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3 years ago

What is the best example of physical contamination?

1 answer:

6 0

I might be wrong but I’d say the first one

Answer:
•when a food service worker touches food

Explanation:
The worker is physically putting his hands on the food, therefore contaminating it which whatever’s on his hands

-YW!! <33

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The water behind Grand Coulee Dam is 1000 m wide and 200 m deep. Find the hydrostatic force on the back of the dam. (Hint: the t

Vika [28.1K]

Answer:

The hydro static force on the back of the dam is $1.96\times10^{11}\ N$

Explanation:

Given that,

Width b= 1000 m

Depth d= 200 m

We need to calculate the average pressure

Using formula of average pressure

$P_{avg}=\rho\times g\times d_{avg}$

Put the value into the formula

$P_{avg}=1000\times9.8\times100$

$P_{avg}=980000\ Pa$

We need to calculate the hydro static force on the back of the dam

Using formula of force

$F = P_{avg}\times A$

Put the value into the formula

$F = 980000\times1000\times200$

$F=1.96\times10^{11}\ N$

Hence, The hydro static force on the back of the dam is $1.96\times10^{11}\ N$

7 0

3 years ago

A parallel-plate vacuum capacitor has 8.60 J of energy stored in it. The separation between the plates is 3.80 mm . If the separ

creativ13 [48]

Answer:

Part a)

$E = \frac{8.60}{2.62} = 3.28 J$

Part b)

$E = 2.62(8.60) = 22.5 J$

Explanation:

As we know that the energy of capacitor when it is not connected to potential source is given as

$U = \frac{Q^2}{2C}$

As we know that initial energy is given as

$8.60 = \frac{Q^2}{2C}$

now we know that capacitance of parallel plate capacitor is given as

$C = \frac{\epsilon_0A}{d}$

now the new capacitance when distance is changed from 3.80 mm to 1.45 mm

$C' = \frac{Cd}{d'}$

$C' = \frac{C(3.80)}{1.45}$

$C' = 2.62 C$

Now the new energy of the capacitor is given as

$E = \frac{Q^2}{2(2.62C)}$

$E = \frac{8.60}{2.62} = 3.28 J$

Part b)

Now if the voltage difference between the plates of capacitor is given constant

now the energy energy of capacitor is

$U = \frac{1}{2}CV^2$

$8.60 = \frac{1}{2}CV^2$

now when capacitance is changed to new value then new energy is given as

$E = \frac{1}{2}C'V^2$

$E = \frac{1}{2}(2.62C)V^2$

$E = 2.62(8.60) = 22.5 J$

6 0

3 years ago

When a system absorbs heat from its surroundings it is a(n) ____ process

Nezavi [6.7K]

C. endothermic

An endothermic process takes heat from the surroundings while an exothermic process gives out heat to the surroundings.

8 0

3 years ago

Big Ben, a large artifact in England, has a mass of 1x10^8 kilograms and the Empire State Building 1x10^9 kilograms. The distanc

TiliK225 [7]

Answer:

The force, exerted by Big Ben on the Empire State Building is 2.66972 × 10⁻⁷ N

Explanation:

The question relates to the force of gravity experienced between two bodies

The given parameters are;

The mass of Big Ben, M₁ = 1 × 10⁸ kg

The mass of the Empire State Building, M₂ = 1 × 10⁹ kg

The distance between the two Big Ben and the Empire State Building, r = 5,000,000 meters

By Newton's Law of gravitation, we have;

$F=G \times \dfrac{M_{1} \times M_{2}}{r^{2}}$

Where;

F = The force exerted by Big Ben on the Empire State Building and vice versa

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M₁, M₂, and r are the given parameters

By plugging in the values of the parameters and the constant into the equation for Newton's Law of gravitation, we have;

$F=6.67430 \times 10^{-11} \times \dfrac{1 \times 10^8 \times 1 \times 10^9}{(5,000,000)^{2}} = 2.66972 \times 10^{-7}$

The force, 'F', exerted by Big Ben on the Empire State Building is F = 2.66972 × 10⁻⁷ N.

3 0

3 years ago

State two applications of electrostatic

kaheart [24]

Atmospheric electricity and storms, electrostatic control filters, and industrial electrostatic seperation as well as spark discharge. these are just a few. hope it helps.

6 0

3 years ago