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jolli1 [7]
4 years ago
14

A gas has an initial volume of 212 cm3 at a temperature of 293 K and a pressure of 0.98 atm. What is the final pressure of the g

as if the volume decreases to 196 cm3 and the temperature of the gas increases to 308 K? 0.86 atm 0.95 atm 1.0 atm 1.1 atm
Physics
2 answers:
PIT_PIT [208]4 years ago
8 0

Answer: 1.1 atm

Explanation:

Using ideal gas equation:

\frac{P_1V_1}{T_1}={P_2V_2}{T_2}

P_1 = initial pressure = 0.98 atm

V_1 = initial volume = 212cm^3

T_1 = initial temperature= 293 K

P_2 = final pressure = ?

V_2 = final volume= 196cm^3

T_2 = final temperature=308 K

\frac{0.98\times 212}{293}={P_2\times 196}{308}

P_2= 1.1atm

Thus final pressure of the gas is 1.1 atm.

Serggg [28]4 years ago
7 0
If the ga has 212 cm3 the. Temperature was at a 2
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6 0
3 years ago
A 1.0-m-tall vertical tube is filled with 20°C water. A tuning fork vibrating at 580 Hz is held just over the top of the tube as
Mademuasel [1]

Answer:

water heights of the tube are 0.851 m  , 0.553 m, 0.255 m

Explanation:

given data

frequency = 580 Hz

temperature = 20°C

tube = 1 m

to find out

water heights of the tube

solution

we will apply here formula for length that is

length L = v ( 2n -1 ) / 4f

here v is velocity o sound that is 343.2 m/s

so for n = 1

L = 343.2 ( 2(1) -1 ) / 4(580) = 0.147931 m

for n = 2

L = 343.2 ( 2(2) -1 ) / 4(580) = 0.443793 m

for n = 3

L = 343.2 ( 2(3) -1 ) / 4(580) = 0.739655 m

for n = 4

L = 343.2 ( 2(4) -1 ) / 4(580) = 1.035517 m is greater than 1

and so here  height is measured less than 1 m

so water heights of the tube are 1 m - 0.147931 m  , 1 m - 0.443793 m, 1 m - 0.739655 m

so water heights of the tube are 0.851 m  , 0.553 m, 0.255 m

3 0
3 years ago
The escape speed from an object is v2 = 2GM/R, where M is the mass of the object, R is the object's starting radius, and G is th
Rom4ik [11]

Answer:

Approximate escape speed = 45.3 km/s

Explanation:

Escape speed

        v=\sqrt{\frac{2GM}{R}}

Here we have

   Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

   R = 1 AU = 1.496 × 10¹¹ m

   M = 2.3 × 10³⁰ kg

Substituting

    v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 2.3\times 10^{30}}{1.496\times 10^{11}}}=4.53\times 10^4m/s=45.3km/s

Approximate escape speed = 45.3 km/s

6 0
4 years ago
What the motion of an object that has an acceleration of 0 m/s
Degger [83]
<span>Everything in the system is stable and therefore the objects motion is stable. That is to say it is not changing what it is already doing. As far as i know zero times zero is still zero. In that case then the motion must be constant or stable.</span>
8 0
3 years ago
A beam of 1.0 MHz ultrasound begins with an intensity of 1000 W/m². After traveling 12 cm through tissue with no significant ref
Ksenya-84 [330]

Answer:

Option C

Explanation:

Given:

- Depth of tissue d = 12 cm

- frequency of ultrasound f = 1 MHz

- Input Intensity I_i = 1000 W/m^2

- attenuation coefficient soft tissue a = 0.54

Find:

- Out-put intensity at the required depth

Solution

- The amount of attenuation in (dB) with the progression of depth is given by:

                                     Attenuation = a*f*d

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- The relation with attenuation and ratio of input and output intensity is given by:

                                     Attenuation = 10*log_10 (I_i / I_o)

                                     6.48 dB = 10*log_10 (I_i / I_o)

                                      I_i / I_o = 10^(0.648)

                                      I_o = 1000 / 10^(0.648)

                                      I_o = 225 W/m^2

- Hence the answer is option C:  I_o = 250 W/m^2  

4 0
4 years ago
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