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asambeis [7]
3 years ago
12

This fluid may be mixed with engine oil for recycling.

Engineering
1 answer:
alisha [4.7K]3 years ago
5 0

Answer: D

Explanation:

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Assume the work done compressing the He gas is -63 kJ and the internal energy change of the gas is 79 kJ. What is the heat loss
klemol [59]

Answer:

Heat gain of 142 kJ

Explanation:

We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.

ΔU = Q + W ⇒ Q = ΔU -W

Q = 79 - (-63) = 142 kJ

Therefore, the gas gained heat by an amount of 142 kJ.

3 0
3 years ago
Identify three material considerations an engineer would need to consider when working on a design process.
Anika [276]

Answer:

Three material considerations are;

1. Identify and appraise the attainment of the goal of the with the design specification

2. Ascertain the required load the product being designed will experience and the suitability of the design specification to that load

3. Review the producibility of the design to ensure that it can be produced with the available technology

Explanation:

1. The appraisal of the design includes the consideration of the factors of the design and the building of reliability and efficiency into the design from the beginning

2. Ascertain if the product will require toughness, elasticity, and if will be subject to sudden or repeated loading conditions

3. Ensure that the design can be readily produced with the accessible manufacturing equipment during the conceptualization stage of the design.

4 0
3 years ago
While playing a game of catch on the quadrangle, you throw a ball at an initial velocity of 17.6 m/s (approximately 39.4 mi/hr),
MAXImum [283]

Answer:

a) The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) The ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

Explanation:

a) The ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical motion at constant acceleration. First, we calculate the time taken by the ball to hit the ground:

y = y_{o} + (v_{o}\cdot \sin \theta) \cdot t+\frac{1}{2}\cdot g\cdot t^{2} (1)

Where:

y_{o}, y - Initial and final vertical position, measured in meters.

v_{o} - Initial speed, measured in meters per second.

\theta - Launch angle, measured in sexagesimal degrees.

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

t - Time, measured in seconds.

If we know that y_{o} = 2\,m, y = 0\,m, v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ} and g = -9.807\,\frac{m}{s^{2}}, then the time taken by the ball is:

-4.904\cdot t^{2}+13.482\cdot t +2 = 0 (2)

This second order polynomial can be solved by Quadratic Formula:

t_{1} \approx 2.890\,s and t_{2} \approx -0.141\,s

Only the first root offers a solution that is physically reasonable. That is, t \approx 2.890\,s.

The vertical velocity of the ball is calculated by this expression:

v_{y} = v_{o}\cdot \sin \theta +g\cdot t (3)

Where:

v_{o,y}, v_{y} - Initial and final vertical velocity, measured in meters per second.

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ}, g = -9.807\,\frac{m}{s^{2}} and t \approx 2.890\,s, then the final vertical velocity is:

v_{y} = -14.860\,\frac{m}{s}

The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) From a) we understand that ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball (x) is determined by the following expression:

x = (v_{o}\cdot \cos \theta)\cdot t (4)

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ} and t \approx 2.890\,s, then the distance covered by the ball is:

x = 32.695\,m

The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before hitting the ground (v), measured in meters per second, is determined by the following Pythagorean identity:

v = \sqrt{(v_{o}\cdot \cos \theta )^{2}+v_{y}^{2}} (5)

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ} and v_{y} = -14.860\,\frac{m}{s}, then the magnitude of the velocity of the ball is:

v \approx 18.676\,\frac{m}{s}.

The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is defined by the following trigonometric relationship:

\tan \theta = \frac{v_{y}}{v_{o}\cdot \cos \theta_{o}}

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta_{o} = 50^{\circ} and v_{y} = -14.860\,\frac{m}{s}, the angle of the total velocity of the ball just before hitting the ground is:

\theta \approx -52.717^{\circ}

The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

3 0
3 years ago
Read 2 more answers
Identify the branch of study each student should pursue.
Paha777 [63]

Answer:

Explanation:

i took the test and screenshot it

3 0
3 years ago
Air, at a pressure of 700 kPa and a temperature of 80°C, flows through a convergent– divergent nozzle. The inlet area is 0.005 m
vodomira [7]

Answer: The complete part of the question is to find the exit velocity

Explanation:

Given the following parameters

Inlet pressure = 700kpa

outlet pressure = 40kpa

Temperature = 80°C = 353k

mass flow rate =  1 kg/s

The application of the continuity and the bernoulli's equation is employed to solve the problem.

The detailed steps and the appropriate formula is as shown in the attached file.

7 0
3 years ago
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