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MatroZZZ [7]
3 years ago
8

how does the graph of displacement versus time look for something moving at a constant positive velocity

Physics
1 answer:
just olya [345]3 years ago
4 0

The answer is: In the graph of displacement versus time, for something moving at a constant positive velocity,  the line is straight and is going up. You can see it in the graph attached.

The explanation is shown below:

In a graph of displacement versus time, the displacement is on the y-axis and the time is on the x-axis. This type of graph that represents an object moving  at a constant positive velocity, shows a linear function where the lines is straight and goes up (As you can see in the graph attached).

Then, the slope of the line is the velocity of the object and the y-intercept is the initial position.

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A car drives over a hilltop that has a radius of curvature 0.120 km at the top of the hill. At what speed would the car be trave
Mashutka [201]

Answer:

34.3 m/s

Explanation:

Newton's Second Law states that the resultant of the forces acting on the car is equal to the product between the mass of the car, m, and the centripetal acceleration a_c (because the car is moving of circular motion). So at the top of the hill the equation of the forces is:

mg-R = m a_c = m\frac{v^2}{r}

where

(mg) is the weight of the car (downward), with m being the car's mass and g=9.8 m/s^2 is the acceleration due to gravity

R is the normal reaction exerted by the road on the car (upward, so with negative sign)

v is the speed of the car

r = 0.120 km = 120 m is the radius of the curve

The problem is asking for the speed that the car would have when it tires just barely lose contact with the road: this means requiring that the normal reaction is zero, R=0. Substituting into the equation and solving for v, we find:

v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(120 m)}=34.3 m/s

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3 years ago
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A cyclist is riding a bicycle at a speed of 22 mph on a horizontal road. The distance between the axles is 42 in., and the mass
stealth61 [152]

Answer:

The shortest distance is  S = 24.86 ft

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

   The speed of the bicycle is v_b = 22\ mph = 22 * \frac{5280}{3600}   =  32.26 ft/s

     The distance between the axial is  d = 42 \ in

The mass center of the cyclist and the bicycle is m_c = 26 \ in  behind the front axle

The mass center of the cyclist and the bicycle is m_h = 40 \ in above the ground

   For the bicycle not to be thrown over the

     Momentum about the back wheel must be zero so

                \sum _B = 0

=>             mg (26) = ma(40)

=>             a = \frac{26}{40} g

Here  g = 32.2 ft/s^2

     So     a =  \frac{26}{40} (32.2)

             a =  20.93 ft/s^2

Apply the equation of motion to this motion we have

       v^2 = u^2 + 2as

 Where  u = 32.26 ft /s

             and v = 0 since the bicycle is coming to a stop

        v^2 = (32.26)^2 - 2(20.93) S

=>      S = 24.86 ft        

                 

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