Each stream in a drainage system drains into a certain area. In a drainage basin the water falling in the basin drain will fall into the same stream. A drainage divides drawing basin from other drainage basins
Because: Some of the work done by the machine is used to overcome the friction created by the use of the machine. ... Work output can never be greater than work input. Machines allow force to be applied over a greater distance, which means that less force will be needed for the same amount of work.
Answer:
The frequency of sound wave created by trumpet is 437.5Hz
Explanation:
Given
the speed of sound wave = 350 m
the wavelength of sound wave = 0.8 m
the frequency of sound wave = ?
All the waves have same relationship among wavelength, frequency and speed, which is given by the equation:
v = fλ, where
v is speed of the wave
f is frequency of the wave
λ is wavelength of the wave
therefore frequency of sound wave is given by
f = v/λ
= 350m
/0.8m
= 437.5
= 437.5Hz (since 1
= 1 Hz (Hertz)
Hence the frequency of sound wave created by trumpet is 437.5Hz
Answer:
Temperature of water leaving the radiator = 160°F
Explanation:
Heat released = (ṁcΔT)
Heat released = 20000 btu/hr = 5861.42 W
ṁ = mass flowrate = density × volumetric flow rate
Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³
ṁ = 1000 × 0.000126 = 0.126 kg/s
c = specific heat capacity for water = 4200 J/kg.K
H = ṁcΔT = 5861.42
ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C
And in change in temperature terms,
10°C= 18°F
11.08°C = 11.08 × 18/10 = 20°F
ΔT = T₁ - T₂
20 = 180 - T₂
T₂ = 160°F
Answer:
A) μ = A.m²
B) z = 0.46m
Explanation:
A) Magnetic dipole moment of a coil is given by; μ = NIA
Where;
N is number of turns of coil
I is current in wire
A is area
We are given
N = 300 turns; I = 4A ; d =5cm = 0.05m
Area = πd²/4 = π(0.05)²/4 = 0.001963
So,
μ = 300 x 4 x 0.001963 = 2.36 A.m².
B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;
B = (μ_o•μ)/(2π•z³)
Let's make z the subject ;
z = [(μ_o•μ)/(2π•B)] ^(⅓)
Where u_o is vacuum permiability with a value of 4π x 10^(-7) H
Also, B = 5 mT = 5 x 10^(-6) T
Thus,
z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)
Solving this gives; z = 0.46m =