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kap26 [50]
3 years ago
6

. . The proper dish washing sequence for a three compartment sink is?. A.scrape or spray, wash, rinse, sanitize, towel dry. b.sc

rape or spray, wash, sanitize, rinse air dry. C.scrape or spray, wash, rinse, sanitize, air dry.. d.scrape or spray, wash, sanitize, rinse, towel dry
Physics
1 answer:
ikadub [295]3 years ago
5 0

The proper dish washing sequence for a three compartment sink is scrape or spray, wash, rinse, sanitize, air dry. The correct answer between all the choices given is the third choice or letter C. I am hoping that this answer has satisfied your query and it will be able to help you, and if you would like, feel free to ask another question.

You might be interested in
3. why is the sum of the maximum voltages across each element in a series r l c circuit usually greater than the maximum applied
Llana [10]

The sum of the maximum voltages across each element in a series RLC circuit is usually greater than the maximum applied voltage because voltages are added by vector addition.

<h3>What is the Kichoff's loop rule?</h3>

Kirchhoff's loop rule states that the algebraic sum of potential differences, as well as the voltage supplied by the voltage sources and resistances, in any loop must be equal to zero.

In a series RLCcircuit, the voltages are not added by scalar addition but by vector addition.

Kirchhoff's loop rule is not violated since the voltages across different elements in the circuit are not at their maximum values.

Therefore, the sum of the maximum voltages across each element in a series RLC circuit is usually greater than the maximum applied voltage because voltages are added by vector addition.

Learn more about Kichoff's loop rule at: https://brainly.in/question/35360816

#SPJ1

8 0
2 years ago
A 50-ohm resistor in a 0.5 Amp circuit produces how much voltage drop?​
ahrayia [7]

25 volts

Explanation:

Use Ohm's law to find the potential drop:

V = IR

= (0.5 A)(50 ohms)

= 25 volts

6 0
3 years ago
2. Two small (green) objects, each of mass m, are separated by a solid, massless rod of length L. They are located so that one o
rusak2 [61]

The rod is stretched due to the force of gravitational attraction.

The magnitude of the deformation force on the rod is  -GMm×((1/r² + 1/(L+r)²)

<h3>What is gravity?</h3>

The force of attraction felt by a person which is directed at the center of a planet or Earth is called as the gravity.

The force of attraction is directly proportional to the product of masses of the object and inversely proportional to the square of distance between them.

F = GMm/R²

Given are two small (green) objects, each of mass m, are separated by a solid, massless rod of length L. They are located so that one of the objects is located at a distance of r away from the center of a uniform spherical planet with mass M.  Assume that m is very small so that you can ignore the gravitational force between the two small (green) objects.

Let the mass closer to the planet is A and the other is B

FA = Force on planet A = -GMm/r²

FB = Force on planet B = -GMm/(L+r)²

Net Force on the rod is given by the addition of the individual forces.

Fnet= FA + FB = -GMm/r² -GMm/(L+r)²

Fnet = -GMm×((1/r² + 1/(L+r)²)

Thus,  the magnitude of the deformation force on the rod is derived above.

A experience more force than B, so A will stretch out from B. Hence the force is stretching the rod.

Learn more about gravity.

brainly.com/question/4014727

#SPJ1

8 0
2 years ago
The value of acceleration due to gravity varies from place to place on the surface of the earth​
Kay [80]

Answer:

True....................

3 0
3 years ago
Read 2 more answers
A cannon shoots a ball at an angle θ above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find
nikitadnepr [17]

Answer:

a)  x = v₀ₓ t ,  y = v_{oy} t - ½ g t²

Explanation:

This is a projectile launch problem, let's use Newton's second law on each axis

X axis

       F = m a

Since there is no acceleration on the x axis, the force on this axis is zero

Y Axis  

       -W = m a_{y}

       -m g = m a_{y}  

       a_{y}  = -g

In this axis the acceleration is the acceleration of gravity

Now we can use science to find the position of the body on each axis

X axis

       x = v₀ₓ t + ½ a  t²

As the acceleration on this axis is zero

      x = v₀ₓ t

Y Axis

       y = v_{oy} t + ½ a_{y}  t²

The acceleration on this axis is –g

        y = v_{oy} t - ½ g t²

B) to find the maximum value of distance r

        r =√ x² + y²

        r = √( v₀ₓ² t² + (v_{oy} t + ½ g t²)²

We can find the maximum value of r using time respect derivatives

      dr / dt = 0

      0 = ½ 1/√( v₀ₓ² t² + (v_{oy} t + ½ g t²)²    (v₀ₓ² 2t + 2 (v_{oy} t - ½ g t²)(v_{oy} - ½ g2t)

We simplify this expression

         0 = v₀ₓ² 2t + 2 (v_{oy} t - ½ g t²)  (v_{oy} - ½ g2t)

       -v₀ₓ² t =( v_{oy} t - ½ g t²)  (v_{oy} - ½ g2t)  

      -v₀ₓ² t = v_{oy}² t -3/2 gt² v_{oy} + 1/2 g² t³  

       ½ g² t²- 3/2 g v_{oy} t  = v_{oy}² + v₀ₓ²

Let's use trigonometry to find go and vox

         sin θ = v_{oy} / v₀

         cos θ = vox / v₀

         v_{oy} = v₀ sin θ

        v₀ₓ = v₀ cos θ

We replace

         ½ g² t² -3/2 g v_{oy} t = v₀ (sin² θ +  cos² θ)

          g t² - 3v₀ sin θ t = 2 v₀/g

 

The time is maximum for the angle is zero

         

8 0
3 years ago
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