Answer:
200
Explanation:
A size sheets (also known as letter size) are 8.5 inches by 11 inches.
B size sheets (also known as ledger size) are 11 inches by 17 inches.
One B size sheet is twice as large as a A size sheet. So if you have 100 B size sheets and cut each one in half, you'll get 200 A size sheets.
Answer:
composition of alpha phase is 27% B
Explanation:
given data
mass fractions = 0.5 for both
composition = 57 wt% B-43 wt% A
composition = 87 wt% B-13 wt% A
solution
as by total composition Co = 57 and by beta phase composition Cβ = 87
we use here lever rule that is
Wα = Wβ ...............1
Wα = Wβ = 0.5
now we take here left side of equation
we will get
= 0.5
= 0.5
solve it we get
Ca = 27
so composition of alpha phase is 27% B
Answer:
250.7mw
Explanation:
Volume of the reservoir = lwh
Length of reservoir = 10km
Width of reservoir = 1km
Height = 100m
Volume = 10x10³x10³x100
= 10⁹m³
Next we find the volume flow rate
= 0.1/100x10⁹x1/3600
= 277.78m³/s
To get the electrical power output developed by the turbine with 92 percent efficiency
= 0.92x1000x9.81x277.78x100
= 250.7MW
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.
