Answer:
baking soda and vinegar dish soap
Explanation:
it will create a bubbles and let it sit for 3 hours and it will go away
Answer:
41.5° C
Explanation:
Given data :
1025 steel
Temperature = 4°C
allowed joint space = 5.4 mm
length of rails = 11.9 m
<u>Determine the highest possible temperature </u>
coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C
Applying thermal strain ( Δl / l ) = ∝ * ΔT
( 5.4 * 10^-3 / 11.9 ) = 12.1 * 10^-6 * ( T2 - 4 )
∴ ( T2 - 4 ) = ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6
hence : T2 = 41.5°C
Answer:
As a worker, it is important to follow the proper set of instructions or emergency plans during an emergent situation. Not carefully following the rules may result to a bigger problem such as further injury and damage to property.
Explanation:
Evacuation Procedure- This is a step-by-step procedure that people follow in order to safely vacate any building or place. This procedure is applicable to any situation, such as the workplace. This is now called the <em>Workplace Evacuation Procedure. </em>This is very important because there are so many unpredictable situations or events that are happening in the world right now, such as fire or earthquake. This procedure is being done through an evacuation plan.
The awareness of the workers regarding the proper way to evacuate during emergency situation is very important. It will be easier for them to know where to locate the nearest exit route. They will also learn to stop any form of device or equipment that could cause a hazzard during the situation. In case of the hospital, which is also a workplace, the employees will also learn how to assist the patients before themselves. They will also know where to assemble if there's a need to do so.
Answer:
Wright Company
Bank Reconciliation
May 31, 2013
Credit side Debit side
Bank statement $26200 | Book balance $27900
<em>Add; </em>
Deposit on May 31 $6400
Bank error $420
Sub-total=$33020
Deductions; | Deduct
ions
Outstanding checks $5800 | Bank service charge $120
Adjusted bank balance $27220 | NSF check $560
Total deduction $680
Adjusted book balance $27220
The correct question;
An object of irregular shape has a characteristic length of L = 1 m and is maintained at a uniform surface temperature of Ts = 400 K. When placed in atmospheric air at a temperature of Tinfinity = 300 K and moving with a velocity of V = 100 m/s, the average heat flux from the surface to the air is 20,000 W/m² If a second object of the same shape, but with a characteristic length of L = 5 m, is maintained at a surface temperature of Ts = 400 K and is placed in atmospheric air at Too = 300 K, what will the value of the average convection coefficient be if the air velocity is V = 20 m/s?
Answer:
h'_2 = 40 W/K.m²
Explanation:
We are given;
L1 = 1m
L2 = 5m
T_s = 400 K
T_(∞) = 300 K
V = 100 m/s
q = 20,000 W/m²
Both objects have the same shape and density and thus their reynolds number will be the same.
So,
Re_L1 = Re_L2
Thus, V1•L1/v1 = V2•L2/v2
Hence,
(h'_1•L1)/k1 = (h'_2•L2)/k2
Where h'_1 and h'_2 are convection coefficients
Since k1 = k2, thus, we now have;
h'_2 = (h'_1(L1/L2)) = [q/(T_s - T_(∞))]• (L1/L2)
Thus,
h'_2 = [20,000/(400 - 300)]•(1/5)
h'_2 = 40 W/K.m²