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melamori03 [73]

3 years ago

14

Effects of biological hazards are widespread. Select the answer options which describe potential effects of coming into contact

with biological hazards.
Select the 2 answer options that apply.

Allergic reactions

Broken limbs

Muscle tears

Life-threatening events

1 answer:

s2008m [1.1K]3 years ago

7 0

Answer:

- Allergic Reactions

- Life Threatinging Events

Explanation:

Sources of biological hazards may include bacteria, viruses, insects, plants, birds, animals, and humans. These sources can cause a variety of health effects ranging from skin irritation and allergies to infections (e.g., tuberculosis, AIDS), cancer and so on.

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A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T

anyanavicka [17]

Answer:

R min = 28.173 ohm

R max = 1.55 × $10^{4}$ ohm

Explanation:

given data

capacitor = 0.227 μF

charged to 5.03 V

potential difference across the plates = 0.833 V

handled effectively = 11.5 μs to 6.33 ms

solution

we know that resistance range of the resistor is express as

V(t) = $V_o \times e^{t\RC}$ ...........1

so R will be

R = $\frac{t}{C\times ln(\frac{V_o}{V})}$ ....................2

put here value

so for t min 11.5 μs

R = $\frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}$

R min = 28.173 ohm

and

for t max 6.33 ms

R max = $\frac{6.33}{11.5} \times 28.173$

R max = 1.55 × $10^{4}$ ohm

4 0

3 years ago

In the first-order process,a blue dye reacts to form a purple dye. The amount of blue dye at the end of 1 hr is 480 g and the en

Tasya [4]

Answer:

The answer is 960 kg

Explanation:

Solution

Given that:

Assume the initial dye concentration as A₀

We write the expression for the dye concentration for one hour as follows:

ln (C₁) = ln (A₀) -kt

Here

C₁ = is the concentration at 1 hour

t =time

Now

Substitute 480 g for C₁ and 1 hour for t

ln (480) = ln (A₀) -k(1) ------- (1)

6.173786 = ln (A₀) -k

Now

We write the expression for the dye concentration for three hours as follows:

ln (C₃) = ln (A₀) -k

Here

C₃ = is the concentration at 3 hour

t =time

Thus

Substitute 480 g for C₃ and 3 hour for t

ln (120) = ln (A₀) -k(3) ------- (2)

4.787492 = ln (A₀) -3k

Solve for the equation 1 and 2

k =0.693

Now

Calculate the amount of blue present initially using the expression:

Substitute 0.693 for k in equation (2)

4.787492 = ln (A₀) -3 (0.693)

ln (A₀) =6.866492

A₀ =e^6.866492

= 960 kg

Therefore, the amount of the blue dye present from the beginning is 960 kg

6 0

4 years ago

A normal shock wave takes place during the flow of air at a Mach number of 1.8. The static pressure and temperature of the air u

Darina [25.2K]

Answer:

The pressure upstream and downstream of a shock wave are related as

$\frac{P_{1}}{P_{o}}=\frac{2\gamma M^{2}-(\gamma -1)}{\gamma +1}$

where,

$\gamma$ = Specific Heat ratio of air

M = Mach number upstream

We know that $\gamma _{air}=1.4$

Applying values we get

$\frac{P_{1}}{100kPa}=\frac{2\times 1.4\times 1.8^{2}-(1.4 -1)}{1.4 +1}\\\\\frac{P_{1}}{100kPa}=3.61\\\\\therefore P_{1}=361.33kPa(Absloute)$

Similarly the temperature downstream is obtained by the relation

$\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}$

Applying values we get

$\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C$

The Mach number downstream is obtained by the relation

$M_{d}^{2}=\frac{(\gamma -1)M^{2}+2}{2\gamma M^{2}-(\gamma -1)}\\\\\therefore M_{d}^{2}=\frac{(1.40-1)\times 1.8^{2}+2}{2\times1.4\times 1.8^{2}-(1.4-1)}\\\\\therefore M_{d}^{2}=0.38\\\\M_{d}=0.616$

3 0

4 years ago

Identify three operational controls and explain<br> how to use them?

ladessa [460]

Answer:

5.1 Personnel Security. ...

5.2 Physical and Environmental Protection. ...

5.3 Production, Input and Output Controls. ...

5.4 Contingency Planning and Disaster Recovery. ...

5.5 System Configuration Management Controls. ...

5.6 Data Integrity / Validation Controls. ...

5.7 Documentation. ...

5.8 Security Awareness and Training.

8 0

3 years ago

An electric power generation station, produce 0.8 MW of power. If the coal releases 8*10^6 k/h of energy, determine the rate at

Fofino [41]

Answer:

The rate at the heat is rejected from the power plant is -802.2 kW

Explanation:

From the first thermodynamics law:

$\DELTA U= \DELTA Q - \DELTA W$

Where ΔU is the change in internal energy, ΔQ is the difference between the inlet heat and the outlet heat and ΔW is the difference between the work doing by the system and the work doing over the system:

$\DELTA Q = Qi - Qo$

$\DELTA W = Wo -Wi$

Considering a system in equilibrium the variation of energy is null:

$\DELTA U=0$

Note that it is not doing work over the system, therefore, Wi is also null.

So, for our purpose we must find the heat rejected, thas is Qo. Take into account that as the power plant is been cooled then the inlet heat is negative:

$Qi=-8*10^{6}\frac{kJ}{h}$

By converting the heat to Watt unit:

$Qi=-8*10^{6}\frac{kJ}{h}*frac{h}{3600 s}$

$Qi=-2222.22 W$

The work calculated in watts is:

$Wo=-0.8 MW$

$Wo=-0.8*10^{6}W$

$Wo=-800000 W$

Solving first thermodynamics law for Qo, and replacing the work and the heat calculated:

$Qo=Qi-Wo$

$Qo=-2222.2W-800000W$

$Qo=-802222.2W$

In kW the heat is:

$Qo=-802.2kW$

7 0

3 years ago