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tatyana61 [14]
3 years ago
5

1. ____________is defined as the distance an object travels per unit of time.

Physics
1 answer:
charle [14.2K]3 years ago
7 0

Answer:

1) Speed

2) X ( maybe acceleration)

3) Motion

4) Vector

5) Distance

6) Scaler

7) Second

8) Velocity

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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
balu736 [363]

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.

5 0
4 years ago
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is
Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s​2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is  c = 28853.78 \ m^2 /s^2

Explanation:

From the question we are told that

         The acceleration is  a =  \frac{c}{s}\   m/s^2

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is s_f =  3 \ m

          The velocity is  v = 200 \ m/s

     Generally  time  =  \frac{ds}{dv}

   and  acceleration is a =  \frac{v}{time }

so

            a = v  \frac{dv}{ds}

 =>        vdv  =  a ds

             vdv  = \frac{c}{s}  ds

integrating both sides

           \int\limits^a_b  vdv  = \int\limits^c_d \frac{c}{s}  ds

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =s_f= 1.5 m

So we have  

           \int\limits^{200}_{0}  vdv  = \int\limits^{3}_{1.5} \frac{c}{s}  ds

              [\frac{v^2}{2} ] \left | 200} \atop {0}} \right.  = c [ln s]\left | 3} \atop {1.5}} \right.

            \frac{200^2}{2}  =  c ln[\frac{3}{1.5} ]

=>           c = \frac{20000}{0.69315}

              c = 28853.78 \ m^2 /s^2

     

5 0
3 years ago
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