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Sati [7]
1 year ago
13

Boiling water is a common method used to sterilize water that may be contaminated. subjecting microbes to such high temperatures

is an example of a(n) __________ technique.
Physics
1 answer:
GaryK [48]1 year ago
7 0

Boiling water is a common method used to sterilize water that may be contaminated. subjecting microbes to such high temperatures is an example of a(n) microbicidal technique.

A substance that eliminates microorganisms (such as bacteria) is known as microbicide and the techniques that involve destruction of microbes are known as microbicidal techniques.

The pathogen biofilm was effectively destroyed by the combined microbicidal effects of hot water (50, 60, or 70 °C) and 2% citric acid. It was discovered that the combination of hot water and citric acid caused sub-lethally harmed cells.

A fairly easy way to disinfect water is to boil it. The majority of dangerous organisms, notably viruses and bacteria that cause waterborne infections, are killed when water is heated to a high temperature of 100°C. The water must boil for at least 20 minutes in order to be most effective.

Hence, water that may be polluted is frequently sterilized by boiling. Using such high temperatures to kill germs is an example of a microbicidal method.

Learn more about Sterilization techniques here brainly.com/question/13062351

#SPJ4

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MOST elements on the periodic table are
lidiya [134]

Answer:

B

Explanation:

6 0
3 years ago
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When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

  = 4.59\times 10^{-19} \ J

or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

  = 2.87 \ eV

(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

5 0
3 years ago
.Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric
maw [93]

Answer:

a) The uncertainty in calculated V, ΔV = 25.3

b) The uncertainty in calculated v, Δv = 0.41 m/s

c) The uncertainty in calculated V, ΔV = 22.2 V

Explanation:

We'll use Upper-Lower Bounds method of uncertainty to estimate the uncertainties.

a) I = 5.1 A, ΔI = 0.3 A

I = (5.1 ± 0.3) A

R = 77.5 ohms, ΔR = 0.4 ohms

R = (77.5 ± 0.4) ohms

V = IR = 5.1 × 77.5 = 395.25 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 5.1 - 0.3 = 4.8 A

Rₗ = 77.5 - 0.4 = 77.1 ohms

Vₗ = 4.8 × 77.1 = 370.08 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 5.1 + 0.3 = 5.4 A

Rᵤ = 77.5 + 0.4 = 77.9 ohms

Vᵤ = 5.4 × 77.9 = 420.66 V

The average of the differences from the mean voltage/true value is 25.3 V

V = 395.25 V, Δ = 25.3V

V = (395.25 ± 25.3) V

b) x = 2.9 m, Δx = 0.3 m

x = (2.9 ± 0.3) m

t = 4.4 s, Δt = 1.8 s

t = (4.4 ± 1.8) ohms

v = x/t = 2.9/4.4 = 0.659 m/s

The lower bound for average speed will be calculated using the lower bounds for distance and upper bounds for time.

xₗ = 2.9 - 0.3 = 2.6 m

tᵤ = 4.4 + 1.8 = 6.2 s

vₗ = 2.6/6.2 = 0.419 m/s

The upper bound for the average speed will be calculated using the upper bound for the distance and lower bound for time

xᵤ = 2.9 + 0.3 = 3.2 m

tₗ = 4.4 - 1.8 = 2.6 s

vᵤ = 3.2/2.6 = 1.231 m/s

The average of the differences from the mean average speed/true value is 0.41 m/s

v = 0.659 m/s, Δv = 0.41 m/s

v = (0.659 ± 0.41) m/s

c) ) I = 9.8 A, ΔI = 0.5 A

I = (9.8 ± 0.5) A

R = 40.5 ohms, ΔR = 0.2 ohms

R = (40.5 ± 0.2) ohms

V = IR = 9.8 × 40.5 = 396.9 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 9.8 - 0.5 = 9.3 A

Rₗ = 40.5 - 0.2 = 40.3 ohms

Vₗ = 9.3 × 40.3 = 374.79 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 9.8 + 0.5 = 10.3 A

Rᵤ = 40.5 + 0.2 = 40.7 ohms

Vᵤ = 10.3 × 40.7 = 419.21 V

The average of the differences from the mean voltage/true value is 22.2 V

V = 396.9 V, Δ = 22.2 V

V = (396.9 ± 22.2) V

7 0
3 years ago
What is the best reason for a student to remove a dangling bracelet when heating test tubes of acidic solutions in a hot water
soldi70 [24.7K]

Answer:

c

Explanation:

this equestion is about the safety

4 0
3 years ago
Read 2 more answers
Which material will displace a volume of water? Which material will displace a volume of water less than its own volume? Which m
horrorfan [7]

Answer:

1. all of them

2. cork and wax

3. iron, lead, and aluminum

4. none of them

Explanation:

1.Which material will displace a volume of water?  all of them

When an object is introduced into a container with a volume of water, a volume of liquid equal to the volume of the object is displaced

2.Which material will displace a volume of water less than its own volume?

cork and wax

because  the density of the object is less than that of the displaced liquid

3.Which material will displace a volume of water equal to its own volume?

iron, lead, and aluminum

because  Arquimedes's principle: any body plunged inside a fluid in this case water experiences an ascending force called push, equivalent to the weight of the fluid removed by the body

4.Which material will displace a volume of water greater than its own volume?

None of them

7 0
3 years ago
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