Answer:
i believe its 26.7
Explanation:
if the runner goes 8.9 m/s each second while accelerating for 3 seconds to reach top speed, the top speed would be 26.7 m/s
FEMA stands for <span>Federal Emergency Management Agency</span>
Answer:
i. Cv =3R/2
ii. Cp = 5R/2
Explanation:
i. Cv = Molar heat capacity at constant volume
Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT
Differentiating U with respect to T, we have
= d(3/2RT)/dT
= 3R/2
ii. Cp - Molar heat capacity at constant pressure
Cp = Cv + R
substituting Cv into the equation, we have
Cp = 3R/2 + R
taking L.C.M
Cp = (3R + 2R)/2
Cp = 5R/2
Answer:
Explanation:
Generally, length of vector means the magnitude of the vector.
So, given a vector
R = a•i + b•j + c•k
Then, it magnitude can be caused using
|R|= √(a²+b²+c²)
So, applying this to each of the vector given.
(a) 2i + 4j + 3k
The length is
L = √(2²+4²+3²)
L = √(4+16+9)
L = √29
L = 5.385 unit
(b) 5i − 2j + k
Note that k means 1k
The length is
L = √(5²+(-2)²+1²)
Note that, -×- = +
L = √(25+4+1)
L = √30
L = 5.477 unit
(c) 2i − k
Note that, since there is no component j implies that j component is 0
L = 2i + 0j - 1k
The length is
L = √(2²+0²+(-1)²)
L = √(4+0+1)
L = √5
L = 2.236 unit
(d) 5i
Same as above no is j-component and k-component
L = 5i + 0j + 0k
The length is
L = √(5²+0²+0²)
L = √(25+0+0)
L = √25
L = 5 unit
(e) 3i − 2j − k
The length is
L = √(3²+(-2)²+(-1)²)
L = √(9+4+1)
L = √14
L = 3.742 unit
(f) i + j + k
The length is
L = √(1²+1²+1²)
L = √(1+1+1)
L = √3
L = 1.7321 unit
Given:
Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min
Required:
Inlet flowrate
Solution:
The problem can be solved by this general formula.
Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
First, we need to convert the units of the accumulation velocity into m/s to be consistent.
Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s
We then calculate the area of the pool and the area of the orifice by:
Area of pool = 3 × 4 m²
Area of pool = 12m²
Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²
Since we have all we need, we plug in the values to the general equation earlier
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²
Transposing terms,
Inlet flowrate = 0.316 m³/s