Divide CFU of Dilution. Divide the CFU of the dilution (the number of colonies you counted) by the result from step 4. For this example, you work out 46 ÷ 1/1000, which is the same as 46 x 1,000. The result is 46,000 CFU in the original sample.
Explanation:
Question 1
1.The wires are made of conducting materials e.g copper
2. A light bulb is made of filament material e.g tungsten with gases enclosed in a glass material
3.The lens is made of transparent glass material
4 The reflector is made of plastic with silver lining
5. The exterior casing of most flash lights are made of plastic
Question 2.
1.Copper is used to conduct electric current from the battery to the bulb
2.The tungsten filament ignite the gases and causes it to glow
3. The lens covers the lamp on your flashlight so that the glass on the lamp
/bulbs does not get broken.
4.The reflector redirects the light rays from the lamp, creating a steady beam of light, which is the light you see emitting from the flashlight.
5. The exterior cases houses the entire assembly and makes the flash light handy for use
Wavelength of X-rays = 10⁻¹⁰ m
Wavelength of UV = 1000 x 10⁻¹⁰
= 10⁻⁷ m
Answer:
The peak-to-peak ripple voltage = 2V
Explanation:
120V and 60 Hz is the input of an unfiltered full-wave rectifier
Peak value of output voltage = 15V
load connected = 1.0kV
dc output voltage = 14V
dc value of the output voltage of capacitor-input filter
where
V(dc value of output voltage) represent V₀
V(peak value of output voltage) represent V₁
V₀ = 1 - (
)V₁
make C the subject of formula
V₀/V₁ = 1 - (1 / 2fRC)
1 / 2fRC = 1 - (v₀/V₁)
C = 2fR ((1 - (v₀/V₁))⁻¹
Substitute for,
f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V
C = 2 * 240 * 1 (( 1 - (14/15))⁻¹
C = 62.2μf
The peak-to-peak ripple voltage
= (1 / fRC)V₁
= 1 / ( (120 * 1 * 62.2) )15V
= 2V
The peak-to-peak ripple voltage = 2V
The most common metals used for permanent magnets are iron, nickel, cobalt and some alloys of rare earth metals