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Kipish [7]
3 years ago
10

Hydraulic systems utilize Pascal's principle by transmitting pressure from one cylinder (called the primary) to another (called

the secondary). Since the pressures will be equal, if the surface areas are different then the forces applied to the cylinders' pistons will be different. Suppose in a hydraulic lift, the piston of the primary cylinder has a 2.05-cm diameter and the piston of the secondary cylinder has a 21.5-cm diameter.

Physics
1 answer:
Pie3 years ago
3 0

Answer:

Pascal's law

According to the Pascal's law ,pressure will be same in the all direction for a static and in-compressible fluid.In-compressible fluid means the density of the fluid is constant through out the volume.

Lets take cylinder 1  and cylinder 2

W= Applied load on the cylinder 1

We know that pressure = Load/Area

A₁=Cross sectional area of first cylinder

P₁=W/A₁

Given that fluid is In-compressible then pressure will be same and the same pressure will transfer to the second cylinder.

P₂=W₂/A₂=P₁

W₂=P₁ A₂

Given that

d₁=2.05 cm

d₂= 21.5 cm

W₂=P₁ A₂=A₂ W/A₁

W₂d₁²=W d₂²

W₂ x 21.5₁²=W x 2.05₂²

W₂=0.0090 W

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mario62 [17]

Answer:

TEJ as this is a thing you wont get

3 0
3 years ago
Directions: Read the following excerpts carefully, then identify whether they employ informative, persuasive, or argumentative w
Savatey [412]

Hi there!

Informative writing has the intent to inform or educate us on a particular topic or event. It gives us more information and insight onto something.

Persuasive writing has the intent of convincing us to believe in a certain idea or to perform a certain action. For instance, advertisements have a persuasive intent; they are persuading us to buy a product or service.

Argumentative writing is similar to persuasive writing in the sense that they are persuading us to believe a certain idea. However, they are often based on logic and fact rather than opinions.

Let's look at the first excerpt.

<em>This morning at 9 a.m., a school bus collided with a car at the intersection of Osmena and Cabrera streets. The passengers were not injured, but the medical personnel checked each student as well as the driver before they were transported to their school.</em>

This text doesn't try to convince us in believing something. It doesn't argue anything and it only tries to give us more insight onto the event, which is a car accident. No opinions are stated and only events are given.

Therefore, this excerpt uses an informative writing technique.

4 0
2 years ago
A glass tube 1 mm in diameter is dipped into glycerin. The density of the glycerin is 1260 kg/m3, surface tension is 6.3x10-2 N/
Sunny_sXe [5.5K]

Answer:

The capillary rise of the glycerin is most nearly  y  =  0.0204 \ m

Explanation:

From the question we are told that

  The diameter of the glass tube is  d =  1 \ mm =  0.001 \ m

   The density of glycerin is  \rho =  1260 \ kg /m^3

   The surface tension of the glycerin is \sigma   =  6.3 *10^{-2} \ N /m

The capillary rise of the glycerin is mathematically represented as

       y  =  \frac{4 * \sigma  *  cos (\theta )}{ \rho * g *  d}

substituting value  

       y  =  \frac{4 * 6.3 *10^{-2}  *  cos (0 )}{ 1260 * 9.8 *  0.001}

      y  =  0.0204 \ m

Therefore the height  of the glass tube  the glycerin was able to cover is

y  =  0.0204 \ m  

4 0
3 years ago
"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e
notka56 [123]

Answer:

The Magnifying power of a telescope is M = 109.26

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective f_{objective} = \frac{R}{2}

⇒ f_{objective} = \frac{590}{2}

⇒ f_{objective} = 295 cm

Focal length of eyepiece f_{eyepiece} = 2.7 cm

Magnifying power of a telescope is given by,

M = \frac{f_{objective} }{f_{eyepiece} }

M = \frac{295}{2.7}

M = 109.26

therefore the Magnifying power of a telescope is M = 109.26

4 0
3 years ago
What happens if :<br> . The test charge is not tiny.
docker41 [41]

The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.

<h3>How does test charge affect electric field?</h3>

As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.

Adjusting the amount of charge on the test charge will not change the electric field force.

<h3>What is a test charge used for?</h3>

The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.

To learn more about test charge, refer

brainly.com/question/16737526

#SPJ9

3 0
2 years ago
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