what is the mass of a pure platinum disk with a volume of 113 cm3? the density of platinum is 21.4 g/cm3

What is one positive outcome of the recent<br>events?

fenix001 [56]

Answer:

Analyze the positive and negative consequences of catastrophic events of the last 40-50 years and the individuals that had an impact on these events. Meaning.

Explanation:

5 0

2 years ago

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What evidence did wegener have to support his theory of plate tectonics?

PSYCHO15rus [73]

Answer:

Wegener first thought of this idea by noticing that the different large landmasses of the Earth almost fit together like a jigsaw puzzle. The continental shelf of the Americas fits closely to Africa and Europe, and other continents showed the same trend. Wegner also analyzed both sides of the Atlantic Ocean for rock type, geological structures and fossils and noticed that there was a significant similarity between matching sides of the continents, especially in fossil plants.

8 0

2 years ago

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.