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SVEN [57.7K]
3 years ago
8

The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.4 square inches. If a force of 6.4 lb

is applied to this piston, what is the force on the brake pad of one wheel if the area of the pad is 1.7 square inches
Physics
1 answer:
Anit [1.1K]3 years ago
7 0

Answer:

The force applied on one wheel during braking = 6.8 lb

Explanation:

Area of the piston (A) = 0.4 in^{2}

Force applied on the piston(F) = 6.4 lb

Pressure on the piston (P) = \frac{F}{A}

⇒ P = \frac{6.4}{0.4}

⇒ P = 16 \frac{lb}{in^{2} }

This is the pressure inside the cylinder.

Let force applied on the brake pad = F_{1}

Area of the brake pad (A_{1})= 1.7 in^{2}

Thus the pressure on the brake pad (P_{1}) =  \frac{F_{1} }{A_{1} }

When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.

⇒ P = P_{1}

⇒ 16 = \frac{F_{1} }{A_{1} }

⇒ F_{1} = 16 × A_{1}

Put the value of A_{1} we get

⇒ F_{1} = 16 × 1.7

⇒ F_{1} = 27.2 lb

This the total force applied during braking.

The force applied on one wheel = \frac{F_{1} }{4} = \frac{27.2}{4} = 6.8 lb

⇒ The force applied on one wheel during braking.

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Answer:

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The formula for kinetic energy <em>K</em> of a particle of mass <em>m</em> moving at velocity <em>v</em> is K=\frac{mv^2}{2} and the formula for the Lorentz force <em>F</em> experimented by a particle of charge <em>q </em>and velocity <em>v</em> under a magnetic field <em>B</em> is (asuming <em>v</em> and <em>B</em> are perpendicular) F=qvB.

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Then we have (<em>q, K, R</em> and <em>B</em> would be what we know):

F=F_{cp}

qvB=\frac{mv^2}{R}=\frac{2K}{R}

v=\frac{2K}{RqB}

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m=\frac{2K}{v^2}=\frac{2K(RqB)^2}{(2K)^2}=\frac{(RqB)^2}{2K}

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4 years ago
The Big Bang is considered a scientific theory. Why is this true?
Kitty [74]

Answer:

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If those models are consistent with our observations then we can make predictions from those models and test them against more observations.

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Einstein's special and general theory of relativity provide more accurate models. So are they true? They predict all sorts of strange behaviour which we have actually observed. They seem to provide useful models, but they do not explain everything. For example, the extreme conditions in a black hole or those we think existed at the start of the universe provide some challenges to General Relativity.

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Answer:

the center of mass is 316,670m bellow the release point.

Explanation:

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We will use the formula for vertical distance in free fall

h=v_{0}t+\frac{1}{2} gt^2

v_{0} is the initial velocity, is 0 for both stones since they were just dropped.

g is acceleration of gravity and t is time (g=9.81m/s^2)

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And finally using the equation for the center of mass:

CM=\frac{m_{1}h_{1}+m_{2}h_{2}}{m_{1}+m_{2}}

We know that the mass of the second stone is twice the mass of the first stone so:

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replacing these values in the equation for the center of mass

CM=\frac{mh_{1}+2mh_{2}}{m+2m}

CM=\frac{m(h_{1}+2h_{2})}{3m}=\frac{h_{1}+2h_{2}}{3}

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the center of mass is 316,670m bellow the release point.

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