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4 years ago

15

A closed, rigid tank is filled with water. Initially, the tank holds 9.9 ft3 of saturated vapor and 0.1 ft3 of saturated liquid,

each at 212°F. The water is heated until the tank contains only saturated vapor. For the water, determine (a) the quality at the initial state, (b) the temperature at the final state, in °F, and (c) the heat transfer, in Btu. Kinetic and potential energy effects can be ignored.

1 answer:

Galina-37 [17]4 years ago

7 0

Answer:

a) x₁ = 0.058

b) T₂ = 416.02°F

c) q = 5047.39 Btu

Explanation:

see the attached file

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PLS ANSWER FAST WILL GIVE BRAINLY!!!! <br><br> Why does the look of an atom keep changing?

kotykmax [81]

Answer:Atomic model keep changing because the electrons around the nucleus are not fixed and they keeps rotating or changing their position in valence orbits around the nucleus.

Explanation:

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3 years ago

The normal formula to find force is F = m* a. What kind of math do you need to do

Oduvanchick [21]

To find the Mass of an object, you need to apply division.
Since Resultant Force = Mass X Acceleration

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3 years ago

Which of the following graph is used for determining the instantaneous velocity from the slope?

hoa [83]

Answer:

B. x - t graph

Explanation:

A position-time (x-t) graph is a graph of the position of an object against (versus) time.

Generally, the slope of the line of a position-time (x-t) graph is typically used to determine or calculate the velocity of an object.

An instantaneous velocity can be defined as the rate of change in position of an object in motion for a short-specified interval of time. Thus, an instantaneous velocity is a quantity that can be found by measuring the slope of a line that is tangent to a point on the graph.

Hence, the x - t graph also referred to as the position-time graph is used for determining the instantaneous velocity from the slope.

<u>For example;</u>

Given that the equation of motion is S(t) = 4t² + 2t + 10. Find the instantaneous velocity at t = 5 seconds.

Solution.

$S(t) = 4t^{2} + 2t + 10$

Differentiating the equation, we have;

$S(t) = 8t + 2$

Substituting the value of "t" into the equation, we have;

$S(5) = 8(5) + 2$

$S(5) = 40 + 2$

S(5) = 42 m/s.

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3 years ago

Read 2 more answers

Models are particularly useful in relativity and quantum mechanics, where conditions are outside those normally encountered by h

mylen [45]

Answer:

A model is defined as a structure used to represent an object, usually of a different scale.

Explanation:

In quantum mechanics and particle physics, many of the particles are subatomic, meaning that they are smaller than atoms. This is where a model would be useful. A model could help people to visualise what the particle looks like, and in general would make it easier to understand the behaviour of such a particle.

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3 years ago

a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^

posledela

Answer:

Approximately $0.8\; \rm s$ (assuming that air resistance is negligible.)

Explanation:

Let $v_0$ denote the initial velocity of this ball. Let $\theta$ denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

Initial vertical velocity: $v_0(\text{vertical}) = v_0 \cdot \sin(\theta)$ .
Initial horizontal velocity: $v_0(\text{vertical}) = v_0 \cdot \cos(\theta)$ .

If air resistance on this ball is negligible, $v_0(\text{vertical})$ alone would be sufficient for finding the time of flight of this ball.

Calculate $v_0(\text{vertical})$ given that $v_0 = 8 \; \rm m \cdot s^{-1}$ and $\theta = 30^\circ$ :

$\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}$ .

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since $v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}$ , the vertical velocity of this ball right before landing would be $v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}$ .

Calculate the change to the vertical velocity of this ball:

$\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}$ .

In other words, the vertical velocity of this ball should have change by $8\; \rm m \cdot s^{-1}$ during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is $g = 10\; \rm m \cdot s^{-2}$ . In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by $10\; \rm m\cdot s^{-1}$ each second. What fraction of a second would it take to change the vertical velocity of this ball by $8\; \rm m \cdot s^{-1}$ ?

$\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}$ .

In other words, it would take $0.8\; \rm s$ to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be $0.8\; \rm s\!$ .

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3 years ago