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marta [7]
3 years ago
11

The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.6 square inches. If a force of 5.6 lb

is applied to this piston, what is the force on the brake pad of one wheel if the area of the pad is 1.8 square inches? (Hint: Since the force on each of the four wheels would be the same, you only need to calculate the force on one.)
Physics
1 answer:
balu736 [363]3 years ago
5 0

Answer:

16.8 lb is the force on the brake pad of one wheel.

Explanation:

Force applied on the piston = F_1=5.6 lb

Area of the piston = A_1=0.6 inches^2

Force applied on the brakes = F_2

Area of the brakes = A_2=1.8 inches^2

Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.

\frac{F_1}{A_2}=\frac{F_2}{A_2}

F_2=\frac{5.6 lb\times 1.8 inhes^2}{0.6 inches^2}=16.8 lb

16.8 lb is the force on the brake pad of one wheel.

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A train accelerates at -1.5 m/s2 for 10 seconds. If the train had an initial
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Answer:

17 m/s

Explanation:

Using formula a = (v-u) /t

acceleration a =  -1.5 m/s2

final velocity v = unknown

initial velocity u = 32 m/s

time t = 10s

-1.5 = (v-32)/10

-15 = v - 32

-15 + 32 = v

v = 17 m/s

7 0
3 years ago
Three diffrent examples of accelerated motion
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Explanation:

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4 0
3 years ago
1. Give three examples, from the lab, where potential energy was converted to kinetic energy: ​
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Answer:

A book on a table before it falls.

A yoyo before it is released.

A raised weight.

Explanation:

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3 0
3 years ago
A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of
ololo11 [35]

Answer:

The initial velocity is 50 m/s.

(C) is correct option.

Explanation:

Given that,

Time = 10 sec

For first half,

We need to calculate the height

Using equation of motion

v^2=u^2+2gh

h =\dfrac{v^2}{2g}....(I)

For second half,

We need to calculate the time

Using equation of motion

h =ut+\dfrac{1}{2}gt_{2}^2

h=0+\dfrac{1}{2}gt_{2}^2

t_{2}=\sqrt{\dfrac{2h}{g}}

Put the value of h from equation (I)

t_{2}=\sqrt{\dfrac{2\times v^2}{g^2}}

t_{2}=\dfrac{v}{g}

According to question,

t_{1}+t_{2}=10

t_{1}=t_{2}

Put the value of t₁ and t₂

\dfrac{v}{g}+\dfrac{v}{g}=10

\dfrac{2v}{g}=10

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Here, g = 10

The initial velocity is

v=\dfrac{10\times10}{2}

v=50\ m/s

Hence, The initial velocity is 50 m/s.

3 0
3 years ago
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