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2 years ago

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Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cook

er has decreased by 0.6 L in 40 min after the steady operating conditions are established, and the cross-sectional area of theexit opening is 8mm2. Determine

1 answer:

Anna71 [15]2 years ago

7 0

Answer:

The mass flow rate is 2.37*10^-4kg/s

The exit velocity is 34.3m/s

The total flow of energy is 0.583 KJ/KgThe rate at which energy leave the cooker is 0.638KW

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Which of the following statements is TRUE about updating the exposure control plan?

iren2701 [21]

Statements that are true as regards exposure control plan and its updating are;

<em>Updates must have the reflection of changes in tasks as well in procedures.</em>

<em>Updates must reflect changes in positions that affect occupational exposure.</em>

<em>Updates must have the cost of PPE that is needed and necessary to reduce exposure</em>

An exposure control plan can be regarded as the framework for compliance between the employer and the workers.

This framework give room for the employer to creates a written plan that will help in protecting their workers from bloodborne pathogens.

This plan gives hope to workers in term of protection when working with their Employer.

There are some elements that is associated with Exposure Control Plan, and theses are;

Health hazards as well as risk that is attributed to each product in the worksite.
Statement of purpose.
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Responsibilities from the Manager, CEO, designated resources and employer.

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2 years ago

State a Newtowns second law

astraxan [27]

Answer:

the second law states that the force F is the product of an object's mass and its acceleration a: F = m * a. For an external applied force, the change in velocity depends on the mass of the object.

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3 years ago

A car is moving at 10 m/s to the right. It accelerates for 10 s after which it is moving at 5 m/s to the left. What was the car'

NNADVOKAT [17]

Answer:

Acceleration, $a=-1.5\ m/s^2$

Explanation:

It is given that,

Initial velocity of the car, u = 10 m/s (in right)

Final velocity of the car, v = -5 m/s (in left)

Time taken, t = 10 s

Let a is the acceleration of the car. It can be calculated using the equation of kinematics. The equation is as :

$v=u+at$

$a=\dfrac{v-u}{t}$

$a=\dfrac{-5-10}{10}$

$a=-1.5\ m/s^2$

So, the acceleration of the car is $-1.5\ m/s^2$ . Hence, this is the required solution.

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3 years ago

In a fireworks display, a rocket is launched from the ground with a speed of 18.0 m/s and a direction of 51.0° above the horizon

Masja [62]

Answer:38.66 m

Explanation:

Given

launch angle $\theta =51^{\circ}$

launch velocity $u=18 m/s$

center of mass continue to travel its original Path so it center of mass will be at a distance of

$R=\frac{u^2\sin 2\theta }{g}$

$R=\frac{18^2\sin 102}{9.8}$

$R=32.33 m$

Center of mass will be at $x=32.33 m$

(b)if one of the piece will be at x=26 m then other will be at

$x_{com}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

$x_{com}=\frac{\frac{m}{2}\cdot 26+\frac{m}{2}\cdot x_0}{\frac{m}{2}+\frac{m}{2}}$

$32.33=\frac{26+x_0}{2}$

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3 years ago

As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an ax

AysviL [449]

Answer:

$I=2.766\ kg.m^2$

Explanation:

We have:

diameter of the wheel, $d=0.88\ m$

weight of the wheel, $w_w=280\ N$

mass of hanging object to the wheel, $m_o=6.32\ kg$

speed of the hanging mass after the descend, $v_o=4\ m.s^{-1}$

height of descend, $h=2.5\ m$

(a)

moment of inertia of wheel about its central axis:

$I=\frac{1}{2} m.r^2$

$I=\frac{1}{2} \frac{w_w}{g}.r^2$

$I=\frac{1}{2} \times \frac{280}{9.8}\times 0.44^2$

$I=2.766\ kg.m^2$

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3 years ago