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makvit [3.9K]
4 years ago
10

-What can you say about the snowboarder’s kinetic energy as he moves?

Physics
1 answer:
Damm [24]4 years ago
7 0

Answer:

  His kinetic energy increases, potential energy decreases

  The sum of kinetic and potential energy is a constant at any instant before he comes to rest.

Explanation:

  Snowboarder is starting from a height and moving to the down direction. As he moves down his velocity increases, we know that kinetic energy is given by the expression \frac{1}{2} mv^2, so as he moves his kinetic energy increases.

  When the snowboarder is starting his potential energy is maximum(Potential energy = mgh), as he comes down his potential energy decreases.

  Based on this we can conclude that the sum of potential energy and kinetic energy is a constant at any instant for a snowboarder before he comes to rest.

                             mgh+\frac{1}{2} mv^2= Constant

 

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Give a specific example of a vector quantity. why this type of quantity would be used to describe this event ?
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4 years ago
A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co
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Answer:

\displaystyle X_T=66.6\ km

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

\displaystyle V_f=V_o+a\ t

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of 0.987 m/s^2 for 182 seconds. Then it brakes with a constant acceleration of -0.321 m/s^2 until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

\displaystyle X=V_o\ t+\frac{a\ t^2}{2}

Our data is

\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec

Let's compute the first distance X1

\displaystyle X_1=0+\frac{0.987\times 182^2}{2}

\displaystyle X_1=16,346.7\ m

Now, we find the speed at the end of the first period of time

\displaystyle V_{f1}=0+0.987\times 182

\displaystyle V_{f1}=179.6\ m/s

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0

\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}

\displaystyle t=559.5\ sec

Computing the second distance

\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}

\displaystyle X_2=50,243.2\ m

The total distance is

\displaystyle X_t=x_1+x_2=16,346.7+50,243.2

\displaystyle X_t=66,589.9\ m

\displaystyle \boxed{X_T=66.6\ km}

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