Answer:
W = 290.7 dynes*cm
Explanation:
d = 1/5 cm = 0.2 cm
The force is in function of the depth x:
F(x) = 1000 * (1 + 2*x)^2
We can expand that as:
F(x) = 1000 * (1 + 4*x + 4x^2)
F(x) = 1000 + 4000*x + 4000*x^2
Work is defined as
W = F * d
Since we have non constant force we integrate

W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2
W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3
W = 200 + 80 + 10.7 = 290.7 dynes*cm
<span>For this example, the value presented would be considered a statistic. The value is a statistic as it represents a numerical measurement of a sample. If it were a parameter, it would need to represent a numerical measurement of a population.</span>
An example of a negative incentive for producers is the
sharp increase in production costs. Producers are the one who manage the production
costs and even the production budget. Anything that relates the production
department is entitled to the management of production producers.
There is what we called positive and negative incentives and
both of these can affect consumers and producers. Positive incentives are those
situations which will give a certain outcome that will benefit the producers,
for example, during the peak season there will be a high demand of products, and
this gives the chance of producers to demand a higher price from the consumers,
in this situation, there will be a big chance of increase sales. A sharp increase in production costs is a
loss for the producers. If there will be
an increase in production costs, the budget will be greatly affective and even
though it is not a peak season, there’s a big chance also to increase prices
which we know, consumers are not fond of.
Answer:
The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R
Explanation:
Given the data in the question;
Using the Clapeyron equation


where
is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu
T is the temperature ( 15 + 460 )R
m is the mass of water ( 0.5 Ibm )
is specific volume ( 1.5 ft³ )
we substitute
/
272.98 Ibf-ft²/R
Now,

where P₁ is the initial pressure ( 50 psia )
P₂ is the final pressure ( 60 psia )
T₁ is the initial temperature ( 15 + 460 )R
T₂ is the final temperature = ?
we substitute;


480.275 R
Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R