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adell [148]
3 years ago
7

A student constructs a simple constant volume gas thermometer and calibrates it using the boiling point of water, 100°C, and the

freezing point of a specific brine solution, -19°C. The pressures measured at the calibration points are 1.366 atm and 0.9267 atm, respectively. From these data, what temperature will the student calculate, in degrees Celsius, for the value of absolute zero?
Physics
1 answer:
just olya [345]3 years ago
4 0

Answer:

The pressure corresponding to the absolute zero temperature is 0.997atm.

Explanation:

To solve this question, you draw a straight vertical line with the boiling point temperature and pressure on top of the line and the freezing point temperature and pressure on the lower part. The absolute temperature somewhere in the middle of the line with the pressure to be obtained.

So, we have;

0- (-19) / 100 - (-19) = P - 0.9267 / 1.366 - 0.9267

19 / 119 = P - 0.9267 / 0.4393

Cross multiply, we have

19 * 0.4393 = 119(P -0.9267)

8.3467 = 119P - 110.2773

119P = 118.624

P = 0.997 atm

So at 0°C, the pressure of the thermometer is 0.997atm.

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I would say Its an example of polygenic inheritance because The determination of a particular characteristic, e.g. height or skin colour, by many genes (polygenes), each having a small effect individually. Characteristics controlled in this way show continuous variation.
3 0
3 years ago
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From a window that is 20 m from the ground a stone with a speed of 10m / s is thrown vertically upwards. Calculate:
Oduvanchick [21]

a)

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the maximum height

v = final velocity at the maximum height = 0 m/s

t = time taken to reach the maximum height

Using the equation

v² = v₀² + 2 a (Y - Y₀)

0² = 10² + 2 (- 9.8) (Y - 20)

Y = 25.1 m


also using the equation

v = v₀ + a t

inserting the values

0 = 10 + (- 9.8) t

t = 1.02 sec


b)

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the ground = 0 m

t = time taken to reach the ground

Using the equation

Y = Y₀ + v₀ t + (0.5) a t²

0 = 20 + 10 t + (0.5) (- 9.8) t²

t = 3.3 sec

3 0
3 years ago
In a car with a mass of 4,000 KG is accelerating at a rate of 2m/s2 and hits a tree what force does it have
Ilia_Sergeevich [38]

According to Newton's second law of motion, the acceleration of a body is directly proportional to the force acting on the body and inversely proportional to its mass. The formula for this law is

F=ma

=4000kg * 2m/s 2 =8000N

3 0
3 years ago
Which process is not required for an animal to obtain energy from food?
Romashka [77]

Complete Question:

Which process is not required for an animal to obtain energy from food?

a. Secretion.

b. Digestion.

c. Excretion.

d. Ingestion.

Answer:

c. Excretion.

Explanation:

Excretion is a metabolic process in which living organisms gets their system rid of metabolic waste, water and excess ions. This is usually done with the aid of kidney, skin and lungs in all vertebrates.

Example of excretions are urination, sweating and defecation.

5 0
3 years ago
The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much
aleksklad [387]

Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}

Upon substituting our given values, we will get:

\frac{(T_1)^2}{(T_2)^2}=\frac{(238,000)^3}{(1,222,000)^3}

\frac{(T_1)^2}{(T_2)^2}=\frac{13481272000000000}{1824793048000000000}

\frac{(T_1)^2}{(T_2)^2}=\frac{13481272}{1824793048}

\frac{(T_1)^2}{(T_2)^2}=0.0073878361246365

Taking square root of both sides, we will get:

\frac{T_1}{T_2}=\sqrt{0.0073878361246365}

\frac{T_1}{T_2}=0.0859525225030452495

\frac{T_2}{T_1}=\frac{1}{0.0859525225030452495}

\frac{T_2}{T_1}=11.634329870476699\approx 11.634

T_2=11.634\cdot T_1

This implies that time period of Titan about Sturn is 11.634 times more compared to time period of Enceladus about Saturn.

So, basically Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

8 0
3 years ago
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