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user100 [1]
2 years ago
13

What is the effect of the following change on the volume of 1 mol of an ideal gas? The initial pressure is 722 torr, the final p

ressure is 0.950 atm, the initial temperature is 32°F, and the final temperature is 273 K. A. Volume is unchanged B. Volume increases by about 10% C. volume decreases by about 92% D. volume increases by 8% E. volume increases by about 90%.
Physics
1 answer:
Ray Of Light [21]2 years ago
4 0

Answer:

A. Volume is unchanged

Explanation:

P_{i} = initial pressure of the gas = 722 torr = 96258.7 pa

P_{f} = final pressure of the gas = 0.950 atm = 96258.75 pa

T_{i} = initial temperature = 32 °F = 272.15 K

T_{f} = final temperature = 273 K

V_{i} = initial volume

V_{f} = final volume

Using the Equation

\frac{P_{i} V_{i}}{T_{i}} = \frac{P_{f} V_{f}}{T_{f}}

Inserting the values

\frac{(96958.7) V_{i}}{272.15} = \frac{(96958.75) V_{f}}{273}

V_{f} = (1.00312) V_{i}

Hence the volume is unchanged.

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Show your work and resoning for the below requirement.
Leno4ka [110]

Answer:

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

Explanation:

We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.

Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn

 They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole

            - T_{y} L + W₂ L + W₁ L / 2 = 0

            T_{y} = W₂ + W₁ / 2

            T_{y} = 120 + 150/2

            T_{y} = 195 lb

we use trigonometry to find the cable tension

             sin 30 = T_{y} / T

             T = T_{y} / sin 30

             T = 195 / sin 30

             T = 390 lb

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

             T < 500 lb

4 0
2 years ago
An empty bottle has a mass of 35.00 grams. When filled with water, it has a mass of 98.44 grams. Of the same bottle is filled wi
sveticcg [70]

Answer:

Specific gravity of other fluid = .854 (Approx)

Explanation:

Given:

Mass of water = 35 g

Mass of filled bottle with water = 98.44 g

Mass of filled bottle with fluid = 89.22 g

Computation:

Mass of water = 98.44g - 35g = 63.44g

Density of water = 1000 g/L

Volume of bottle = 63.44/1000 = 0.06344L

Mass of other liquid = 89.22g - 35g = 54.22g

Density of other liquid = 54.22g/0.06344L = 854.665826 g/L

Water has a specific gravity = 1

So ,  specific gravity of other fluid

1000 / 854.665826 = 1 / specific gravity of other fluid

Specific gravity of other fluid = .854 (Approx)

5 0
2 years ago
The study of alternating electric current requires the solutions of equations of the form i equals Upper I Subscript max Baselin
KiRa [710]

Answer:

Explanation:

i = Imax sin2πft

given i = 180 , Imax = 200 , f = 50  , t = ?

Put the give values in the equation above

180 = 200 sin 2πft

sin 2πft = .9

sin2π x 50t = .9

sin 360 x 50 t = sin ( 360n + 64 )

360 x 50 t = 360n + 64

360 x 50 t =  64 ,  ( putting n = 0 for least value of t )

18000 t = 64

t = 3.55 ms  .

8 0
2 years ago
An electric clock is hanging on a wall. As you are watching the second hand rotate, the clock's battery stops functioning, and t
Setler [38]

Answer:

B. W is positive and a is negative

Explanation:

As we know that the angular speed of the second clock is in positive direction so as it comes to halt from its initial direction of motion then we have

initial angular velocity is termed as positive angular velocity

\omega = positive

now it comes to stop so angular acceleration is taken in opposite to the direction of angular speed

so we will have

\alpha = negative

so here correct answer is

B. W is positive and a is negative

8 0
2 years ago
3. A 4.1 x 10-15 C charge is able to pick up a bit of paper when it is initially 1.0 cm above the paper. Assume an induced charg
Anni [7]

Answer:

\mathbf{1.51\times10^{-15}N}

Explanation:

The computation of the weight of the paper in newtons is shown below:

On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.

Therefore the paper charge is

q_{paper}=-4.1\times10^{-15}C

Now the distance from the charge is

r=1cm=0.01m

Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.

mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}

\Rightarrow W=mg

=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}

=\mathbf{1.51\times10^{-15}N}

6 0
1 year ago
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