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1 year ago

A skier’s most important piece of equipment

Physics

2 answers:

leva [86]1 year ago

4 0

Answer: A skier’s most important piece of equipment is Ski boots.

Explanation: To find the answer, we need to know about the ski boots.

<h3>Why ski boots are the most important equipment for a skier?</h3>

Skiing can be defined as the action of travelling over snow on skis.
There are different equipment's we use during skiing, among them, the most important equipment's is ski boot.

Ski boots are not meant for walking, but they provide the skis an extension for their legs.
The other equipment's are ski goggles, ski helmets, ski jackets and so on.

Thus, we can conclude that, a skier's most important equipment is Ski boots.

Learn more about ski boots here:

brainly.com/question/28108459

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lilavasa [31]1 year ago

3 0

Ski boots are a skier's most essential piece of gear.

We must understand the ski boots in order to determine the solution.

<h3>Why are ski boots the most crucial piece of gear for a skier?</h3>

Skiing is the activity of moving over snow on a pair of skis.

While we utilize a variety of equipment when skiing, the ski boot is by far the most crucial piece of gear.

Although ski boots are not designed for walking, they provide the skis an extra leg.

The other equipment includes ski coats, ski helmets, and ski goggles.

So, it stands to reason that ski boots are a skier's most essential piece of gear.

Learn more about the ski boots here:

brainly.com/question/28108459

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When enough ____ is added to the substance, the solid reaches its ____ point and becomes a liquid

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Answer: When enough __energy__ is added to the substance, the solid reaches its _melting_ point and becomes a liquid

Explanation: since energy is being added the substance changes phase into a liquid .

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3 years ago

Which pair of graphs represent the same motion?

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3 years ago

a 70 kg desk sits on the floor motionless If gravity is pulling it with an acceleration of 9.8 m/s/s how much force is gravity e

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Since force is mass*acceleration,

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3 years ago

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

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3 years ago

What would happen if you tried to use a prism to disperse a beam that contained only green light？

Margaret [11]

It is determined by the nature of the green light. Because lasers create light at almost a single frequency, green laser light would appear as a thin line of pure green. Other sources of "green" light emit light at a variety of frequencies, including yellow and blue, resulting in a strong green band in the center that fades into blue-green and yellow-green at the borders.

For example, here’s a graph of the spectrum of a green LED, showing the color range: Attachment #1

and here’s a graph of the transmission spectra of several standard photographic filters, including green: Attachment #2

Learn more about the color spectrum:

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1 year ago