Answer:
fb = 240.35 Hz
Explanation:
In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.
Open tube:
(1)
vs: speed of sound = 343m/s
L: length of the open tube = 0.47328m
You replace in the equation (1):
Closed tube:
L': length of the closed tube = 0.702821m
Next, you use the following formula for the beat frequency:
The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz
Answer:
6.38 x 10^4 J
Explanation:
d = 0.33 cm = 0.33 x 10^-2 m, Area = 87 x 36 cm^2 = 0.87 x 0.36 m^2
ΔT = 14 degree C, t = 1 min = 60 second
K = 0.8 W / m K
Heat = K A ΔT t / d
H = 0.8 x 0.87 x 0.36 x 14 x 60 / (0.33 x 10^-2)
H = 6.38 x 10^4 J
Answer:
the needle will direct its North South according to the magnetic field of current carrying wire.
Explanation:
A current carrying wire always has a magnetic field around it, in circular loops. This magnetic field will be either clockwise or anticlockwise depending on the direction of current.
Right hand rule tells the direction. Place the current carrying wire in your right hand with thumb pointing the direction of current. Curl of the fingers tell the direction of current.
When the needle gets in the vicinity of the field, its poles aligns itself with the field. (previous position of the compass needle has no effect on its position in the field). The north pole and south pole will be set in the direction of magnetic field.
The distance between the needle and wire does effect the strength (accuracy) of the needle position. Strong field will create strong deflection of the needle whereas when the distance from wire increases, field weakens, thus the deflection of needle will be weak.
Answer: equation for the reaction is given below
PCL2+PCL3=PCL5
Where pcl2=0.40atm,pcl3=0.27atm
Pcl5=0.0029atm
Using ∆G=-RTin(PCL5/PCl2*PCL3)
Where R=8.314J/K/mol and T=298K
∆G=-8.314*298in(0.0029/0.40*.27)
∆G=8962.6J/mol
Explanation:
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.
