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ElenaW [278]
1 year ago
8

Two moles of an ideal gas at 3.0 atm and 10 °C are heated up to

Physics
1 answer:
ololo11 [35]1 year ago
5 0

Answer:

Explanation:

Two moles of an ideal gas at 3.0 atm and 10°C are heated up to 150 °C. If the volume is held constant during this heating, what is the final pressure? a. 4.5 atm.

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"For a first order instrument with a sensitivity of .4 mV/K and a time" constant of 25 ms, find the instrument’s response as a f
ELEN [110]

Answer:

Explanation:

Given that:

For a first order instrument with a sensitivity of .4 mV/K

constant c  = 25 ms = 25 × 10⁻³ s

The initial temperature T_1 = 273 K

The final temperature T_2 = 473 K

The initial volume = 0.4 mV/K × 273 K = 109.2 V

The final volume =  0.4 mV/K × 473 K =  189.2 V

the instrument’s response as a function of time for a sudden temperature increase can be computed as follows:

Let consider y to be the function of time i.e y(t).

So;

y(t) = 109.2  + (189.2 - 109.2)( 1 - \mathbf{e^{-t/c}})mV

y(t) = (109.2 +  80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

Plot the response y(t) as a function of time.

The plot of y(t) as a function of time can be seen in the diagram  attached below.

What are the units for y(t)?

The unit for y(t) is mV.

Find the 90% rise time for y(t90) and the error fraction,

The 90% rise time for y(t90) is as follows:

Initially 90% of 189.2 mV = 0.9 ×  189.2 mV

=  170.28 mV

170.28 mV = (109.2 +  80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

170.28 mV - 109.2 mV = 80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

61.08 mV =  80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

0.7635  mV = ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

t = 1.44 × 25  × 10⁻³ s

t = 0.036 s

t = 36 ms

The error fraction = \dfrac{189.2-170.28  }{189.2}

The error fraction = 0.1

The error fraction = 10%

8 0
2 years ago
You are driving on the highway, and you come to a steep downhill section. As you roll down the hill, you take your foot off the
Natasha2012 [34]

Answer:

air

Explanation:

The car is being slowed down by air.

5 0
2 years ago
Read 2 more answers
An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0
2 years ago
For nitrogen feel like with its temperature must be within 12.78 Fahrenheit of -333.22 Fahrenheit which equation can be used to
photoshop1234 [79]

Answer:

The following equation can be used.

(32°F − 32) × 5/9=C

7 0
2 years ago
Read 2 more answers
Which of the following are inertial reference frames? A. A car driving at steady speed on a straight and level road. B. A car dr
Aloiza [94]

Answers:

A. A car driving at steady speed on a straight and level road.

B. A car driving at steady speed up a 10∘ incline.

Explanation:

An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:

F=ma

Since F=0, a=0 as well.

Let's now analyze each case.

A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.

D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.

So the only two choices which are correct are A and B.

8 0
2 years ago
Read 2 more answers
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