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sammy
2 months ago
11

Marble A is placed in a hollow tube, and the tube is swung in a horizontal plane causing the marble to be thrown out. As viewed

from the top, which of the following choices best describes the path of the marble after leaving the tube?

Engineering
1 answer:
grin007 [323]2 months ago
6 0

Response:

The path of the ball will correspond to Number 4 (as indicated in the attached image) ---- the ball emerges from the tube at an angle.

Clarification:

The problem outlined is not fully articulated. Therefore, the accompanying image, which is absent in the question, is provided for your reference.

As depicted in the image, the marble exiting the tube will possess two velocity components. One component aligns with the tube's length (as the ball is propelled in this direction as it exits through the opening), while the other component is directed perpendicularly to the tube (attributable to the ball being pushed in that direction due to the tube's motion across the surface).

Thus, when determining the resultant of the two velocity vectors, the ball's directional path will be Number 4 (as illustrated in the attached image) ---- it will form an angle as it exits the tube.

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6.15. In an attempt to conserve water and to be awarded LEED (Leadership in Energy and Environmental Design) certification, a 20
Viktor [391]

Explanation:

At a temperature of 33^{\circ} C and relative humidity of 86%, the humidity ratio stands at 0.0223 with a specific volume of 14.289.

At a temperature of 33^{\circ} C and relative humidity of 40%, the humidity ratio is 0.0066 while the specific volume is 13.535.

To determine the mass of air, the following formula can be used:

\begin{aligned}m _{1} &=\frac{ v }{ v }(1- w ) \\&=\frac{1 \times 10^{5}}{13.535}(1-0.0066) \\&=7339.49 lb / min \\v _{ a } &=\frac{ m _{1} v }{(1- w )} \\v _{ a } &=\frac{7339.49 \times 14.289}{(1-0.0223)} \\v _{ a } &=107266.0 ft ^{3} / min\end{aligned}

Now, we will calculate the volume

\begin{aligned}m _{ w } &=\frac{ v _{ a }}{ v _{ a }} w _{ a }-\frac{ v _{ i }}{ v _{ i }} w _{ i } \\&=\frac{107266.0}{14.289} \times 0.0223-\frac{100000}{13.535} \times 0.0066 \\&=118.64 lb / min\end{aligned}

The duration required to fill the cistern can be determined with the equation:

Time \(=\frac{\text { cistern volume }}{\text { removal water perminute volume }}\)

By substituting the values into the preceding formula, we find:

\(\frac{\left(15 \times 10^{3} L\right) \times\left(0.0353147 ft ^{3} / L \right)}{(118.641 b / min ) \times\left(\frac{1}{62.41 lb / ft ^{3}}\right)}\)\\\(=279.09\) minutes\\\(=4.65\) hours.

Thus, the hours necessary to fill the cistern amount to 4.65 hours.

3 0
2 months ago
The critical resolved shear stress for iron is 27 MPa (4000 psi). Determine the maximum possible yield strength for a single cry
grin007 [323]

Answer:

The following represents the answer.

Explanation:

Calculation for the maximum yield strength of a single crystal of Fe subjected to tension can be found in the attached image.

The maximum yield strength value is 54 MPa.

7 0
3 months ago
The manufacturer of a 1.5 V D flashlight battery says that the battery will deliver 9 mA for 40 continuous hours. During that ti
pantera1 [306]

Response:

a) 144.000 seconds

b) and c) Battery voltage and power graphs are in the attached image.

   V=-\frac{0.5}{144000} t + 1.5 V[tex] [tex]P(t)=-(31.25X10^{-9}) t+0.0135  where D:{0<t h="" />

d) 1620 J

Description:

a) The initial response is derived via a rule of three

s=\frac{3600s * 40h}{1h} = 144000s

b) Using the line equation from the starting point (0 seconds, 1.5 V)

m=\frac{1-1.5}{144000-0} = \frac{-0.5}{144000}

where m denotes the slope.

V-V_{1}=m(x-x_{1})

where V represents voltage in volts and t signifies time in seconds

V=m(t-t_{1}) + V_{1} along with P and m.

V=-\frac{0.5}{144000} t + 1.5 V[tex] c) Using the equation VPOWER IS DEFINED AS:[tex] P(t) = v(t) * i(t) [tex]so.[tex] P(t) = 9mA * (-\frac{0.5}{144000} t + 1.5) [tex][tex]P(t) = - (31.25X10^{-9}) t + 0.0135

d) By evaluating that count.

E = \int\limits^{144000}_{0} {P(t)} \, dt = \int\limits^{144000}_{0} {v(t)*i(t)} \, dt

E = \int\limits^{144000}_{0} {-\frac{0.5}{144000} t + 1.5*0.009} \, dt = 1620 J

4 0
2 months ago
The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O. Find the general
Daniel [329]

Answer:

  M_o = 18.84 N*m clockwise.  

Explanation:

Given:

- Force F = 120 N

- Length b = 610 mm

- Height h = 330 mm

Required:

Calculate the moment M_o at the origin and its direction:

Solution:

- The force is divided into components F_x and F_y along the base b and height h, respectively:

                    F_x = F*cos(Q)

                    F_x = F*(h / sqrt(h² + b²))

                    F_x = 120*(330 / sqrt(330² + 610²))

                    F_x = 57.098 N

- The F_y component can be excluded as it passes through the origin, resulting in zero moment.

- The moment at point O is calculated as:

                     M_o = F_x * h

                     M_o = 57.098*.33

                    M_o = 18.84 N*m clockwise.  

3 0
2 months ago
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