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kipiarov [429]
3 years ago
11

Why does the liquid rise up through the dip tube when the valve is open

Physics
2 answers:
Aleonysh [2.5K]3 years ago
6 0

The liquid rise up through the dip tube when the valve is opened. The propellant gas wants to expand as much as it can, so if the valve is open, the propellant expands and pushes the spray up out of the can making more room for itself to expand


mark brainliest   :)

IceJOKER [234]3 years ago
5 0

Answer:

A...........................................

Explanation:

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A wave is a:
liraira [26]

Answer:

wave

Explanation:

5 0
3 years ago
What is the horse power of an electric motor which can do by 1250 joule of work in 5 seconds​
Ksju [112]
1250 J in 5 sec= 250 Joule(s) per second (1250/5 0

250 Joules per second = 250 Watts ( 1J/s = 1 Watt per definition)

250 Watts output = 250/0.65 efficiency = 384 Watts input

1 Horsepower = 732 Watts

Motors 1 Horsepower and under are made in certain step sizes like

3/4 , 1/2 , 1/3, 1/4, 1/16 1/20 of a Horsepower.

3/4 Horsepower is 549 Watts

1/2 Horsepower is 366 Watts

so you need to 3/4 horsepower motor to achieve 1250 J of work in 5 seconds.
5 0
2 years ago
A object with a mass of 1.5 kg is lifted from the ground to a height of 0.22 m what is the objects potential energy
svet-max [94.6K]

Answer:

<h2>3.3 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 1.5 × 10 × 0.22

We have the final answer as

<h3>3.3 J</h3>

Hope this helps you

5 0
2 years ago
Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it
jek_recluse [69]

Answer:

a)    I= I₀ (cos²θ - cos⁴θ)    b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

         I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

         I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

              θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

          I = I₁ cos² θ'

       

we substitute

         I = (I₀ cos² tea) cos² (θ - 90)

        cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

         I = Io cos² θ sin² θ

        1= cos²θ+ sin²θ

       sin²θ = 1 - cos²θ

        I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

          \frac{dI}{d \theta} = 0

          \frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

            cos θ - 2 cos³ θ = 0

            cos θ ( 1 - 2 cos² θ) = 0  

The zeros of this function are in

           θ = 90º

           1-2cos²θ =0       cos θ = 0.25  θ =  75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

5 0
2 years ago
Which statement describes the horizontal acceleration of a projectile? It is –9.8 m/s2. It is 9.8 m/s2. It is constant. It is ze
lina2011 [118]
In a projectile, the horizontal acceleration is zero. The velocity remains constant at all times. However, the <u>vertical acceleration</u> is -9.81m/s^2.

Hope this helps!
6 0
3 years ago
Read 2 more answers
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