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Ronch [10]
3 years ago
11

On a trip to the Colorado Rockies, you notice that when the freeway goes steeply down a hill, there are emergency exits every fe

w miles. These emergency exits are straight ramps which leave the freeway and are sloped uphill. They are designed to stop runaway trucks and cars that lose their brakes on downhill stretches of the freeway even if the road is covered with ice. You are curious, so you stop at the next emergency exit to take some measurements. You determine that the exit rises at an angle of 10o from the horizontal and is 100m long. What is the maximum speed of a truck that you are sure will be stopped by this road, even if the frictional force of the road surface is negligible?
Physics
1 answer:
Zanzabum3 years ago
4 0

Answer:

The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.

Explanation:

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A piece of wood has a mass of 25g and a volume of <br> 10cm3 What is its density?
Tasya [4]

Answer:

<h3>The answer is 2.5 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question we have

density =  \frac{25}{10}  =  \frac{5}{2}  \\

We have the final answer as

<h3>2.5 g/cm³</h3>

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3 0
3 years ago
What causes the earths plates to move
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A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a ho
Aneli [31]

Answer:

h=2.86m

Explanation:

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f

There is no kinetic energy in the initial state, nor potential energy in the end,

mgh+0=0+KE_f

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

mgh = (\frac{1}{2}mv^2+\frac{1}{2}I\omega^2)

The inertia of the bodies is given by the equation,

I=\frac{m(R_1^2+R^2_2)}{2}

I=\frac{2(0.2^2+0.1^2)}{2}

I=0.05Kgm^2

On the other hand the angular velocity is given by

\omega =\frac{v}{R_2}=\frac{4}{1/5} = 2rad/s

Replacing these values in the equation,

(0.5)(9.8)(h) =\frac{1}{2}*0.5*4^2+\frac{1}{2}*0.05*20^2

Solving for h,

h=2.86m

5 0
2 years ago
If something is traveling at 20 m/s constant velocity, is it in equilibrium? If a projectile is launched at a velocity of 20 m/s
Irina18 [472]

If something is traveling at 20 m/s constant speed AND its direction isn't changing, then its velocity is constant.  Another way to say that is: Its acceleration is zero.  Zero acceleration means zero NET force acting on the object, or a group of BALANCED forces acting on it, also called EQUILIBRIUM.  The required answer is: YES.

If a real projectile is launched, the force of gravity acts on it vertically downward.  There's no upward force acting on it to balance gravity.  Therefore, the forces on the projectile are NOT balanced, there IS a net vertical force on it, and it's NOT in equilibrium.  Too bad.


3 0
2 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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