Answer:
A) 0.0614 inches
b) The standard steel paper clip should float on water
Explanation:
The maximum diameter that the rod can have before it will sink
we can calculate this using this formula :
D = ----- 1
∝ = value of surface tension of water at 60⁰f = 5.03×10^−3 lb/ft
y = 490 Ib/ft^3
input the given values into equation 1 above
D =
= 5.11 * 10^-3 ft convert to inches
= 5.11 *10^-3 ( 12 in/ 1 ft ) = 0.0614 inches
B) The diameter of a standard paper Cliphas = 0.036 inches
and the diameter of the rod = 0.0614. Hence the standard steel paper clip should float on water
Answer: This is EASY!
Explanation:
To make it easy, you would convert those binary numbers and to denary. And this gives:
84 104 105 115 32 105 115 32 69 65 83 89 33
Then, the denary numbers generated can be converted to ASCII code using this ASCII Table in the attachment below. And the result is: This is EASY!
again = "false";
result = 0;
a. choice
b. again
c. result
d. none of these are Boolean variables
Answer:
C
Explanation:
Boolean Algebra deals with either a one or a zero and how to manipulate them in computers or elsewhere. The "choice" option may not work, since for text it must be enclosed in quotation marks, usually. For "again," it's text and not a 1 or 0. So, the answer is C, since this is a 0.
Answer:
A) approximate alkalinity = 123.361 mg/l
B) exact alkalinity = 124.708 mg/l
Explanation:
Given data :
A) determine approximate alkalinity first
Bicarbonate ion = 120 mg/l
carbonate ion = 15 mg/l
Approximate alkalinity = ( carbonate ion ) * 50/30 + ( bicarbonate ion ) * 50/61
= 15 * (50/30) + 120*( 50/61 ) = 123.361 mg/l as CaCO3
B) calculate the exact alkalinity of the water if the pH = 9.43
pH + pOH = 14
9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57
[OH^- ] = 10^-4.57 = 2.692*10^-5 moles/l
[ OH^- ] = 2.692*10^-5 * 179/mole * 10^3 mg/g = 0.458 mg/l
[ H^+ ] = 10^-9.43 * 1 * 10^3 = 3.7154 * 10^-7 mg/l
therefore the exact alkalinity can be calculated as
= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )
= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )
= 124.708 mg/l
Answer:
Heat loss=85.9W/m^2
ΔT1(Steel)=0.04C
ΔT2(Brick1)=110.13C
ΔT3(Mwood)=343.6C
ΔT1(Brick2)=154.18C
Explanation:
raise the heat transfer equation from the air inside the wall to the outside air from the wall, because that is where you have the temperature data, to find the heat.
To find the temperatures you use the heat found in the previous step, and you use the conduction and convection equations in each wall layer.
I attached the procedure
Answer:
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Explanation:
please
Answer:
Personal computers:
Personal computers may be useful and lead to productivity as using a computer an employee familiars with is a good thing. However, the disadvantages in some facilities especially ones dealing with customer and information security can include data theft, unauthorized data sharing, uses of internet connection for personal purposes, as this can slow down internet connection at the facility, distraction at work place etc.
Hard drive:
Due to large amount of data that can be stored on a hard drive, it might not be allowed in some facilities to avoid data theft and unauthorized transfer.
Music players:
This might be restricted to avoid distraction at work. Noice in places such as libraries would cause abnormality and poor service delivery. An employee with loud speaker at work would not only distracts himself but also other staffs and customers.
PSP Game Device and other game devices:
Playing games during working hour may jeopardize the productivity and therefore might be resctrited in some facilities and working places.
Electronic digital notepad:
Carrying a handheld electronic digital notepad to the work place can cause lack of concentration and division of attention on work and other personal activities. These can harm working harmony and business productivity.
Video recorder:
In some facilities, this device might not be allowed due to facility privacy and protection from unwanted navigation.
Explanation:
Answer:
3 m/s²
Explanation:
Acceleration is calculated as :
a= Δv/ t
where ;
Δv = change in velocity
Δv = 45 - 0 = 45 m/s
t= 15 s
a= 45 /15
a= 3 m/s²