Answer:
a-1 Graph is attached. The relation is linear.
a-2 The corresponding height for 68 kPa Pressure is 7.54 m
a-3 The corresponding weight for 68 kPa Pressure is 1394726kg
b The original height of the column is 5.98 m
Explanation:
Part a
a-1
The graph is attached with the solution. The relation is linear as indicated by the line.
a-2
By the equation

Here
- P is the pressure which is given as 68 kPa.
- ρ is the density of the oil whose SG is 0.92. It is calculated as

- g is the gravitational constant whose value is 9.8 m/s^2
- h is the height which is to be calculated

So the height of column is 7.54m
a-3
By the relation of volume and density

Here
- ρ is the density of the oil which is 920 kg/m^3
- V is the volume of cylinder with diameter 16m calculated as follows

Mass is given as

So the mass of oil leading to 68kPa is 1394726kg
Part b
Pressure variation is given as

Now corrected pressure is as

Finding the value of height for this corrected pressure as

The original height of column is 5.98m
Answer:
There is a loss of fluid in the container of 0.475L
Explanation:
To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.
The formula that describes this thermal expansion process is given by:

Where,
Change in volume
Initial Volume
Change in temperature
coefficient of volume expansion (Coefficient of copper and of the liquid for this case)
There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

Where,
= Change in the volume of liquid
= Change in the volume of copper
Then replacing with the previous equation we have:


Our values are given as,
Thermal expansion coefficient for copper and the liquid to 20°C is




Replacing we have that,



Therefore there is a loss of fluid in the container of 0.475L
Answer:
graph A
Explanation:
the slope of the distance-time graph is speed, speed is a scalar (with magnitudes but no direction)
but the slope for the velocity time graph is acceleration, acceleration is vector quantity ( has magnitude and direction)
The coefficient of expansion is 13 * 10^-6 m per meter length.per oK
The temperature difference = 42 - - 8 = 50 oC
delta T = (42 + 273) - (-8 + 273) = 50 oK
delta L = L * 13* 10^6 m/oK
oK = 50 oK delta L = 19.5 cm = 19.5 cm [1m / 100 cm] = 0.195m
So we need to find the length and it is computed by:
0.195= L * 13 * 10^-6 * 50 L = 0.195 / (13*10^-6*50) L = 300 m
The frequency of rotation of Mars is 0.0000113 Hertz.
<u>Given the following data:</u>
- Period = 1 day and 37 minutes.
To find the frequency of rotation in Hertz:
First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.
<u>Conversion:</u>
1 day = 24 hours
24 hours to minutes =
×
=
minutes

1 minute = 60 seconds
1477 minute = X seconds
Cross-multiplying, we have:
× 
X = 88620 seconds
Now, we can find the frequency of rotation of Mars by using the formula:

<em>Frequency </em><em>of rotation</em> = <em>0.0000113 Hertz</em>
Therefore, the frequency of rotation of Mars is 0.0000113 Hertz.
Read more: brainly.com/question/14708169