Which property of a star can help us deduce its surface temperature?

In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a

matrenka [14]

Answer:

Approximately $\displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right]$ .

Explanation:

Consider this $45^{\circ}$ slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin $(0, 0)$ .

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at $45^{\circ}$ to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as $y = -x$ .

Convert the initial speed of this diver to SI units:

$\displaystyle v = \rm 110\; km\cdot h^{-1} = 110 \times \frac{1}{3.6} = 30.556\; m\cdot s^{-1}$ .

The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at $g$ ( $g \approx \rm -9.81\; m\cdot s^{-2}$ near the surface of the earth.) At $t$ seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

$x$ -coordinate: $30.556t$ meters (constant velocity;)
$y$ -coordinate: $\displaystyle -\frac{1}{2}g\cdot t^{2} = -\frac{9.81}{2}\cdot t^{2}$ meters (constant acceleration with an initial vertical velocity of zero.)

To eliminate $t$ from this expression, solve the equation between $t$ and $x$ for $t$ . That is: express $t$ as a function of $x$ .

$x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}$ .

Replace the $t$ in the equation of $y$ with this expression:

$\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}$ .

Plot the two functions:

$y = -x$ ,
$\displaystyle y= -0.0052535\;x^{2}$ ,

and look for their intersection. Refer to the diagram attached.

Alternatively, equate the two expressions of $y$ (right-hand side of the equation, the part where $y$ is expressed as a function of $x$ .)

$-0.0052535\;x^{2} = -x$ ,

$\implies x = 190.35$ .

The value of $y$ can be found by evaluating either equation at this particular $x$ -value: $x = 190.35$ .

$y = -190.35$ .

The position vector of a point $(x, y)$ on a cartesian plane is $\displaystyle \left[\begin{array}{l}x \\ y\end{array}\right]$ . The coordinates of this skier is approximately $(190.35, -190.35)$ . The position vector of this skier will be $\displaystyle\rm \left[ \begin{array}{c}\rm191\\\rm-191\end{array}\right]$ . Keep in mind that both numbers in this vectors are in meters.

4 0

4 years ago

Which current is produced in homes

Natali [406]

Batteries Produce DC current while homes produce AC current.
Hope this helped xxxx

6 0

3 years ago

Read 2 more answers

3. What is the potential energy of a 8 Newton book sitting on a shelf that is 12 meters high?

Svetradugi [14.3K]

Answer:

P = 96 J

Explanation:

Given that,

Weight of the book, W = mg = 8 N

It is placed at a height of 12 m

We need to find the potential energy of the book. The potential energy of an object is given by the formula as follows :

E = mgh

mg = Weight

$E=8\ N\times 12\ m\\E=96\ J$

So, the potential energy of the book is 96 J.

8 0

4 years ago

What is the buoyant force of a dog displacing 10 pounds of water?

scZoUnD [109]

Answer:

find V using d = M/V

Explanation:

F = d*V*g

d = density of fluid (in this case, 1000)

V = volume of object

g = gravity

3 0

3 years ago

At what point in its path does a baseball that is hit to the outfield have its minimum vertical speed?

ioda

Answer:

At the point when it reaches its maximum height (answer C)

Explanation:

As soon as the ball leaves the bat is has not only horizontal component of the velocity, but also vertical, which allows it travel upwards in its trajectory.

When it reaches its maximum elevation, and the trajectory starts curving down, is when the velocity of the baseball is just horizontal (the vertical component has been reduced to zero due to the constant action of the acceleration of gravity which was slowing down it ascending pattern.

After that point, the ball starts gaining now vertical velocity being accelerated towards the center of the Earth due to gravity. At the maximum height of its path is when the vertical component of the velocity switches from positive (pointing up) to negative (pointing down) going through zero magnitude.

5 0

3 years ago