height, we can neglect changes in gravitational potential energy. By Principle of Energy Conservation, we describe the motion of the ball by the following expression:

$U_{k, 1}+K_{1}=U_{k,2}+K_{2}$ (Eq. 1)

Where:

$U_{k,1}$ , $U_{k,2}$ - Initial and final elastic potential energies of spring, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After using definitions of elastic potential and translational kinetic energies, we expand the equation above as:

$\frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2}) = \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})$

And the final velocity is cleared:

$m\cdot (v_{2}^{2}-v_{1}^{2}) = k\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2}-v_{1}^{2} =\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2} =v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }$ (Eq. 2)

Where:

$v_{1}$ , $v_{2}$ - Initial and final velocities of the ball, measured in meters per second.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the ball, measured in kilograms.

$x_{1}$ , $x_{2}$ - Initial and final position of spring, measured in meters.

If we know that $v_{1} = 0\,\frac{m}{s}$ , $k = 667\,\frac{N}{m}$ , $m = 1.50\,kg$ , $x_{1} = -0.25\,m$ and $x_{2} = 0\,cm$ , the muzzle velocity of the ball is:

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\left(\frac{667\,\frac{N}{m} }{1.50\,kg} \right)\cdot [(-0.25\,m)^{2}-(0\,m)^{2}]}$

$v_{2}\approx 5.272\,\frac{m}{s}$

The muzzle velocity of the ball is approximately 5.272 meters per second.

3) After leaving the toy gun, the ball is solely decelerated by gravity. We construct this model by Principle of Energy Conservation:

$U_{g,2}+K_{2} = U_{g,3}+K_{3}$ (Eq. 3)

Where:

$U_{g,2}$ , $U_{g,3}$ - Initial and gravitational potential energies of the ball, measured in joules.

$K_{2}$ , $K_{3}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After applying definitions of gravitational potential and translational kinetic energies, we expand the equation above and solve the resulting for the final height:

$m\cdot g \cdot (h_{3}-h_{2}) = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{3}^{2})$

$h_{3}-h_{2}=\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$

$h_{3} = h_{2} +\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$ (Eq. 4)

$h_{2}$ , $h_{3}$ - Initial and final heights of the ball, measured in meters.

$v_{2}$ , $v_{3}$ - Initial and final velocities of the ball, measured in meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

If we get that $v_{2} = 5.272\,\frac{m}{s}$ , $v_{3} = 0\,\frac{m}{s}$ , $h_{2} = 0\,m$ and $g = 9.807\,\frac{m}{s^{2}}$ , the maximum height of the ball is:

$h_{3} = 0\,m+\frac{\left(5.272\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h_{3} = 1.417\,m$

The maximum height of the ball is 1.417 meters.

5 0

2 years ago

What is the linear distance traveled in one revolution of a 36-inch wheel

Afina-wow [57]

Answer:

36π inches ≈ 113.0973 inches

Explanation:

The circumference of the wheel is pi times its diameter.

C = πd

C = π(36 in) = 36π in ≈ 113.0973 in

The distance around the wheel is 36π inches, about 113.0973 inches.

7 0

2 years ago

The electrical energy used by an air conditioner for 2 minutes is 180 kJ. Calculate the power of this air conditioner in the fol

ELEN [110]

Answer:

I hope it is correct.....

5 0

2 years ago

A converging-diverging nozzle is designed to operate with an exit Mach number of 1.75 . The nozzle is supplied from an air reser

Flura [38]

Answer:

a. 4.279 MPa

b. 3.198 MPa to 4.279 MPa

c. 0.939 MPa

d. Below 3.198 MPa

Explanation:

From the given parameters

M $_{exit}$ = 1.75 MPa

M at 1.6 MPa gives A $_{exit}$ /A* = 1.2502

M at 1.8 MPa gives A $_{exit}$ /A* = 1.4390

Therefore, by interpolation, we have M $_{exit}$ = 1.75 MPa gives A

However, we shall use M $_{exit}$ = 1.75 MPa and A

Similarly,

P $_{exit}$ /P₀ = 0.1878

a) Where the nozzle is choked at the throat there is subsonic flow in the following diverging part of the nozzle. From tables, we have

A $_{exit}$ /A* = 1.387. by interpolation M

Therefore P $_{exit}$ = P₀ × P

Which shows that the nozzle is choked for back pressures lower than 4.279 MPa

b) Where there is a normal shock at the exit of the nozzle, we have;

M₁ = 1.75 MPa, P₁ = 0.1878 × 5 = 0.939 MPa

Where the normal shock is at M₁ = 1.75 MPa, P₂/P₁ = 3.406

Where the normal shock occurs at the nozzle exit, we have

$P_b = 3.406\times 0.939 = 3.198 MPa$

Where the shock occurs t the section prior to the nozzle exit from the throat, the back pressure was derived as $P_b$ = 4.279 MPa

Therefore the back pressure value ranges from 3.198 MPa to 4.279 MPa

c) At M $_{exit}$ = 1.75 MPa and P

d) Where the back pressure is less than 3.198 MPa according to isentropic flow relations supersonic flow will exist at the exit plane

8 0

2 years ago

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the

morpeh [17]

The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.

<u>Explanation</u>:

<u>Given</u>:

tensile stress is applied parallel to the [100] direction

Shear stress is 0.5 MPA.

<u>To calculate</u>:

The magnitude of applied stress in the direction of [101] and [011].

<u>Formula</u>:

zcr=σ cosФ cosλ

<u>Solution</u>:

For in the direction of 101

cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)

cos λ = 1/√2

The magnitude of stress in the direction of 101 is 12.25 MPA

In the direction of 011

We have an angle between 100 and 011

cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)

cosλ = 0

Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.

5 0

2 years ago