Answer:
(a). 14.4 lbf/in^2.
(b). 27.8 in, AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.
Explanation:
So, from the question above we are given the following parameters which are going to help us in solving this particular Question;
=> The "barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a vacuum at the top of the barometer)"
=> "On a day when the temperature is 70oF, the mercury column height is 28.35 inches (corrected for thermal expansion)."
With these knowledge, let us delve right into the solution;
(a). The barometric pressure = water vapor pressure + acceleration due to gravity (ft/s^2) × water density(slug/ft^3) × {ft/12 in}^3 × [ height of mercury column + specific gravity of mercury × height of water column].
The barometric pressure= 0.363 + {(62.146) ÷ (12^3) × 390.6425}. = 14.4 lbf/in^2.
(b). { (13.55 × length of mercury) + 6.5 } × (62.15÷ 12^3) = 14.4 - 0.603.
Length of mercury = 27.8 in.
AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.
Choice c it's confusing to understand but it's c
Answer: -4.4 m/s
Explanation:
This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum
must be equal to the final momentum
:
(1)
Where:
(2)
(3)
is the mass of the child
is the initial velocity of the child
is the mass of the adult
is the initial velocity of the adult (it is sitting still)
is the final velocity of the child
is the final velocity of the adult
Substituting (2) and (3) in (1):
(4)
Isolating
:
(5)
(6)
Finally:
This means the velocity of the child is in the opposite direction
Answer:
1. decreases
2. inverse proportionality
3. 1 A
Explanation
Just did it
Answer
given,
SAT is 500 with a standard deviation of 100.
a sample of 400 students whose family income was between $70,000 and $80,000 had an average verbal SAT score of 511.
sample mean = 
= 
= 5
95% confidence level is achieved within +/- 1.960 standard deviations.
1.960 standard deviations x 5 is equal to +/- 9.8
confidence interval = 511 - 9.8 --- 511 + 9.8
= 501.2-----520.8