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2 years ago

Can someone please help me this is urgent!?

Engineering

2 answers:

Thepotemich [5.8K]2 years ago

8 0

Answer:

it would be C

Explanation:

if you look at the times 0-2 you see it goes to 2.0 so now it cant be A or D now if you look at times 2-4 you see it goes to 3.0 so now B isnt right and then 4-6 its stays at 3.0 so now its definitely C

Vlad1618 [11]2 years ago

5 0

Answer: my answer would be C.

Explanation:

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A piston-cylinder device contains 0.58 kg of steam at 300°C and 0.5 MPa. Steam is cooled at constant pressure until one-half of

Mumz [18]

Answer:

a) Tբ = 151.8°C

b) ΔV = - 0.194 m³

c) The T-V diagram is sketched in the image attached.

Explanation:

Using steam tables,

At the given pressure of 0.5 MPa, the saturation temperature is the final temperature.

Right from the steam tables (A-5) with a little interpolation, Tբ = 151.793°C

b) The volume change

Using data from A-5 and A-6 of the steam tables,

The volume change will be calculated from the mass (0.58 kg), the initial specific volume (αᵢ) and the final specific volume

(αբ) (which is calculated from the final quality and the consituents of the specific volumes).

ΔV = m(αբ - αᵢ)

αբ = αₗ + q(αₗᵥ) = αₗ + q (αᵥ - αₗ)

q = 0.5, αₗ = 0.00109 m³/kg, αᵥ = 0.3748 m³/kg

αբ = 0.00109 + 0.5(0.3748 - 0.00109)

αբ = 0.187945 m³/kg

αᵢ = 0.5226 m³/kg

ΔV = 0.58 (0.187945 - 0.5226) = - 0.194 m³

c) The T-V diagram is sketched in the image attached

3 0

3 years ago

Many households in developing countries prepare food over indoor cook stoves with no ducting system to exhaust the combustion pr

lorasvet [3.4K]

Answer:

C = 0.22857 ng / m³

Explanation:

Let's solve this problem for part the total time in the kitchen is

t = 2h (60 min / 1h) = 120 min

The concentration (C) quantity of benzol pyrene is the initial quantity plus the quantity generated per area minus the quantity eliminated by the air flow. The amount removed can be calculated assuming that an amount of extra air that must be filled with the pollutant

amount generated

C = co + time_generation rate / (area_house + area_flow)

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3 years ago

The minimum requirements for engineering documents are enumerated in

vladimir1956 [14]

Answer:

The answer will be Rule 61G15-23 F.A.C, relating to Seals.

Explanation:

According to the description given by: Florida administrative code&Florida administrative register the Minimum requirements for engineering documents are in the section 'Final 61G15-23' from 11/3/2015. This document provides specifications of materials required for the safe operation of the system that is the result of engineering calculations, knowledge and experience.

8 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: