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marysya [2.9K]
3 years ago
14

You have 0.5 l of air at a pressure of 203 kpa and -70°c in a rigid, sealed contaainer. what is the absolute tempertaure of the

air
Physics
1 answer:
ANTONII [103]3 years ago
5 0
From the information given and if the question is complete then;
Absolute temperature is the temperature in Kelvin
To convert degree Celsius to kelvin we normally add 273
that is Kelvin = deg Celsius + 273
Thus since we have been given that the air was at -70 degrees celcius;
then; - 70° C  + 273 = 203 K
 Therefore; the absolute temperature is 203 K 

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Dr. Glover's office has one vendor for their practice management software and another for their electronic health record, but th
Sliva [168]

Answer:

Interface

Explanation:

This is a classic example of Interface technology.

An interface allows different software packages to communicate without re-entering data.

Here in this case also systems are able to communicate with one another without duplicating data entry. For example, practice management software and another for their electronic health record.

5 0
3 years ago
The proper order of the cycle of addiction is​
Viktor [21]

Answer:

The answer would be drug use, addiction, dependence, tolerance, and withdrawal.

4 0
3 years ago
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Consider the following distribution of objects: a 2.00-kg object with its center of gravity at (0, 0) m, a 2.20-kg object at (0,
adelina 88 [10]

Answer:

body position 4 is (-1,133, -1.83)

Explanation:

The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by

               x_cm = 1 /M   ∑ x_{i} m_{i}

               y_cm = 1 /M   ∑ y_{i} mi

Where M is the total mass of the body, mi is the mass of each element

give us the mass and position of this masses

body 1

m1 = 2.00 ka

x1 = 0 me

y1 = 0 me

body 2

m2 = 2.20 kg

x2 = 0m

y2 = 5 m

body 3

m3 = 3.4 kg

x3 = 2.00 m

y3 = 0

body 4

m4 = 6 kg

    x4=?

   y4=?

mass center position

x_cm = 0

y_cm = 0

let's apply to the equations of the initial part

X axis

    M = 2.00 + 2.20 + 3.40

    M = 7.6 kg

    0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)

     x4 = -6.8 / 6

     x4 = -1,133 m

Axis y

    0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)

    y4 = -11/6

    y4 = -1.83 m

body position 4 is (-1,133, -1.83)

7 0
3 years ago
A car's velocity as a function of time is given by vx(t)=α+βt2, where α=3.00m/s and β=0.100m/s3.
Mice21 [21]

1) Analyze the equation

Vₓ = α + β t²

Vₓ = 3.00 + 0.100 t²

That is a quadratic equation, so the graph is a parabola.

2) Therefore, take into account that the main points to draw a parabola are:

i) vertex

ii) concavity

iii) y-intercepts

iv) x - intercepts

Also, for all graphs you need the domain and the range.

3) Find the y-intercept (t = 0)

t = 0 ⇒ Vₓ = 3.00 + 0.100 (0)² = 3.00

4) Find the x-intercepts (Vₓ = 0)

Vₓ = 0 ⇒ 0 = 3.00 + 0.100 t²

⇒ t² = - 3.00 / 0.100 = -30.0. Since t² cannot be negative, there is not x-intercepts.

5) Concavity

Since, the coefficient of t² is positive, the parabola open upwards.

6) Vertex

It is the local minimum of the equation. You can find it by the first derivative

Vₓ' = 2×0.100 t = 0 ⇒ t = 0

Vₓ = 3.00 + 0.100 (0)² = 3.00 m/s

⇒ vertex = (0,3.00)

7) The domain is given t ∈ [0,5.00]

8) You can also build a table with several points in the domain

t =0; Vₓ = 3.00 + 0.100 (0)² = 0

t = 1; Vₓ = 3.00 + 0.100 (1)² = 3.10

t = 2; Vₓ = 3.00 + 0.100 (2)² = 3.40

t = 3; Vₓ = 3.00 + 0.100 (3)² = 3.90

t = 4; Vₓ = 3.00 + 0.100 (4)² = 4.60

t = 5; Vₓ = 3.00 + 0.100 (5)² = 5.50

9) Range: Vₓ ∈ [ 3.00, 5.50]

10) All that information permits you to put several points in a coordinate sysment and sketch the same graph as the shown in the figure attached.

4 0
3 years ago
Read 2 more answers
In the data table , distance is measured in meters and time is in seconds. Calculate the mans average velocity using the equatio
Artist 52 [7]

Answer:

3.626 m/s

Explanation:

v=d/t

1. -0.02/0 = 0 m/s

2. 0.86/0.2 = 4.3 m/s

3. 1.71/0.4 = 4.275 m/s

4. 2.54/0.6 = 4.23 m/s

5. 3.32/0.8 = 4.15 m/s

6. 4.08/1.0 = 4.08 m/s

7. 4.79/1.2 = 3.99 m/s

8. 5.48/1.4 = 3.91 m/s

9. 6.15/1.6 = 3.84 m/s

10. 6.76/1.8 = 3.76 m/s

11. 7.37/2.0 = 3.66 m/s

12. 7.92/2.2 = 3.6 m/s

13. 8.45/2.4 = 3.52 m/s

14. 8.96/2.6 = 3.45 m/s

the mean of these numbers is 3.626

his average velocity ks 3.626 m/s

6 0
3 years ago
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