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oksano4ka [1.4K]
3 years ago
5

Bernoulli’s principle helps explain the hydraulics of brake fluid in cars. true or false

Physics
2 answers:
IceJOKER [234]3 years ago
8 0
This statement is true.

<span>Bernoulli’s principle explains the relation between the speed of the fluid and the pressure or potential energy of the fluid.
Bernoulli explained that </span>an increase in the speed of a fluid is accompanied with a decrease in pressure<span> and potential energy of the fluid.  

This principle can be applied to the hydraulics </span><span>of brake fluid in cars. </span>
gavmur [86]3 years ago
7 0

Explanation :

The principle behind Hydraulic brakes is Pascal's law.

According to Bernoulli's principle, the speed of a fluid depends inversely on the pressure or on the potential energy of the fluid.

Mathematically,

P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2

According to Pascal's law, there is an equal distribution of the pressure in the container having fluids i.e.

P_1=P_2

Hence, the given statement is false.

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Magma from the mantle rises and flows out through a vent onto Earth's surface as lava. This molten material, sometimes along with ash, cinders, and rock, builds up a mountain around the vent. The vent and its mountain together are called a sediment.


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Water (rhoH20 = 1000.0 kg/m3 ) flows through a garden hose that goes up a step 20.0 cm high. The cross-sectional area of the hos
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Answer:

 P₂ = 138.88 10³ Pa

Explanation:

This is a problem of fluid mechanics, we must use the continuity and Bernoulli equation

Let's start by looking for the top speed

        Q = A₁ v₁ = A₂ v₂

We will use index 1 for the lower part and index 2 for the upper part, let's look for the speed in the upper part (v2)

         v₂ = A₁ / A₂ v₁

They indicate that A₂ = ½ A₁ and give the speed at the bottom (v₁ = 1.20 m/s)

         v₂ = 2  1.20

         v₂ = 2.40 m / s

Now let's write the Bernoulli equation

        P₁ + ½ ρ v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂

Let's clear the pressure at point 2

       P₂ = P₁ + ½ ρ (v₁² - v₂²) + ρ g (y₁-y₂)

we put our reference system at the lowest point

        y₁ - y₂ = -20 cm

Let's calculate

       P₂ = 143 10³ + ½ 1000 (1.20² - 2.40²) + 1000 9.8 (-0.200)

       P₂ = 143 103 - 2,160 103 - 1,960 103

       P₂ = 138.88 10³ Pa

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3 years ago
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Answer:

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A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotating t
koban [17]

Answer:

The coefficient of static friction between the object and the disk is 0.087.

Explanation:

According to the statement, the object on the disk experiments a centrifugal force due to static friction. From 2nd Newton's Law, we can represent the object by the following formula:

\Sigma F_{r} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R} (1)

\Sigma F_{y} = N - m\cdot g = 0 (2)

Where:

N - Normal force from the ground on the object, measured in newtons.

m - Mass of the object, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v - Linear speed of rotation of the disk, measured in meters per second.

R - Distance of the object from the center of the disk, measured in meters.

By applying (2) on (1), we obtain the following formula:

\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}

\mu_{s} = \frac{v^{2}}{g\cdot R}

If we know that v = 0.8\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}} and R = 0.75\,m, then the coefficient of static friction between the object and the disk is:

\mu_{s} = \frac{\left(0.8\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.75\,m)}

\mu_{s} = 0.087

The coefficient of static friction between the object and the disk is 0.087.

5 0
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