Answer:
Explanation:
There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.
We are given the following information:
y₁ = 3 m
y₂ = 3 m − 2.4 m = 0.6 m
t₂ − t₁ = 0.4 s
a = -9.8 m/s²
t₀ = 0 s
v₀ = 0 m/s
We need to find y₀.
Use a constant acceleration equation:
y = y₀ + v₀ t + ½ at²
Evaluated at point 1:
3 = y₀ + (0) t₁ + ½ (-9.8) t₁²
3 = y₀ − 4.9 t₁²
Evaluated at point 2:
0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²
0.6 = y₀ − 4.9 t₂²
Solve for y₀ in the first equation and substitute into the second:
y₀ = 3 + 4.9 t₁²
0.6 = (3 + 4.9 t₁²) − 4.9 t₂²
0 = 2.4 + 4.9 (t₁² − t₂²)
We know t₂ = t₁ + 0.4:
0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)
0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))
0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)
0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)
0 = 2.4 − 3.92 t₁ − 0.784
0 = 1.616 − 3.92 t₁
t₁ = 0.412
Now we can plug this into the original equation and find y₀:
3 = y₀ − 4.9 t₁²
3 = y₀ − 4.9 (0.412)²
3 = y₀ − 0.83
y₀ = 3.83
Rounded to two significant figures, the height of the tree is 3.8 meters.
Answer:
<em>Python code is as follows:
</em>
********************************************************************************
#function to get number up to any number of decimal places
def toFixed(value, digits):
return "%.*f" % (digits, value)
print("Enter the price: ", end='', flush=True) #prompt for the input of price
price = float(input()) #taken input
totalCost = price + 0.05 * price #calculating cost
print("Total Cost after the sales tax of 5% is applied: " + toFixed(totalCost,2)) #print and output totalCost
************************************************************************************
Answer:
diameter of the sprue at the bottom is 1.603 cm
Explanation:
Given data;
Flow rate, Q = 400 cm³/s
cross section of sprue: Round
Diameter of sprue at the top 
= 3.4 cm
Height of sprue, h = 20 cm = 0.2 m
acceleration due to gravity g = 9.81 m/s²
Calculate the velocity at the sprue base
= √2gh
we substitute
= √(2 × 9.81 m/s² × 0.2 m )
= 1.98091 m/s
= 198.091 cm/s
diameter of the sprue at the bottom will be;
Q = AV = (π
/4) ×
= √(4Q/π)
we substitute our values into the equation;
= √(4(400 cm³/s) / (π×198.091 cm/s))
= 1.603 cm
Therefore, diameter of the sprue at the bottom is 1.603 cm
Answer:
Concentration = 10.33 kg/m³
Explanation:
We are given;
Mass of solids; 10,000 kg
Volume; V = 440,000 L = 440 m³
Rate at which water is pumped out = 40,000 liter/h
Thus, at the end of 5 hours we amount of water that has been replaced with fresh water is = 40,000 liter/h x 5 hours = 200,000 L = 200 m³
Now, since the tank is perfectly mixed, therefore we can calculate a ratio of fresh water to sewage water as;
200m³/440m³ = 5/11
Thus, the amount left will be calculated by multiplying that ratio by the amount of solids;
Thus,
Amount left; = 10000 x (5/11) = 4545 kg
The concentration would be calculated by:
Concentration = amount left/initial volume
Thus,
Concentration = 4545/440 = 10.3 kg/m³
Answer:
For detailed answer of "
In subsea oil and natural gas production, hydrocarbon fluids may leave the reservoir with a temperature of 70°C and flow in subsea surrounding of S°C. As a result of the temperature difference between the reservoir and the subsea surrounding, the knowledge of heat transfer is critical to prevent gas hydrate and wax deposition blockages. Consider a subsea pipeline with inner diameter of O.S m and wall thickness of 8 mm is used for transporting liquid hydrocarbon at an average temperature of 70°C, and the average convection heat transfer coefficient on the inner pipeline surface is estimated to be 2SO W/m2.K. The subsea surrounding has a temperature of soc and the average convection heat transfer coefficient on the outer pipeline surface is estimated to be ISO W /m2 .K. If the pipeline is made of material with thermal conductivity of 60 W/m.K, by using the heat conduction equation (a) obtain the temperature variation in the pipeline wall, (b) determine the inner surface temperature of the pipeline wall, (c) obtain the mathematical expression for the rate of heat loss from the liquid hydrocarbon in the pipeline, and (d) determine the heat flux through the outer pipeline surface."
see attachment.
Explanation:
