Classify each model A-D as either an element, a compound, or a mixture. Explain your reasoning for each answer

A uniform solid cylindrical log begins rolling without slipping down a ramp that rises 28.0° above the horizontal. After it has

Novay_Z [31]

Answer:

3.07 m/s

Explanation:

6 0

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.)

7 0

3 years ago

Most of the earth,s surface is covered by what ?

GenaCL600 [577]

Around 70-72% of earth’s surface is covered in water (most of it is salt water).
Hope this helps.

4 0

3 years ago

If a car is traveling at an average speed of 60 kilometers per hour how long does it take to travel 12 kilometers

lara31 [8.8K]

Answer:

The time taken to travel is, t = 12 minutes

Explanation:

Given data,

The speed of the car, v = 60 km/h

The distance of travel, d = 12 km

The time taken for the travel is t = ?

The speed is defined as the distance divided by the time taken to travel. The formula for speed is,

v = d/t

∴ t = d/v

t = 12 km / 60 km/h

t = 0.2 h

t = 12 minutes

Hence, the time taken to travel is, t = 12 minutes.

5 0

3 years ago

The tire above has a recommended tire pressure of 35 PSI, however, its current pressure is only 26 PSI. Which of the following c

Dima020 [189]

Answer:

option B

Explanation:

It is given that in the tire the recommended pressure is 35 PSI however the current pressure is only 26 PSI which means that pressure in the tire is less than the recommended so the chances of blowout of the tire gets eliminated hence option A is not correct.

Having pressure less in the tire can lead to the Unstable handling of the vehicle.

so correct answer is option B

7 0

3 years ago